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(if needed) Find the Probability The Problem Exactly 2 More than 2 2 or more Less than 2 Between 2 and 4 (inclusive) 2 or less BASIC TECHNIQUES b. Find the probability that during a given month there 5.39 Let x be a Poisson random variable with mean m 2. Calculate these probabilities: b. P(x 1) a. P(x 0) d. P(x 5) c. P(x 1) 5.40 Let x be a Poisson random variable with mean m 2.5. Use Table 2 in Appendix I to calculate these probabilities: a. P(x 5) c. P(x 2) 5.41 Poisson vs. Binomial Let x be a binomial random variable with n 20 and p.1. a. Calculate P(x 2) using Table 1 in Appendix I to b. P(x 6) d. P(1 x 4) obtain the exact binomial probability. b. Use the Poisson approximation to calculate P(x 2). c. Compare the results of parts a and b. Is the approxi- mation accurate? 5.42 Poisson vs. Binomial II To illustrate how well the Poisson probability distribution approximates the binomial probability distribution, calculate the Poisson approximate values for p(0) and p(1) for a binomial probability distribution with n 25 and p.05. Compare the answers with the exact values obtained from Table 1 in Appendix I. APPLICATIONS 5.43 Airport Safety The increased number of small commuter planes in major airports has heightened concern over air safety. An eastern airport has recorded a monthly average of ﬁve near-misses on landings and takeoffs in the past 5 years. a. Find the probability that during a given month there are no near-misses on landings and takeoffs at the airport. are ﬁve near-misses. c. Find the probability that there are at least ﬁve near- misses during a particular month. 5.44 Intensive Care The number x of people entering the intensive care unit at a particular hospital on any one day has a Poisson probability distribution with mean equal to ﬁve persons per day. a. What is the probability that the number of people entering the intensive care unit on a particular day is two? Less than or equal to two? b. Is it likely that x will exceed 10? Explain. 5.45 Acc
ident Prone Parents who are concerned that their children are “accident prone” can be reassured, according to a study conducted by the Department of Pediatrics at the University of California, San Francisco. Children who are injured two or more times tend to sustain these injuries during a relatively limited time, usually 1 year or less. If the average number of injuries per year for school-age children is two, what are the probabilities of these events? a. A child will sustain two injuries during the year. b. A child will sustain two or more injuries during the year. c. A child will sustain at most one injury during the year. 5.46 Accident Prone, continued Refer to Exercise 5.45. a. Calculate the mean and standard deviation for x, the number of injuries per year sustained by a schoolage child. b. Within what limits would you expect the number of injuries per year to fall? 5.47 Bacteria in Water Samples If a drop of water is placed on a slide and examined under a microscope, the number x of a particular type of bacteria 5.4 THE HYPERGEOMETRIC PROBABILITY DISTRIBUTION ❍ 205 present has been found to have a Poisson probability distribution. Suppose the maximum permissible count per water specimen for this type of bacteria is ﬁve. If the mean count for your water supply is two and you test a single specimen, is it likely that the count will exceed the maximum permissible count? Explain. 5.48 E. coli Outbreak Increased research and discussion have focused on the number of illnesses involving the organism Escherichia coli (01257:H7), which causes a breakdown of red blood cells and intestinal hemorrhages in its victims.5 According to the Center for Disease Control, an estimated 73,000 cases of E. coli infection and 61 deaths occur in the United States each year. A 2006 outbreak traced to wild pigs, who spread the bacteria into a spinach ﬁeld in California, sickened 204 people in 26 states and 1 Canadian province.6 Outbreaks have occurred at a rate of 2.5 per 100,000. Let us suppose that this rate has not changed. a. What is the probability that at most ﬁve cases of E.coli per 100,000 are reported in California this year? b. What is the probability that more than ﬁve cases of E. coli are reported in California this year? c. Approximately 95% of occurrences of E
. coli involve at most how many cases? THE HYPERGEOMETRIC PROBABILITY DISTRIBUTION 5.4 Suppose you are selecting a sample of elements from a population and you record whether or not each element possesses a certain characteristic. You are recording the typical “success” or “failure” data found in the binomial experiment. The sample survey of Example 5.1 and the sampling for defectives of Example 5.2 are practical illustrations of these sampling situations. If the number of elements in the population is large relative to the number in the sample (as in Example 5.1), the probability of selecting a success on a single trial is equal to the proportion p of successes in the population. Because the population is large in relation to the sample size, this probability will remain constant (for all practical purposes) from trial to trial, and the number x of successes in the sample will follow a binomial probability distribution. However, if the number of elements in the population is small in relation to the sample size (n/N.05), the probability of a success for a given trial is dependent on the outcomes of preceding trials. Then the number x of successes follows what is known as a hypergeometric probability distribution. It is easy to visualize the hypergeometric random variable x by thinking of a bowl containing M red balls and N M white balls, for a total of N balls in the bowl. You select n balls from the bowl and record x, the number of red balls that you see. If you now deﬁne a “success” to be a red ball, you have an example of the hypergeometric random variable x. The formula for calculating the probability of exactly k successes in n trials is given next. THE HYPERGEOMETRIC PROBABILITY DISTRIBUTION A population contains M successes and N M failures. The probability of exactly k successes in a random sample of size n is M P(x k) C M N k C nk N C n 206 ❍ CHAPTER 5 SEVERAL USEFUL DISCRETE DISTRIBUTIONS for values of k that depend on N, M, and n with C N N! n n)! n!(N The mean and variance of a hypergeometric random variable are very similar to those of a binomial random variable with a correction for the ﬁnite population size: N m nM N s 2 nM N MN N n 1 N EXAMPLE 5
.11 A case of wine has 12 bottles, 3 of which contain spoiled wine. A sample of 4 bottles is randomly selected from the case. 1. Find the probability distribution for x, the number of bottles of spoiled wine in the sample. 2. What are the mean and variance of x? Solution For this example, N 12, n 4, M 3, and (N M) 9. Then p(x draw 4 3S 9G 9 9 1. The possible values for x are 0, 1, 2, and 3, with probabilities 3 0C p(0) C 6).25 1 2 4 1( 12 C 5 9 4 4 3 1C p(1) C ).51 (8 3 3 4 12 C 9 4 5 4 3 2C p(2) C ).22 2 3 (3 6 12 C 9 4 5 4 3 3C p(3) C ).02 ( 9 1 1 12 C 5 9 4 4 9 9 2. The mean is given by m 4 1 3 2 1 and the variance is s 2 4 3 2 1 9 2 1 12 4.5455 1 1 EXAMPLE 5.12 A particular industrial product is shipped in lots of 20. Testing to determine whether an item is defective is costly; hence, the manufacturer samples production rather than using a 100% inspection plan. A sampling plan constructed to minimize the number of defectives shipped to customers calls for sampling ﬁve items from each lot and rejecting the lot if more than one defective is observed. (If the lot is rejected, each 5.4 THE HYPERGEOMETRIC PROBABILITY DISTRIBUTION ❍ 207 item in the lot is then tested.) If a lot contains four defectives, what is the probability that it will be accepted? Solution Let x be the number of defectives in the sample. Then N 20, M 4, (N M) 16, and n 5. The lot will be rejected if x 2, 3, or 4. Then 4 4 16 1C 5 C 0C P(accept the lot) P(x 1) p(0) p(1) C 4 20 20 C C 5 5 16! 4!!4 0! 6 1 1!1 5 0! 2! 5!1 5!! 4!!3 1! 6 1 2!1 4 0! 2! 5!1 5! 5 5 4 1 9.2817.4696.7513 9 6 9
2 3 3 5.4 EXERCISES BASIC TECHNIQUES 5.49 Evaluate these probabilities: 2 3 1C a. C 1 5 C 2 3 4 2C b. C 1 7 C 3 3 5 4C c. C 0 8 C 4 5.50 Let x be the number of successes observed in a sample of n 5 items selected from N 10. Suppose that, of the N 10 items, 6 are considered “successes.” a. Find the probability of observing no successes. b. Find the probability of observing at least two successes. c. Find the probability of observing exactly two successes. 5.51 Let x be a hypergeometric random variable with N 15, n 3, and M 4. a. Calculate p(0), p(1), p(2), and p(3). b. Construct the probability histogram for x. c. Use the formulas given in Section 5.4 to calculate m E(x) and s 2. d. What proportion of the population of measurements fall into the interval (m 2s)? Into the interval (m 3s)? Do these results agree with those given by Tchebysheff’s Theorem? a. What is the probability that there are two blue and one red candies in the selection? b. What is the probability that the candies are all red? c. What is the probability that the candies are all blue? APPLICATIONS 5.53 Defective Computer Chips A piece of electronic equipment contains six computer chips, two of which are defective. Three computer chips are randomly chosen for inspection, and the number of defective chips is recorded. Find the probability distribution for x, the number of defective computer chips. Compare your results with the answers obtained in Exercise 4.90. 5.54 Gender Bias? A company has ﬁve applicants for two positions: two women and three men. Suppose that the ﬁve applicants are equally qualiﬁed and that no preference is given for choosing either gender. Let x equal the number of women chosen to ﬁll the two positions. a. Write the formula for p(x), the probability distribu- tion of x. b. What are the mean and variance of this distribution? c. Construct a probability histogram for x. 5.52 Candy Choices A candy dish contains ﬁve blue and three red candies. A child reaches up and selects three cand
ies without looking. 5.55 Teaching Credentials In southern California, a growing number of persons pursuing a teaching credential are choosing paid internships over traditional 208 ❍ CHAPTER 5 SEVERAL USEFUL DISCRETE DISTRIBUTIONS student teaching programs. A group of eight candidates for three local teaching positions consisted of ﬁve candidates who had enrolled in paid internships and three candidates who had enrolled in traditional student teaching programs. Let us assume that all eight candidates are equally qualiﬁed for the positions. Let x represent the number of internship-trained candidates who are hired for these three positions. a. Does x have a binomial distribution or a hypergeo- metric distribution? Support your answer. b. Find the probability that three internship-trained candidates are hired for these positions. c. What is the probability that none of the three hired 5.56 Seed Treatments Seeds are often treated with a fungicide for protection in poor-draining, wet environments. In a small-scale trial prior to a large-scale experiment to determine what dilution of the fungicide to apply, ﬁve treated seeds and ﬁve untreated seeds were planted in clay soil and the number of plants emerging from the treated and untreated seeds were recorded. Suppose the dilution was not effective and only four plants emerged. Let x represent the number of plants that emerged from treated seeds. a. Find the probability that x 4. b. Find P(x 3). c. Find P(2 x 3). was internship-trained? d. Find P(x 1). CHAPTER REVIEW Key Concepts and Formulas I. The Binomial Random Variable c. Individual and cumulative probabilities 1. Five characteristics: n identical independent trials, each resulting in either success (S) or failure (F); probability of success is p and remains constant from trial to trial; and x is the number of successes in n trials 2. Calculating binomial probabilities kpkqnk a. Formula: P(x k) Cn b. Cumulative binomial tables using MINITAB 3. Mean of the Poisson random variable: E(x) m 4. Variance and standard deviation: s 2 m and s m 5. Binomial probabilities can be approximated with Poisson probabilities when np 7, using m np. c. Individual and cumulative probabilities III. The Hypergeometric Random using MINITAB Variable 3. Mean of the binomial random variable: m np 4. Vari
ance and standard deviation: s 2 npq and s npq 1. The number of successes in a sample of size n from a ﬁnite population containing M successes and N M failures 2. Formula for the probability of k successes in II. The Poisson Random Variable 1. The number of events that occur in a period of time or space, during which an average of m such events are expected to occur 2. Calculating Poisson probabilities m mk e a. Formula: P(x k) k! b. Cumulative Poisson tables n trials: M N P(x k) CM C nk N C n k 3. Mean of the hypergeometric random variable: m nM N 4. Variance and standard deviation: s 2 nM N MN and s s 2 n 1 N N N Binomial and Poisson Probabilities MY MINITAB ❍ 209 For a random variable that has either a binomial or a Poisson probability distribution, MINITAB has been programmed to calculate either exact probabilities—P(x k)—for a given value of k or the cumulative probabilities—P(x k)—for a given value of k. You must specify which distribution you are using and the necessary parameters: n and p for the binomial distribution and m for the Poisson distribution. Also, you have the option of specifying only one single value of k or several values of k, which should be stored in a column (say, C1) of the MINITAB worksheet. Consider a binomial distribution with n 16 and p.25. Neither n nor p appears in the tables in Appendix I. Since the possible values of x for this binomial random variable range from 0 to 16, we can generate the entire probability distribution as well as the cumulative probabilities by entering the numbers 0 to 16 in column C1. One way to quickly enter a set of consecutive integers in a column is to do the fol- lowing: • Name columns C1 and C2 as “x” and “p(x)”, respectively. • Enter the ﬁrst two values of x—0 and 1—to create a pattern in column C1. • Use your mouse to highlight the ﬁrst two integers. • Use your mouse to grab the square handle in the lower right corner of the highlighted area. Drag the handle down to continue the pattern. • You will see an integer appear in a small yellow square. Release the mouse
when you have the desired number of integers—in this case, 16. Once the necessary values of x have been entered, use Calc Probability Distributions Binomial to generate the Dialog box shown in Figure 5.7. Type the number of trials and the value of p (Event probability) in the appropriate boxes, and select FI GUR E 5. 7 ● 210 ❍ CHAPTER 5 SEVERAL USEFUL DISCRETE DISTRIBUTIONS “x” for the input column. (If you do not type a column number for storage, MINITAB will display the results in the Session window. If you type C2 or p(x) in the box marked “Optional storage,” the results will appear in column C2 rather than in the Session window.) Make sure that the radio button marked “Probability” is selected. The probability density function appears in the Session window when you click OK (a portion is shown in Figure 5.8). What is the probability that x equals 4? That x is either 3 or 4? To calculate cumulative probabilities, make sure that the dot marked “Cumulative probability” is selected, and enter the appropriate values of x in C1. If you have only one value of x, it is simpler to select the Input constant box and enter the appropriate value. For example, for a Poisson random variable with m 5, use Calc Probability Distributions Poisson and enter a mean of 5. Enter the number 6 into the Input constant box, click OK, and the probability that x is less than or equal to 6 will appear in the Session window (see Figure 5.9). What value k is such that only 5% of the values of x exceed this value (and 95% are less than or equal to k)? If you enter the probability.95 into the Input constant box, select the option marked “Inverse cumulative probability”, and click OK (see Figure 5.10), then the values of x on either side of the “.95 mark” are shown in the Session window as in Figure 5.11. Hence, if you observed a value of x 10, this would be an unusual observation because P(x 9) 1.968172.031828. F IG URE 5. 8 ● F IG URE 5. 9 ● SUPPLEMENTARY EXERCISES ❍ 211 FI GUR E 5. 10 ● FI GUR
E 5. 11 ● Supplementary Exercises 5.57 List the ﬁve identifying characteristics of the binomial experiment. 5.58 Under what conditions can the Poisson random variable be used to approximate the probabilities associated with the binomial random variable? What application does the Poisson distribution have other than to estimate certain binomial probabilities? 5.59 Under what conditions would you use the hypergeometric probability distribution to evaluate the probability of x successes in n trials? 5.60 Tossing a Coin A balanced coin is tossed three times. Let x equal the number of heads observed. a. Use the formula for the binomial probability distribution to calculate the probabilities associated with x 0, 1, 2, and 3. b. Construct the probability distribution. c. Find the mean and standard deviation of x, using these formulas: m np s npq 212 ❍ CHAPTER 5 SEVERAL USEFUL DISCRETE DISTRIBUTIONS d. Using the probability distribution in part b, ﬁnd the fraction of the population measurements lying within one standard deviation of the mean. Repeat for two standard deviations. How do your results agree with Tchebysheff’s Theorem and the Empirical Rule? 5.61 Coins, continued Refer to Exercise 5.60. Suppose the coin is deﬁnitely unbalanced and the probability of a head is equal to p.1. Follow the instructions in parts a, b, c, and d. Note that the probability distribution loses its symmetry and becomes skewed when p is not equal to 1/2. 5.62 Cancer Survivor Rates The 10-year survival rate for bladder cancer is approximately 50%. If 20 people who have bladder cancer are properly treated for the disease, what is the probability that: a. At least 1 will survive for 10 years? b. At least 10 will survive for 10 years? c. At least 15 will survive for 10 years? 5.63 Garbage Collection A city commissioner claims that 80% of all people in the city favor garbage collection by contract to a private concern (in contrast to collection by city employees). To check the theory that the proportion of people in the city favoring private collection is.8, you randomly sample 25 people and ﬁnd that x, the number of people who support the commissioner’s claim, is 22. a. What is the probability of observing at least 22 who support the commissioner’s claim if, in fact, p.8? b. What
is the probability that x is exactly equal to 22? c. Based on the results of part a, what would you conclude about the claim that 80% of all people in the city favor private collection? Explain. 5.64 Integers If a person is given the choice of an integer from 0 to 9, is it more likely that he or she will choose an integer near the middle of the sequence than one at either end? a. If the integers are equally likely to be chosen, ﬁnd the probability distribution for x, the number chosen. b. What is the probability that a person will choose a 4, 5, or 6? c. What is the probability that a person will not choose a 4, 5, or 6? 5.65 Integers II Refer to Exercise 5.64. Twenty people are asked to select a number from 0 to 9. Eight of them choose a 4, 5, or 6. a. If the choice of any one number is as likely as any other, what is the probability of observing eight or more choices of the numbers 4, 5, or 6? b. What conclusions would you draw from the results of part a? 5.66 Job Security A USA Today Snapshot reports that among people 35 to 65 years old, nearly twothirds say they are not concerned about being forced into retirement.7 Suppose that we randomly select n 15 individuals that in this age category and approximate the value of p as p.7. Let x be the number that say they are not concerned with forced retirement. a. What is the probability distribution for x? b. What is P(x 8)? c. Find the probability that x exceeds 8? d. What is the largest value of c for which P(x c).10? 5.67 Teen Magazines Although teen magazines Teen People, Hachette Filipacche, and Elle Girl folded in 2006, 70% of people in a phone-in poll said teens are still a viable market for print, but they do not want titles that talk to them like they are teens.8 They read more sophisticated magazines. A sample of n 400 people are randomly selected. a. What is the average number in the sample who said that teenagers are still a viable market for print? b. What is the standard deviation of this number? c. Within what range would you expect to ﬁnd the number in the sample who said that there is a viable market for teenage print? d. If only 225
in a sample of 400 people said that teenagers are still a viable market for print, would you consider this unusual? Explain. What conclusions might you draw from this sample information? 5.68 Checking In We look forward to the time when we will leave it all behind and take a vacation. However, fewer Americans are really getting away while on vacation. Among small business owners, more than half (51%) say they check in with the office at least once a day while on vacation; only 27% say they cut the cord completely.9 If 20 small business owners are randomly selected, and we assume that exactly half check in with the office at least once a day, then n 20 and p.5. Find the following probabilities. a. Exactly 16 say that they check in with the office at least once a day while on vacation. b. Between 15 and 18 (inclusive) say they check in with the office at least once a day while on vacation. c. Five or fewer say that they check in with the office at least once a day while on vacation. Would this be an unlikely occurrence? 5.69 Psychosomatic Problems A psychiatrist believes that 80% of all people who visit doctors have problems of a psychosomatic nature. She decides to select 25 patients at random to test her theory. a. Assuming that the psychiatrist’s theory is true, what is the expected value of x, the number of the 25 patients who have psychosomatic problems? b. What is the variance of x, assuming that the theory is true? c. Find P(x 14). (Use tables and assume that the theory is true.) d. Based on the probability in part c, if only 14 of the 25 sampled had psychosomatic problems, what conclusions would you make about the psychiatrist’s theory? Explain. 5.70 Student Fees A student government states that 80% of all students favor an increase in student fees to subsidize a new recreational area. A random sample of n 25 students produced 15 in favor of increased fees. What is the probability that 15 or fewer in the sample would favor the issue if student government is correct? Do the data support the student government’s assertion, or does it appear that the percentage favoring an increase in fees is less than 80%? 5.71 Gray Hair on Campus College campuses are graying! According to a recent article, one in four college students is aged 30 or older. Many of these students are women updating their job skills. Assume that
the 25% ﬁgure is accurate, that your college is representative of colleges at large, and that you sample n 200 students, recording x, the number of students age 30 or older. a. What are the mean and standard deviation of x? b. If there are 35 students in your sample who are age 30 or older, would you be willing to assume that the 25% ﬁgure is representative of your campus? Explain. SUPPLEMENTARY EXERCISES ❍ 213 5.72 Probability of Rain Most weather forecasters protect themselves very well by attaching probabilities to their forecasts, such as “The probability of rain today is 40%.” Then, if a particular forecast is incorrect, you are expected to attribute the error to the random behavior of the weather rather than to the inaccuracy of the forecaster. To check the accuracy of a particular forecaster, records were checked only for those days when the forecaster predicted rain “with 30% probability.” A check of 25 of those days indicated that it rained on 10 of the 25. a. If the forecaster is accurate, what is the appropriate value of p, the probability of rain on one of the 25 days? b. What are the mean and standard deviation of x, the number of days on which it rained, assuming that the forecaster is accurate? c. Calculate the z-score for the observed value, x 10. [HINT: Recall from Section 2.6 that z-score (x m)/s.] d. Do these data disagree with the forecast of a “30% probability of rain”? Explain. 5.73 What’s for Breakfast? A packaging experiment is conducted by placing two different package designs for a breakfast food side by side on a supermarket shelf. The objective of the experiment is to see whether buyers indicate a preference for one of the two package designs. On a given day, 25 customers purchased a package from the supermarket. Let x equal the number of buyers who choose the second package design. a. If there is no preference for either of the two de- signs, what is the value of p, the probability that a buyer chooses the second package design? b. If there is no preference, use the results of part a to calculate the mean and standard deviation of x. c. If 5 of the 25 customers choose the ﬁrst package design and 20 choose the second design, what do you conclude about the
customers’ preference for the second package design? 5.74 Plant Density One model for plant competition assumes that there is a zone of resource depletion around each plant seedling. Depending on the size of the zones and the density of the plants, the zones of resource depletion may overlap with those of other seedlings in the vicinity. When the seeds are randomly dispersed over a wide area, the number of neighbors that a seedling may have usually follows a Poisson 214 ❍ CHAPTER 5 SEVERAL USEFUL DISCRETE DISTRIBUTIONS distribution with a mean equal to the density of seedlings per unit area. Suppose that the density of seedlings is four per square meter (m2). a. What is the probability that a given seedling has no neighbors within 1 m2? b. What is the probability that a seedling has at most three neighbors per m2? c. What is the probability that a seedling has ﬁve or more neighbors per m2? d. Use the fact that the mean and variance of a Poisson random variable are equal to ﬁnd the proportion of neighbors that would fall into the interval m 2s. Comment on this result. 5.75 Plant Genetics A peony plant with red petals was crossed with another plant having streaky petals. The probability that an offspring from this cross has red ﬂowers is.75. Let x be the number of plants with red petals resulting from ten seeds from this cross that were collected and germinated. a. Does the random variable x have a binomial distribution? If not, why not? If so, what are the values of n and p? b. Find P(x 9). c. Find P(x 1). d. Would it be unusual to observe one plant with red petals and the remaining nine plants with streaky petals? If these experimental results actually occurred, what conclusions could you draw? 5.76 Dominant Traits The alleles for black (B) and white (b) feather color in chickens show incomplete dominance; individuals with the gene pair Bb have “blue” feathers. When one individual that is homozygous dominant (BB) for this trait is mated with an individual that is homozygous recessive (bb) for this trait, 1/4 of the offspring will carry the gene pair BB, 1/2 will carry the gene pair Bb, and 1/4 will carry the gene pair bb.
Let x be the number of chicks with “blue” feathers in a sample of n 20 chicks resulting from crosses involving homozygous dominant chickens (BB) with homozygous recessive chickens (bb). a. Does the random variable x have a binomial distribution? If not, why not? If so, what are the values of n and p? b. What is the mean number of chicks with “blue” feathers in the sample? d. What is the probability that the number of chicks with “blue” feathers is greater than or equal to 10 but less than or equal to 12? 5.77 Football Coin Tosses During the 1992 football season, the Los Angeles Rams (now the St. Louis Rams) had a bizarre streak of coin-toss losses. In fact, they lost the call 11 weeks in a row.10 a. The Rams’ computer system manager said that the odds against losing 11 straight tosses are 2047 to 1. Is he correct? b. After these results were published, the Rams lost the call for the next two games, for a total of 13 straight losses. What is the probability of this happening if, in fact, the coin was fair? 5.78 Diabetes in Children Insulin-dependent diabetes (IDD) is a common chronic disorder of children. This disease occurs most frequently in persons of northern European descent, but the incidence ranges from a low of 1–2 cases per 100,000 per year to a high of more than 40 per 100,000 in parts of Finland.11 Let us assume that an area in Europe has an incidence of 5 cases per 100,000 per year. a. Can the distribution of the number of cases of IDD in this area be approximated by a Poisson distribution? If so, what is the mean? b. What is the probability that the number of cases of IDD in this area is less than or equal to 3 per 100,000? c. What is the probability that the number of cases is greater than or equal to 3 but less than or equal to 7 per 100,000? d. Would you expect to observe 10 or more cases of IDD per 100,000 in this area in a given year? Why or why not? 5.79 Defective Videotapes A manufacturer of videotapes ships them in lots of 1200 tapes per lot. Before shipment, 20 tapes are randomly selected from each lot and tested. If none is defective, the lot is shipped
. If one or more are defective, every tape in the lot is tested. a. What is the probability distribution for x, the num- ber of defective tapes in the sample of 20? b. What distribution can be used to approximate prob- abilities for the random variable x in part a? c. What is the probability that a lot will be shipped c. What is the probability of observing fewer than ﬁve chicks with “blue” feathers? if it contains 10 defectives? 20 defectives? 30 defectives? 5.80 Dark Chocolate Despite reports that dark chocolate is beneficial to the heart, 47% of adults still prefer milk chocolate to dark chocolate.12 Suppose a random sample of n 5 adults is selected and asked whether they prefer milk chocolate to dark chocolate. a. What is the probability that all ﬁve adults say that they prefer milk chocolate to dark chocolate? b. What is the probability that exactly three of the ﬁve adults say they prefer milk chocolate to dark chocolate? c. What is the probability that at least one adult prefers milk chocolate to dark chocolate? 5.81 Tay–Sachs Disease Tay–Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that their offspring will develop the disease is approximately.25. Suppose a husband and wife are both carriers of the disease and the wife is pregnant on three different occasions. If the occurrence of Tay–Sachs in any one offspring is independent of the occurrence in any other, what are the probabilities of these events? a. All three children will develop Tay–Sachs disease. b. Only one child will develop Tay–Sachs disease. c. The third child will develop Tay–Sachs disease, given that the ﬁrst two did not. 5.82 Playing Hooky Many employers provide workers with sick/personal days as well as vacation days. Among workers who have taken a sick day when they were not sick, 49% say that they needed a break!13 Suppose that a random sample of n 12 workers who took a sick day is selected. Rounding 49% to p.5, ﬁnd the probabilities of the following events. a. What is the probability that more than six workers say that they took a sick day because they needed a break? b. What is the probability that fewer than ﬁve of the workers needed a break? c. What is
the probability that exactly 10 of the workers took a sick day because they needed a break? 5.83 The Triangle Test A procedure often used to control the quality of name-brand food products utilizes a panel of five “tasters.” Each member of the panel tastes three samples, two of which are from batches of the product known to have the desired SUPPLEMENTARY EXERCISES ❍ 215 taste and the other from the latest batch. Each taster selects the sample that is different from the other two. Assume that the latest batch does have the desired taste, and that there is no communication between the tasters. a. If the latest batch tastes the same as the other two batches, what is the probability that the taster picks it as the one that is different? b. What is the probability that exactly one of the tasters picks the latest batch as different? c. What is the probability that at least one of the tasters picks the latest batch as different? 5.84 Do You Return Your Questionnaires? The president of a company specializing in public opinion surveys claims that approximately 70% of all people to whom the agency sends questionnaires respond by ﬁlling out and returning the questionnaire. Twenty such questionnaires are sent out, and assume that the president’s claim is correct. a. What is the probability that exactly ten of the ques- tionnaires are ﬁlled out and returned? b. What is the probability that at least 12 of the ques- tionnaires are ﬁlled out and returned? c. What is the probability that at most ten of the questionnaires are ﬁlled out and returned? 5.85 Questionnaires, continued Refer to Exercise 5.84. If n 20 questionnaires are sent out, a. What is the average number of questionnaires that will be returned? b. What is the standard deviation of the number of questionnaires that will be returned? c. If x 10 of the 20 questionnaires are returned to the company, would you consider this to be an unusual response? Explain. 5.86 Poultry Problems A preliminary investigation reported that approximately 30% of locally grown poultry were infected with an intestinal parasite that, though not harmful to those consuming the poultry, decrease the usual weight growth rates in the birds. A diet supplement believed to be effective against this parasite was added to the bird’s food. Twenty-ﬁve birds were examined after having the supplement for at least two weeks
, and three birds were still found to be infested with the parasite. a. If the diet supplement is ineffective, what is the probability of observing three or fewer birds infected with the intestinal parasite? 216 ❍ CHAPTER 5 SEVERAL USEFUL DISCRETE DISTRIBUTIONS b. If in fact the diet supplement was effective and reduced the infection rate to 10%, what is the probability observing three or fewer infected birds? in earthquake insurance. If 15 homeowners are randomly chosen to be interviewed, a. What is the probability that at least one had earth- 5.87 Machine Breakdowns In a food processing and packaging plant, there are, on the average, two packaging machine breakdowns per week. Assume the weekly machine breakdowns follow a Poisson distribution. a. What is the probability that there are no machine breakdowns in a given week? b. Calculate the probability that there are no more than two machine breakdowns in a given week. 5.88 Safe Drivers? Evidence shows that the probability that a driver will be involved in a serious automobile accident during a given year is.01. A particular corporation employs 100 full-time traveling sales reps. Based on this evidence, use the Poisson approximation to the binomial distribution to ﬁnd the probability that exactly two of the sales reps will be involved in a serious automobile accident during the coming year. quake insurance? b. What is the probability that four or more have earthquake insurance? c. Within what limits would you expect the number of homeowners insured against earthquakes to fall? 5.92 Bad Wiring Improperly wired control panels were mistakenly installed on two of eight large automated machine tools. It is uncertain which of the machine tools have the defective panels, and a sample of four tools is randomly chosen for inspection. What is the probability that the sample will include no defective panels? Both defective panels? 5.93 Eating on the Run How do you survive when there’s no time to eat—fast food, no food, a protein bar, candy? A Snapshot in USA Today indicates that 36% of women aged 25–55 say that, when they are too busy to eat, they get fast food from a drive-thru.14 A random sample of 100 women aged 25–55 is selected. 5.89 Stressed Out A subject is taught to do a task in two different ways. Studies have shown that when subjected to mental strain and asked to perform the task, the subject most often reverts to the method �
�rst learned, regardless of whether it was easier or more difficult. If the probability that a subject returns to the ﬁrst method learned is.8 and six subjects are tested, what is the probability that at least ﬁve of the subjects revert to their ﬁrst learned method when asked to perform their task under stress? 40% 30% 20% 10% 0% How Women Eat on the Run Drive-thru Skip a meal Protein bar or shake Candy/ snack food 5.90 Enrolling in College A West Coast university has found that about 90% of its accepted applicants for enrollment in the freshman class will actually enroll. In 2007, 1360 applicants were accepted to the university. Within what limits would you expect to ﬁnd the size of the freshman class at this university in the fall of 2007? 5.91 Earthquakes! Suppose that one out of every ten homeowners in the state of California has invested a. What is the average number of women who say they eat fast food when they’re too busy to eat? b. What is the standard deviation for the number of women who say they eat fast food when they’re too busy to eat? c. If 49 of the women in the sample said they eat fast food when they’re too busy to eat, would this be an unusual occurrence? Explain. Exercises Use the Calculating Binomial Probabilities applet for the following set of exercises. 5.94 Refer to Exercise 5.8 and 5.9. a. Use the applet to construct the probability histogram for a binomial random variable x with n 6 and p.2. b. Use the applet to construct the probability histogram for a binomial random variable x with n 6 and p.8. How would you describe the shapes of the distributions in parts a and b? c. Use the applet to construct the probability histogram for a binomial random variable x with n 6 and p.5. How would you describe the shape of this distribution? 5.95 Use the applet to ﬁnd the following: a. P(x 6) for n 22, p.65 b. P(x 8) for n 12, p.4 c. P(x 14) for n 20, p.5 d. P(2 x 6) for n 15, p.3 e. P(x 6) for n 50, p.7 5.96 Success
ful Surgeries A new surgical procedure is said to be successful 80% of the time. Suppose the operation is performed ﬁve times and the results are assumed to be independent of one another. What are the probabilities of these events? a. All ﬁve operations are successful. b. Exactly four are successful. c. Less than two are successful. 5.97 Surgery, continued Refer to Exercise 5.96. If less than two operations were successful, how would you feel about the performance of the surgical team? MYAPPLET EXERCISES ❍ 217 5.98 Engine Failure Suppose the four engines of a commercial aircraft are arranged to operate independently and that the probability of in-ﬂight failure of a single engine is.01. What is the probability of the following events on a given ﬂight? a. No failures are observed. b. No more than one failure is observed. 5.99 McDonald’s or Burger King? Suppose that 50% of all young adults prefer McDonald’s to Burger King when asked to state a preference. A group of 100 young adults were randomly selected and their preferences recorded. a. What is the probability that more than 60 preferred McDonald’s? b. What is the probability that between 40 and 60 (inclusive) preferred McDonald’s? c. What is the probability that between 40 and 60 (inclusive) preferred Burger King? 5.100 Vacation Destinations High gas prices may keep some American vacationers closer to home. However, when given a choice of getaway spots, 66% of U.S. leisure travelers indicated that they would like to visit national parks.15 A random sample of n 100 leisure travelers is selected. a. What is the average of x, the number of travelers in the sample who indicate they would like to visit national parks? What is the standard deviation of x? b. Would it be unlikely to ﬁnd only 50 or fewer of those sampled who indicated they would like to visit national parks? Use the applet to ﬁnd the probability of this event. c. How many standard deviations from the mean is the value x 50? Does this conﬁrm your answer in part b? 218 ❍ CHAPTER 5 SEVERAL USEFUL DISCRETE DISTRIBUTIONS CASE STUDY A Mystery: Cancers Near a Reactor How safe is it to live near a nuclear reactor? Men who lived in a coastal strip that extends 20
miles north from a nuclear reactor in Plymouth, Massachusetts, developed some forms of cancer at a rate 50% higher than the statewide rate, according to a study endorsed by the Massachusetts Department of Public Health and reported in the May 21, 1987, edition of the New York Times.16 The cause of the cancers is a mystery, but it was suggested that the cancer was linked to the Pilgrim I reactor, which had been shut down for 13 months because of management problems. Boston Edison, the owner of the reactor, acknowledged radiation releases in the mid-1970s that were just above permissible levels. If the reactor was in fact responsible for the excessive cancer rate, then the currently acknowledged level of radiation required to cause cancer would have to change. However, confounding the mystery was the fact that women in this same area were seemingly unaffected. In his report, Dr. Sidney Cobb, an epidemiologist, noted the connection between the radiation releases at the Pilgrim I reactor and 52 cases of hematopoietic cancers. The report indicated that this unexpectedly large number might be attributable to airborne radioactive effluents from Pilgrim I, concentrated along the coast by wind patterns and not dissipated, as assumed by government regulators. How unusual was this number of cancer cases? That is, statistically speaking, is 52 a highly improbable number of cases? If the answer is yes, then either some external factor (possibly radiation) caused this unusually large number, or we have observed a very rare event! The Poisson probability distribution provides a good approximation to the distributions of variables such as the number of deaths in a region due to a rare disease, the number of accidents in a manufacturing plant per month, or the number of airline crashes per month. Therefore, it is reasonable to assume that the Poisson distribution provides an appropriate model for the number of cancer cases in this instance. 1. If the 52 reported cases represented a rate 50% higher than the statewide rate, what is a reasonable estimate of m, the average number of such cancer cases statewide? 2. Based on your estimate of m, what is the estimated standard deviation of the num- ber of cancer cases statewide? 3. What is the z-score for the x 52 observed cases of cancer? How do you interpret this z-score in light of the concern about an elevated rate of hematopoietic cancers in this area? The Normal Probability Distribution 6 © AFP/Getty Images GENERAL OBJECTIVES In Chapters 4 and 5, you learned about discrete random variables and their probability distributions. In
this chapter, you will learn about continuous random variables and their probability distributions and about one very important continuous random variable—the normal. You will learn how to calculate normal probabilities and, under certain conditions, how to use the normal probability distribution to approximate the binomial probability distribution. Then, in Chapter 7 and in the chapters that follow, you will see how the normal probability distribution plays a central role in statistical inference. CHAPTER INDEX ● Calculation of areas associated with the normal probabil- ity distribution (6.3) ● The normal approximation to the binomial probability dis- tribution (6.4) ● The normal probability distribution (6.2) ● Probability distributions for continuous random variables (6.1) How Do I Use Table 3 to Calculate Probabilities under the Standard Normal Curve? How Do I Calculate Binomial Probabilities Using the Normal Approximation? The Long and Short of It If you were the boss, would height play a role in your selection of a successor for your job? Would you purposely choose a successor who was shorter than you? The case study at the end of this chapter examines how the normal curve can be used to investigate the height distribution of Chinese men eligible for a very prestigious job. 219 220 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION PROBABILITY DISTRIBUTIONS FOR CONTINUOUS RANDOM VARIABLES 6.1 When a random variable x is discrete, you can assign a positive probability to each value that x can take and get the probability distribution for x. The sum of all the probabilities associated with the different values of x is 1. However, not all experiments result in random variables that are discrete. Continuous random variables, such as heights and weights, length of life of a particular product, or experimental laboratory error, can assume the inﬁnitely many values corresponding to points on a line interval. If you try to assign a positive probability to each of these uncountable values, the probabilities will no longer sum to 1, as with discrete random variables. Therefore, you must use a different approach to generate the probability distribution for a continuous random variable. Suppose you have a set of measurements on a continuous random variable, and you create a relative frequency histogram to describe their distribution. For a small number of measurements, you could use a small number of classes; then as more and more measurements are collected, you can use more classes and reduce the class width. The outline of the histogram will change slightly
, for the most part becoming less and less irregular, as shown in Figure 6.1. As the number of measurements becomes very large and the class widths become very narrow, the relative frequency histogram appears more and more like the smooth curve shown in Figure 6.1(d). This smooth curve describes the probability distribution of the continuous random variable. ● F IG URE 6. 1 Relative frequency histograms for increasingly large sample sizes acbd) x x x x 6.1 PROBABILITY DISTRIBUTIONS FOR CONTINUOUS RANDOM VARIABLES ❍ 221 How can you create a model for this probability distribution? A continuous random variable can take on any of an inﬁnite number of values on the real line, much like the inﬁnite number of grains of sand on a beach. The probability distribution is created by distributing one unit of probability along the line, much as you might distribute a handful of sand. The probability—grains of sand or measurements—will pile up in certain places, and the result is the probability distribution shown in Figure 6.2. The depth or density of the probability, which varies with x, may be described by a mathematical formula f(x), called the probability distribution or probability density function for the random variable x. ● f(x) FI GUR E 6. 2 The probability distribution f (x); P (a x b) is equal to the shaded area under the curve P(a < x < b) a b x For continuous random variables, area probability. Several important properties of continuous probability distributions parallel their discrete counterparts. Just as the sum of discrete probabilities (or the sum of the relative frequencies) is equal to 1, and the probability that x falls into a certain interval can be found by summing the probabilities in that interval, continuous probability distributions have the characteristics listed next. • The area under a continuous probability distribution is equal to 1. • The probability that x will fall into a particular interval—say, from a to b—is equal to the area under the curve between the two points a and b. This is the shaded area in Figure 6.2. Area under the curve equals 1. There is also one important difference between discrete and continuous random variables. Consider the probability that x equals some particular value—say, a. Since there is no area above a single point—say, x a—in the probability distribution for a continuous random variable, our deﬁnition implies
that the probability is 0. • P(x a) 0 for continuous random variables. • This implies that P(x a) P(x a) and P(x a) P(x a). • This is not true in general for discrete random variables. How do you choose the model—that is, the probability distribution f (x)— appropriate for a given experiment? Many types of continuous curves are available for modeling. Some are mound-shaped, like the one in Figure 6.1(d), but others are not. In general, try to pick a model that meets these criteria: • • It ﬁts the accumulated body of data. It allows you to make the best possible inferences using the data. 222 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION EXAMPLE 6.1 The uniform random variable is used to model the behavior of a continuous random variable whose values are uniformly or evenly distributed over a given interval. For example, the error x introduced by rounding an observation to the nearest inch would probably have a uniform distribution over the interval from.5 to.5. The probability density function f (x) would be “ﬂat” as shown in Figure 6.3. The height of the rectangle is set at 1, so that the total area under the probability distribution is 1. F IG URE 6. 3 A uniform probability distribution ● 1.50 1.25 1.00 ) x ( f 0.75 0.50 EXAMPLE 6.2 0.5 0.2 0.0 x 0.2 0.5 What is the probability that the rounding error is less than.2 in magnitude? Solution This probability corresponds to the area under the distribution between x.2 and x.2. Since the height of the rectangle is 1, P(.2 x.2) [.2 (.2)] 1.4 The exponential random variable is used to model continuous random variables such as waiting times or lifetimes associated with electronic components. For example, the waiting time at a supermarket checkout counter has an exponential distribution with an average waiting time of 5 minutes. The probability density function f (x).2e.2x is shown in Figure 6.4. To ﬁnd areas under this curve, you can use the fact that P(x a) e.2a for a 0. What is the probability that you have to wait 10 minutes or more at the checkout counter? Solution The probability to
be calculated is the area shaded in Figure 6.4. Use the general formula for P(x a) to ﬁnd P(x 10) e.2(10).135 F IG URE 6. 4 An exponential probability distribution ● 0.20 0.15 ) x ( f 0.10 0.05 0.00 0 5 10 x 15 20 6.2 THE NORMAL PROBABILITY DISTRIBUTION ❍ 223 Your model may not always ﬁt the experimental situation perfectly, but you should try to choose a model that best ﬁts the population relative frequency histogram. The better the model approximates reality, the better your inferences will be. Fortunately, many continuous random variables have mound-shaped frequency distributions, such as the data in Figure 6.1(d). The normal probability distribution provides a good model for describing this type of data. THE NORMAL PROBABILITY DISTRIBUTION 6.2 Continuous probability distributions can assume a variety of shapes. However, a large number of random variables observed in nature possess a frequency distribution that is approximately mound-shaped or, as the statistician would say, is approximately a normal probability distribution. The formula that generates this distribution is shown next. NORMAL PROBABILITY DISTRIBUTION 1 e(xm)2/(2s 2) f (x) 2p s x The symbols e and p are mathematical constants given approximately by 2.7183 and 3.1416, respectively; m and s (s 0) are parameters that represent the population mean and standard deviation, respectively. The graph of a normal probability distribution with mean m and standard deviation s is shown in Figure 6.5. The mean m locates the center of the distribution, and the distribution is symmetric about its mean m. Since the total area under the normal probability distribution is equal to 1, the symmetry implies that the area to the right of m is.5 and the area to the left of m is also.5. The shape of the distribution is determined by s, the population standard deviation. As you can see in Figure 6.6, large values of s reduce the height of the curve and increase the spread; small values of s increase the height of the curve and reduce the spread. Figure 6.6 shows three normal probability distributions with different means and standard deviations. Notice the differences in shape and location. FI GUR E 6. 5 Normal probability distribution ● f(x) Area to left of mean equals.5
Area to right of mean equals.5 µ – σ µ µ + σ x 224 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION F IG URE 6. 6 Normal probability distributions with differing values of m and s ● f(x) x The Java applet called Visualizing Normal Curves gives a visual display of the normal distribution for values of m between 10 and 8 and for values of s between.5 and 1.8. The dark blue curve is the standard normal z with mean 0 and standard deviation 1. You can use this applet to compare its shape to the shape of other normal curves (the red curve on your monitor, light blue in Figure 6.7) by moving the sliders to change the mean and standard deviation. What happens when you change the mean? When you change the standard deviation? F IG URE 6. 7 Visualizing Normal Curves applet ● You rarely ﬁnd a variable with values that are inﬁnitely small () or inﬁnitely large (). Even so, many positive random variables (such as heights, weights, and times) have distributions that are well approximated by a normal distribution. According to the Empirical Rule, almost all values of a normal random variable lie in the interval m 3s. As long as the values within three standard deviations of the mean are positive, the normal distribution provides a good model to describe the data. 6.3 TABULATED AREAS OF THE NORMAL PROBABILITY DISTRIBUTION ❍ 225 TABULATED AREAS OF THE NORMAL PROBABILITY DISTRIBUTION 6.3 To ﬁnd the probability that a normal random variable x lies in the interval from a to b, we need to ﬁnd the area under the normal curve between the points a and b (see Figure 6.2). However (see Figure 6.6), there are an inﬁnitely large number of normal distributions—one for each different mean and standard deviation. A separate table of areas for each of these curves is obviously impractical. Instead, we use a standardization procedure that allows us to use the same table for all normal distributions. The Standard Normal Random Variable A normal random variable x is standardized by expressing its value as the number of standard deviations (s) it lies to the left or right of its mean m. This is really just a change in the units of measure that we use, as if we were measuring in
inches rather than in feet! The standardized normal random variable, z, is deﬁned as m z x s or equivalently, x m zs Area under the z-curve equals 1. From the formula for z, we can draw these conclusions: • When x is less than the mean m, the value of z is negative. • When x is greater than the mean m, the value of z is positive. • When x m, the value of z 0. The probability distribution for z, shown in Figure 6.8, is called the standardized normal distribution because its mean is 0 and its standard deviation is 1. Values of z on the left side of the curve are negative, while values on the right side are positive. The area under the standard normal curve to the left of a speciﬁed value of z—say, z0—is the probability P(z z0). This cumulative area is recorded in Table 3 of Appendix I and is shown as the shaded area in Figure 6.8. An abbreviated version of Table 3 is given in Table 6.1. Notice that the table contains both positive and negative values of z. The left-hand column of the table gives the value of z correct to the tenth place; the second decimal place for z, corresponding to hundredths, is given across the top row. FI GUR E 6. 8 Standardized normal distribution ● f(z) (–) 0 z0 (+) z 226 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION TABLE 6.1 ● Table 3. Areas Under the Normal Curve Abbreviated Version of Table 3 in Appendix I....09.0010.01.0003.0005.0007.0009.0013.02.0003.0005.0006.0009.0013.03.0003.0004.0006.0009.0012 z 3.4 3.3 3.2 3.1 3.0 2.9 2.8 2.7 2.6 2.5 2.0.00.0003.0005.0007.0010.0013.0019.0026.0035.0047.0062.0228 ● Table 3. Areas Under the Normal Curve (continued).04.5160.5557.5948.6331.6700....09.6879.01.
5040.5438.5832.6217.6591.02.5080.5478.5871.6255.6628.03.5120.5517.5910.6293.6664 z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 2.0.00.5000.5398.5793.6179.6554.6915.7257.7580.7881.8159.9772 EXAMPLE 6.3 Find P(z 1.63). This probability corresponds to the area to the left of a point z 1.63 standard deviations to the right of the mean (see Figure 6.9). P(z 1.63) P(z 1.63) Solution The area is shaded in Figure 6.9. Since Table 3 in Appendix I gives areas under the normal curve to the left of a speciﬁed value of z, you simply need to ﬁnd the tabled value for z 1.63. Proceed down the left-hand column of the table to z 1.6 and across the top of the table to the column marked.03. The intersection of this row and column combination gives the area.9484, which is P(z 1.63). 6.3 TABULATED AREAS OF THE NORMAL PROBABILITY DISTRIBUTION ❍ 227 FI GUR E 6. 9 Area under the standard normal curve for Example 6.3 ● f(z).9484 0 1.63 z Areas to the left of z 0 are found using negative values of z. EXAMPLE 6.4 Find P(z.5). This probability corresponds to the area to the right of a point z.5 standard deviation to the left of the mean (see Figure 6.10). ● FI GUR E 6. 10 Area under the standard normal curve for Example 6.4 A1 =.3085 –0.5 0 z Solution The area given in Table 3 is the area to the left of a speciﬁed value of z. Indexing z.5 in Table 3, we can ﬁnd the area A1 to the left of.5 to be.3085. Since the area under the curve is 1, we ﬁnd P(z.5) 1 A1 1.3085
.6915. EXAMPLE 6.5 Find P(.5 z 1.0). This probability is the area between z.5 and z 1.0, as shown in Figure 6.11. FIGURE 6.11 Area under the standard normal curve for Example 6.5 ● f(z) A1 =.3085 A2 –.5 0 1.0 z Solution The area required is the shaded area A2 in Figure 6.11. From Table 3 in Appendix I, you can ﬁnd the area to the left of z.5 (A1.3085) and the area to 228 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION the left of z 1.0 (A1 A2.8413). To ﬁnd the area marked A2, we subtract the two entries: A2 (A1 A2) A1.8413.3085.5328 That is, P(.5 z 1.0).5328. How Do I Use Table 3 to Calculate Probabilities under the Standard Normal Curve? • To calculate the area to the left of a z-value, ﬁnd the area directly from Table 3. • To calculate the area to the right of a z-value, ﬁnd the area in Table 3, and subtract from 1. • To calculate the area between two values of z, ﬁnd the two areas in Table 3, and subtract one area from the other. Exercise Reps Consider a standard random variable with mean m 0 and standard deviation s 1. Use Table 3 and ﬁll in the probabilities below. The third probability is calculated for you. Rewrite the Probability (if needed) Find the Probability 1 P(x 2.33) 1.9901.0099 The Interval Less than 1.5 Greater than 2 Greater than 2.33 Between 1.96 and 1.96 Between 1.24 and 2.37 Less than or equal to 1 Write the Probability P(z _____) P(z _____) P(z _____ 2.33 ) P(_____ z _____) P(_____ z _____) P(z _____) Progress Report • Still having trouble? Try again using the Exercise Reps at the end of this section. • Mastered the z-table? You can skip the Exercises Reps at the end of this section! Answers are located on the
perforated card at the back of this book. EXAMPLE 6.6 Find the probability that a normally distributed random variable will fall within these ranges: 1. One standard deviation of its mean 2. Two standard deviations of its mean 6.3 TABULATED AREAS OF THE NORMAL PROBABILITY DISTRIBUTION ❍ 229 Solution 1. Since the standard normal random variable z measures the distance from the mean in units of standard deviations, you need to ﬁnd P(1 z 1).8413.1587.6826 Remember that you calculate the area between two z-values by subtracting the tabled entries for the two values. 2. As in part 1, P(2 z 2).9772.0228.9544. These probabilities agree with the approximate values of 68% and 95% in the Empirical Rule from Chapter 2. EXAMPLE 6.7 We know the area. Work from the inside of the table out. Find the value of z—say z0—such that.95 of the area is within z0 standard deviations of the mean. Solution The shaded area in Figure 6.12 is the area within z0 standard deviations of the mean, which needs to be equal to.95. The “tail areas” under the curve are not shaded, and have a combined area of 1.95.05. Because of the symmetry of the normal curve, these two tail areas have the same area, so that A1.05/2.025 in Figure 6.12. Thus, the entire cumulative area to the left of z0 to equal A1 A2.95.025.9750. This area is found in the interior of Table 3 in Appendix I in the row corresponding to z 1.9 and the.06 column. Hence, z0 1.96. Note that this result is very close to the approximate value, z 2, used in the Empirical Rule. FI GUR E 6. 12 Area under the standard normal curve for Example 6.7 ● f(z) A1 =.025.95 = A2 –z0 0 z0 z Calculating Probabilities for a General Normal Random Variable Most of the time, the probabilities you are interested in will involve x, a normal random variable with mean m and standard deviation s. You must then standardize the interval of interest, writing it as the equivalent interval in terms of z, the standard normal random variable.
Once this is done, the probability of interest is the area that you ﬁnd using the standard normal probability distribution. EXAMPLE 6.8 Let x be a normally distributed random variable with a mean of 10 and a standard deviation of 2. Find the probability that x lies between 11 and 13.6. 230 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION Solution The interval from x 11 to x 13.6 must be standardized using the formula for z. When x 11, 11 z x m 10.5 s 2 Always draw a picture— it helps! and when x 13.6, 10 1.8 z x m 13.6 s 2 The desired probability is therefore P(.5 z 1.8), the area lying between z.5 and z 1.8, as shown in Figure 6.13. From Table 3 in Appendix I, you ﬁnd that the area to the left of z.5 is.6915, and the area to the left of z 1.8 is.9641. The desired probability is the difference between these two probabilities, or P(.5 z 1.8).9641.6915.2726 F IG URE 6. 13 Area under the standard normal curve for Example 6.8 ● f(z) A1 A2 10 0 11.5 13.6 1.8 x z The Java applet called Normal Distribution Probabilities allows you to calculate areas under a normal distribution for any values of m and s you select. Simply type the appropriate mean and standard deviation into the boxes at the top of the applet, type the interval of interest into the boxes at the bottom of the applet, and press “Enter” at each step to record your changes. (The “Tab” key will move your cursor from box to box.) The necessary area will be shaded in red on your monitor (light blue in Figure 6.14) and the probability is given to the left of the curve. If you need an area under the standard normal distribution, use m 0 and s 1. In Example 6.8, we need an area under a normal distribution with m 10 and s 2. Notice the values of x and z located along the horizontal axis. Find the probability, P(11 x 13.6) P(0.5 z 1.8).2726, in Figure 6.14. • • 6.3 TABULATED AREAS
OF THE NORMAL PROBABILITY DISTRIBUTION ❍ 231 FI GUR E 6. 14 Normal Distribution Probabilities applet ● EXAMPLE 6.9 Studies show that gasoline use for compact cars sold in the United States is normally distributed, with a mean of 25.5 miles per gallon (mpg) and a standard deviation of 4.5 mpg. What percentage of compacts get 30 mpg or more? Solution The proportion of compacts that get 30 mpg or more is given by the shaded area in Figure 6.15. To solve this problem, you must ﬁrst ﬁnd the z-value corresponding to x 30. Substituting into the formula for z, you get 30 z x m 25.5 1.0 s 4.5 The area A1 to the left of z 1.0, is.8413 (from Table 3 in Appendix I). Then the proportion of compacts that get 30 mpg or more is equal to P(x 30) 1 P(z 1) 1.8413.1587 The percentage exceeding 30 mpg is 100(.1587) 15.87% FI GUR E 6. 15 Area under the standard normal curve for Example 6.9 ● f(x) A1 1 – A1 =.1587 25.5 0 30 1 x z 232 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION EXAMPLE 6.10 Refer to Example 6.9. In times of scarce energy resources, a competitive advantage is given to an automobile manufacturer who can produce a car that has substantially better fuel economy than the competitors’ cars. If a manufacturer wishes to develop a compact car that outperforms 95% of the current compacts in fuel economy, what must the gasoline use rate for the new car be? Solution The gasoline use rate x has a normal distribution with a mean of 25.5 mpg and a standard deviation of 4.5 mpg. You need to ﬁnd a particular value— say, x0—such that P(x x0).95 This is the 95th percentile of the distribution of gasoline use rate x. Since the only information you have about normal probabilities is in terms of the standard normal random variable z, start by standardizing the value of x0: z0 x0 25.5 4.5 Since the value of z0 corresponds to x0, it must also have area.95 to its left, as
shown in Figure 6.16. If you look in the interior of Table 3 in Appendix I, you will ﬁnd that the area.9500 is exactly halfway between the areas for z 1.64 and z 1.65. Thus, z0 must be exactly halfway between 1.64 and 1.65, or z0 x0 25.5 1.645 4.5 Solving for x0, you obtain x0 m z0s 25.5 (1.645)(4.5) 32.9 F IG URE 6. 16 Area under the standard normal curve for Example 6.10 ● f(z).95.05 x0 x0 – 25.5 4.5 z0 = x z The manufacturer’s new compact car must therefore get 32.9 mpg to outperform 95% of the compact cars currently available on the U.S. market. The Java applet called Normal Probabilities and z-Scores allows you to calculate areas under a normal distribution for any values of m and s you select. Once you specify one value for x, the applet calculates the value of z, and one of four types of areas, which you can select from the dropdown list at the bottom of the applet: 6.3 TABULATED AREAS OF THE NORMAL PROBABILITY DISTRIBUTION ❍ 233 • Area to Left ⇒ area to the left of z • Area to Right ⇒ area to the right of z • Two Tails ⇒ area in two tails cut off by z and z • Middle ⇒ area between z and z You can also work backward as we did to solve the problem in Example 6.10. We entered the mean and standard deviation, and then selected “Area to Left” with a probability of.95. If the boxes for x and z are left blank, pressing “Enter” will solve for these values, as shown in Figure 6.17. What is the value of x, correct to 5 decimal places? FI GUR E 6. 17 Normal Probabilities and z-Scores applet ● Remember that the applet is programmed to calculate probabilities using a value of z with full decimal accuracy. If you choose to round off a calculated z to the nearest hundredth, so that you can use Table 3 in Appendix I, you may obtain a slightly different probability than the one calculated using the applet. 6.3 EXERCISES EXERCISE REPS
These exercises refer back to the MyPersonal Trainer section on page 228. 6.1 Consider a standard random variable with m 0 and standard deviation s 1. Use Table 3 and ﬁll in the probabilities below. Rewrite the Probability (if needed) Find the Probability The Interval Less than 2 Greater than 1.16 Greater than 1.645 Between 2.33 and 2.33 Between 1.24 and 2.58 Less than or equal to 1.88 Write the Probability P(z _____) P(z _____) P(z _____) P(_____ z _____) P(_____ z _____) P(z _____) 234 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION 6.2 Repeat Exercise 6.1. Use Table 3 and ﬁll in the probabilities below. Rewrite the Probability (if needed) Find the Probability The Interval Greater than 5 Between 3 and 3 Between.5 and 1.5 Less than or equal to 6.7 Less than 2.81 Greater than 2.81 Write the Probability P(z _____) P(_____ z _____) P(_____ z _____) P(z _____) P(z _____) P(z _____) BASIC TECHNIQUES b. P(.58 z 1.74) b. z 1.83 d. z 4.18 6.3 Calculate the area under the standard normal curve to the left of these values: a. z 1.6 c. z.90 6.4 Calculate the area under the standard normal curve between these values: b. z 3.0 and z 3.0 a. z 1.4 and z 1.4 6.5 Find the following probabilities for the standard normal random variable z: a. P(1.43 z.68) c. P(1.55 z.44) d. P(z 1.34) e. P(z 4.32) 6.6 Find these probabilities for the standard normal random variable z: a. P(z 2.33) c. P(z 1.96) 6.7 a. Find a z0 such that P(z z0).025. b. Find a z0 such that P(z z0).9251. 6.8 Find a z0 such that P(z0 z z0).8262. 6.
9 a. Find a z0 that has area.9505 to its left. b. Find a z0 that has area.05 to its left. 6.10 a. Find a z0 such that P(z0 z z0).90. b. Find a z0 such that P(z0 z z0).99. 6.11 Find the following percentiles for the standard normal random variable z: a. 90th percentile c. 98th percentile 6.12 A normal random variable x has mean m 10 and standard deviation s 2. Find the probabilities of these x-values: a. x 13.5 b. P(z 1.645) d. P(2.58 z 2.58) b. 95th percentile d. 99th percentile b. x 8.2 c. 9.4 x 10.6 6.13 A normal random variable x has mean m 1.20 and standard deviation s.15. Find the probabilities of these x-values: a. 1.00 x 1.10 c. 1.35 x 1.50 6.14 A normal random variable x has an unknown mean m and standard deviation s 2. If the probability that x exceeds 7.5 is.8023, ﬁnd m. b. x 1.38 6.15 A normal random variable x has mean 35 and standard deviation 10. Find a value of x that has area.01 to its right. This is the 99th percentile of this normal distribution. 6.16 A normal random variable x has mean 50 and standard deviation 15. Would it be unusual to observe the value x 0? Explain your answer. 6.17 A normal random variable x has an unknown mean and standard deviation. The probability that x exceeds 4 is.9772, and the probability that x exceeds 5 is.9332. Find m and s. APPLICATIONS 6.18 Hamburger Meat The meat department at a local supermarket speciﬁcally prepares its “1-pound” packages of ground beef so that there will be a variety of weights, some slightly more and some slightly less than 1 pound. Suppose that the weights of these “1pound” packages are normally distributed with a mean of 1.00 pound and a standard deviation of.15 pound. a. What proportion of the packages will weigh more than 1 pound? b. What proportion of the packages will weigh between.95 and 1.05 pounds? c. What
is the probability that a randomly selected package of ground beef will weigh less than.80 pound? 6.3 TABULATED AREAS OF THE NORMAL PROBABILITY DISTRIBUTION ❍ 235 d. Would it be unusual to ﬁnd a package of ground beef that weighs 1.45 pounds? How would you explain such a large package? 6.19 Human Heights Human heights are one of many biological random variables that can be modeled by the normal distribution. Assume the heights of men have a mean of 69 inches with a standard deviation of 3.5 inches. a. What proportion of all men will be taller than 60? (HINT: Convert the measurements to inches.) b. What is the probability that a randomly selected man will be between 58 and 61 tall? c. President George W. Bush is 511 tall. Is this an unusual height? d. Of the 42 presidents elected from 1789 through 2006, 18 were 60 or taller.1 Would you consider this to be unusual, given the proportion found in part a? 6.20 Christmas Trees The diameters of Douglas ﬁrs grown at a Christmas tree farm are normally distributed with a mean of 4 inches and a standard deviation of 1.5 inches. a. What proportion of the trees will have diameters between 3 and 5 inches? b. What proportion of the trees will have diameters less than 3 inches? c. Your Christmas tree stand will expand to a diameter of 6 inches. What proportion of the trees will not ﬁt in your Christmas tree stand? 6.21 Cerebral Blood Flow Cerebral blood ﬂow (CBF) in the brains of healthy people is normally distributed with a mean of 74 and a standard deviation of 16. a. What proportion of healthy people will have CBF readings between 60 and 80? a. If you apply your brakes, what is the probability that you will brake to a stop within 40 feet or less? Within 50 feet or less? b. If the only way to avoid a collision is to brake to a stop, what is the probability that you will avoid the collision? 6.23 Elevator Capacities Suppose that you must establish regulations concerning the maximum number of people who can occupy an elevator. A study of elevator occupancies indicates that if eight people occupy the elevator, the probability distribution of the total weight of the eight people has a mean equal to 1200 pounds and a standard deviation of 99 pounds. What is the probability that the total weight of eight
people exceeds 1300 pounds? 1500 pounds? (Assume that the probability distribution is approximately normal.) 6.24 A Phosphate Mine The discharge of suspended solids from a phosphate mine is normally distributed, with a mean daily discharge of 27 milligrams per liter (mg/l) and a standard deviation of 14 mg/l. What proportion of days will the daily discharge exceed 50 mg/l? 6.25 Sunﬂowers An experimenter publishing in the Annals of Botany investigated whether the stem diameters of the dicot sunﬂower would change depending on whether the plant was left to sway freely in the wind or was artiﬁcially supported.2 Suppose that the unsupported stem diameters at the base of a particular species of sunﬂower plant have a normal distribution with an average diameter of 35 millimeters (mm) and a standard deviation of 3 mm. a. What is the probability that a sunﬂower plant will have a basal diameter of more than 40 mm? b. If two sunﬂower plants are randomly selected, what is the probability that both plants will have a basal diameter of more than 40 mm? b. What proportion of healthy people will have CBF c. Within what limits would you expect the basal readings above 100? c. If a person has a CBF reading below 40, he is classiﬁed as at risk for a stroke. What proportion of healthy people will mistakenly be diagnosed as “at risk”? 6.22 Braking Distances For a car traveling 30 miles per hour (mph), the distance required to brake to a stop is normally distributed with a mean of 50 feet and a standard deviation of 8 feet. Suppose you are traveling 30 mph in a residential area and a car moves abruptly into your path at a distance of 60 feet. diameters to lie, with probability.95? d. What diameter represents the 90th percentile of the distribution of diameters? 6.26 Breathing Rates The number of times x an adult human breathes per minute when at rest depends on the age of the human and varies greatly from person to person. Suppose the probability distribution for x is approximately normal, with the mean equal to 16 and the standard deviation equal to 4. If a person is selected at random and the number x of breaths per minute while at rest is recorded, what is the probability that x will exceed 22? 236 ❍ CHAPTER 6 THE NORMAL PROBABILITY
DISTRIBUTION 6.27 Economic Forecasts One method of arriving at economic forecasts is to use a consensus approach. A forecast is obtained from each of a large number of analysts, and the average of these individual forecasts is the consensus forecast. Suppose the individual 2008 January prime interest rate forecasts of economic analysts are approximately normally distributed with the mean equal to 8.5% and a standard deviation equal to 0.2%. If a single analyst is randomly selected from among this group, what is the probability that the analyst’s forecast of the prime rate will take on these values? a. Exceed 8.75% b. Be less than 8.375% 6.28 Tax Audit How does the IRS decide on the percentage of income tax returns to audit for each state? Suppose they do it by randomly selecting 50 values from a normal distribution with a mean equal to 1.55% and a standard deviation equal to.45%. (Computer programs are available for this type of sampling.) a. What is the probability that a particular state will have more than 2.5% of its income tax returns audited? b. What is the probability that a state will have less than 1% of its income tax returns audited? 6.29 Bacteria in Drinking Water Suppose the numbers of a particular type of bacteria in samples of 1 milliliter (ml) of drinking water tend to be approximately normally distributed, with a mean of 85 and a standard deviation of 9. What is the probability that a given 1-ml sample will contain more than 100 bacteria? 6.30 Loading Grain A grain loader can be set to discharge grain in amounts that are normally distributed, with mean m bushels and standard deviation equal to 25.7 bushels. If a company wishes to use the loader to ﬁll containers that hold 2000 bushels of grain and wants to overﬁll only one container in 100, at what value of m should the company set the loader? 6.31 How Many Words? A publisher has discovered that the numbers of words contained in a new manuscript are normally distributed, with a mean equal to 20,000 words in excess of that speciﬁed in the author’s contract and a standard deviation of 10,000 words. If the publisher wants to be almost certain (say, with a probability of.95) that the manuscript will have less than 100,000 words, what number of words should the publisher specify in the contract? 6.32 Tennis Anyone? A
stringer of tennis rackets has found that the actual string tension achieved for any individual racket stringing will vary as much as 6 pounds per square inch from the desired tension set on the stringing machine. If the stringer wishes to string at a tension lower than that speciﬁed by a customer only 5% of the time, how much above or below the customer’s speciﬁed tension should the stringer set the stringing machine? (NOTE: Assume that the distribution of string tensions produced by the stringing machine is normally distributed, with a mean equal to the tension set on the machine and a standard deviation equal to 2 pounds per square inch.) 6.33 Mall Rats An article in American Demographics claims that more than twice as many shoppers are out shopping on the weekends than during the week.3 Not only that, such shoppers also spend more money on their purchases on Saturdays and Sundays! Suppose that the amount of money spent at shopping centers between 4 P.M. and 6 P.M. on Sundays has a normal distribution with mean $85 and with a standard deviation of $20. A shopper is randomly selected on a Sunday between 4 P.M. and 6 P.M. and asked about his spending patterns. a. What is the probability that he has spent more than $95 at the mall? b. What is the probability that he has spent between $95 and $115 at the mall? c. If two shoppers are randomly selected, what is the probability that both shoppers have spent more than $115 at the mall? 6.34 Pulse Rates Your pulse rate is a measure of the number of heartbeats per minute. It can be measured in several places on your body, where an artery passes close to the skin. Once you ﬁnd the pulse, count the number of beats per minute, or, count for 30 seconds and multiply by two. What’s a normal pulse rate? That depends on a variety of factors. Pulse rates between 60 and 100 beats per minute are considered normal for children over 10 and adults.4 Suppose that these pulse rates are approximately normally distributed with a mean of 78 and a standard deviation of 12. a. What proportion of adults will have pulse rates between 60 and 100? b. What is the 95th percentile for the pulse rates of adults? c. Would a pulse rate of 110 be considered unusual? Explain. 6.4 THE NORMAL APPROXIMATION TO THE BINOMIAL
PROBABILITY DISTRIBUTION (OPTIONAL) ❍ 237 THE NORMAL APPROXIMATION TO THE BINOMIAL PROBABILITY DISTRIBUTION (OPTIONAL) 6.4 In Chapter 5, you learned three ways to calculate probabilities for the binomial random variable x: • Using the binomial formula, P(x k) Cn • Using the cumulative binomial tables • Using the Java applets k pkq nk The binomial formula produces lengthy calculations, and the tables are available for only certain values of n and p. There is another option available when np 7; the Poisson probabilities can be used to approximate P(x k). When this approximation does not work and n is large, the normal probability distribution provides another approximation for binomial probabilities. THE NORMAL APPROXIMATION TO THE BINOMIAL PROBABILITY DISTRIBUTION Let x be a binomial random variable with n trials and probability p of success. The probability distribution of x is approximated using a normal curve with m np and s npq This approximation is adequate as long as n is large and p is not too close to 0 or 1. Since the normal distribution is continuous, the area under the curve at any single point is equal to 0. Keep in mind that this result applies only to continuous random variables. Because the binomial random variable x is a discrete random variable, the probability that x takes some speciﬁc value—say, x 11—will not necessarily equal 0. Figures 6.18 and 6.19 show the binomial probability histograms for n 25 with p.5 and p.1, respectively. The distribution in Figure 6.18 is exactly symmetric. ● p(x).2 FI GUR E 6. 18 The binomial probability distribution for n 25 and p.5 and the approximating normal distribution with m 12.5 and s 2.5.1 0 5 10 15 20 25 x 7.5 10.5 238 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION F IG URE 6. 19 The binomial probability distribution and the approximating normal distribution for n 25 and p.1 ● p(x).2.1 EXAMPLE 6.11 0 5 10 15 20 25 x If you superimpose a normal curve with the same mean, m np, and the same standard deviation, s npq, over the top of the bars, it “ﬁts
” quite well; that is, the areas under the curve are almost the same as the areas under the bars. However, when the probability of success, p, gets small and the distribution is skewed, as in Figure 6.19, the symmetric normal curve no longer ﬁts very well. If you try to use the normal curve areas to approximate the area under the bars, your approximation will not be very good. Use the normal curve to approximate the probability that x 8, 9, or 10 for a binomial random variable with n 25 and p.5. Compare this approximation to the exact binomial probability. Solution You can ﬁnd the exact binomial probability for this example because there are cumulative binomial tables for n 25. From Table 1 in Appendix I, P(x 8, 9, or 10) P(x 10) P(x 7).212.022.190 To use the normal approximation, ﬁrst ﬁnd the appropriate mean and standard deviation for the normal curve: m np 25(.5) 12.5 s npq 25(.5)(.5) 2.5 Only use the continuity correction if x has a binomial distribution! The probability that you need corresponds to the area of the three rectangles lying over x 8, 9, and 10. The equivalent area under the normal curve lies between x 7.5 (the lower edge of the rectangle for x 8) and x 10.5 (the upper edge of the rectangle for x 10). This area is shaded in Figure 6.18. 6.4 THE NORMAL APPROXIMATION TO THE BINOMIAL PROBABILITY DISTRIBUTION (OPTIONAL) ❍ 239 To ﬁnd the normal probability, follow the procedures of Section 6.3. First you stan- dardize each interval endpoint: 7.5 z x m 12.5 2.0 s 2.5 z x m 12.5.8 10.5 s.5 2 Then the approximate probability (shaded in Figure 6.20) is found from Table 3 in Appendix I: P(2.0 z.8).2119.0228.1891 You can compare the approximation,.1891, to the actual probability,.190. They are quite close! FI GUR E 6. 20 Area under the normal curve for Example 6.11 ● f(x) 7.5 –2.0
10.5 12.5 –.8 0 x z You can use the Java applet called Normal Approximation to Binomial Probabilities shown in Figure 6.21 to compare the actual and approximate probabilities for the binomial distribution in Example 6.11. Enter the appropriate values of n and p in the boxes at the top left corner of the applet, and press “Enter” to record each entry. The exact binomial distribution on the left of the applet will change depending on the value of n you have entered. Now change the value of k in the box at the bottom left corner of the applet, and press “Enter.” The applet will calculate the exact binomial probability P(x k) in the box marked “Prob:” It will also calculate the approximate probability using the area under the normal curve. The z-value, with the continuity correction is shown at the top right, and the approximate probability is shown to the left of the normal curve. For Example 6.11, the applet calculates the normal approximation as P(x 10).2119. What is the exact value of P(x 10)? If you change k to 7 and press “Enter,” what is the approximate value for P(x 7)? Now calculate P(8 x 10). Does it match the answer we got in Example 6.11? You will use this applet again for the MyApplet Exercises section at the end of the chapter. 240 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION ● F IG URE 6. 21 Normal Approximation to Binomial Probabilities applet You must be careful not to exclude half of the two extreme probability rectangles when you use the normal approximation to the binomial probability distribution. This adjustment, called the continuity correction, helps account for the fact that you are approximating a discrete random variable with a continuous one. If you forget the correction, your approximation will not be very good! Use this correction only for binomial probabilities; do not try to use it when the random variable is already continuous, such as a height or weight. How can you tell when it is appropriate to use the normal approximation to binomial probabilities? The normal approximation works well when the binomial histogram is roughly symmetric. This happens when the binomial distribution is not “bunched up” near 0 or n—that is, when it can spread out at least two
standard deviations from its mean without exceeding its limits, 0 and n. Using this criterion, you can derive this simple rule of thumb: RULE OF THUMB The normal approximation to the binomial probabilities will be adequate if both np 5 and nq 5 How Do I Calculate Binomial Probabilities Using the Normal Approximation? • Find the necessary values of n and p. Calculate m np and s npq. • Write the probability you need in terms of x and locate the appropriate area on the curve. • Correct the value of x by.5 to include the entire block of probability for that value. This is the continuity correction. 6.4 THE NORMAL APPROXIMATION TO THE BINOMIAL PROBABILITY DISTRIBUTION (OPTIONAL) ❍ 241 • Convert the necessary x-values to z-values using x np.5 z q np • Use Table 3 in Appendix I to calculate the approximate probability. Exercise Reps Consider a binomial random variable with n 30 and p.4. Fill in the blanks below to ﬁnd some probabilities using the normal approximation. A. Preliminary Steps: 1. Can we use the normal approximation? Calculate np 2. Are np and nq both greater than 5? 3. If the answer to Question 2 is yes, calculate m np s npq Yes and nq No and B. Calculate the Probability: 1. To ﬁnd the probability of 20 or more successes, what values of x should be in- cluded? x 2. To include the entire block of probability for the ﬁrst value of x, start at. np.5 3. Calculate z q np x. 4. Calculate P(x 20) P(z ) 1. Progress Report • Still having trouble? Try again using the Exercise Reps at the end of this section. • Mastered the normal approximation? You can skip the Exercise Reps at the end of this section! Answers are located on the perforated card at the back of this book. EXAMPLE 6.12 The reliability of an electrical fuse is the probability that a fuse, chosen at random from production, will function under its designed conditions. A random sample of 1000 fuses was tested and x 27 defectives were observed. Calculate the approximate probability of observing 27 or more defectives, assuming that the fuse reliability is.98. Solution The probability of observing a defective when a single fuse is tested is p.02
, given that the fuse reliability is.98. Then If np and nq are both greater than 5, you can use the normal approximation. m np 1000(.02) 20 s npq 1000(.02)(.98) 4.43 The probability of 27 or more defective fuses, given n 1000, is P(x 27) p(27) p(28) p(29) p(999) p(1000) 242 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION It is appropriate to use the normal approximation to the binomial probability because np 1000(.02) 20 and nq 1000(.98) 980 are both greater than 5. The normal area used to approximate P(x 27) is the area under the normal curve to the right of 26.5, so that the entire rectangle for x 27 is included. Then, the z-value corresponding to x 26.5 is z x m 20 26. 5. 5 6 1.47 s 3.4 4 3 4. 4 and the area to the left of z 1.47 is equal to.9292, as shown in Figure 6.22. Since the total area under the curve is 1, you have P(x 27) P(z 1.47) 1.9292.0708 F IG URE 6. 22 Normal approximation to the binomial for Example 6.12 ● f(x).9292 20 0 26.5 1.47 x z EXAMPLE 6.13 A producer of soft drinks was fairly certain that her brand had a 10% share of the soft drink market. In a market survey involving 2500 consumers of soft drinks, x 211 expressed a preference for her brand. If the 10% ﬁgure is correct, ﬁnd the probability of observing 211 or fewer consumers who prefer her brand of soft drink. Solution If the producer is correct, then the probability that a consumer prefers her brand of soft drink is p.10. Then m np 2500(.10) 250 s npq 2500(.10)(.90) 15 The probability of observing 211 or fewer who prefer her brand is P(x 211) p(0) p(1) p(210) p(211) The normal approximation to this probability is the area to the left of 211.5 under a normal curve with a mean of 250 and a standard deviation of 15. First calculate z x m 250 2.57 211.5 s 5
1 Then P(x 211) P(z 2.57).0051 The probability of observing a sample value of 211 or less when p.10 is so small that you can conclude that one of two things has occurred: Either you have observed 6.4 THE NORMAL APPROXIMATION TO THE BINOMIAL PROBABILITY DISTRIBUTION (OPTIONAL) ❍ 243 an unusual sample even though really p.10, or the sample reﬂects that the actual value of p is less than.10 and perhaps closer to the observed sample proportion, 211/2500.08. 6.4 EXERCISES EXERCISE REPS These exercises refer back to the MyPersonal Trainer section on page 240. 6.35 Consider a binomial random varible with n 25 and p.6. Fill in the blanks below to ﬁnd some probabilities using the normal approximation. a. Can we use the normal approximation? Calculate np b. Are np and nq both greater than 5? c. If the answer to part b is yes, calculate m np d. To ﬁnd the probability of more than 9 successes, what values of x should be included? and nq No and s npq Yes x e. To include the entire block of probability for the ﬁrst value of x, start at. np.5 f. Calculate z q np x. ) 1 g. Calculate P(x 9) P(z 6.36 Consider a binomial random variable with n 45 and p.05. Fill in the blanks below to ﬁnd some probabilities using the normal approximation. a. Can we use the normal approximation? Calculate np b. Are np and nq both greater than 5? c. If the answer to part b is yes, calculate m np d. To ﬁnd the probability of 10 or fewer successes, what values of x should be included? and nq No and s npq Yes. x e. To include the entire block of probability for the ﬁrst value of x, start at. np.5 f. Calculate z q np x. g. Calculate P(x 10) P(z ). BASIC TECHNIQUES 6.37 Let x be a binomial random variable with n 25 and p.3. a. Is the normal approximation appropriate for this binomial random variable? b. Find the mean and standard
deviation for x. c. Use the normal approximation to ﬁnd P(6 x 9). d. Use Table 1 in Appendix I to ﬁnd the exact probability P(6 x 9). Compare the results of parts c and d. How close was your approximation? 6.38 Let x be a binomial random variable with n 15 and p.5. a. Is the normal approximation appropriate? b. Find P(x 6) using the normal approximation. c. Find P(x 6) using the normal approximation. d. Find the exact probabilities for parts b and c, and compare these with your approximations. 6.39 Let x be a binomial random variable with n 100 and p.2. Find approximations to these probabilities: 244 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION b. P(x 22) d. P(x 25) a. P(x 22) c. P(20 x 25) 6.40 Let x be a binomial random variable for n 25, p.2. a. Use Table 1 in Appendix I to calculate P(4 x 6). b. Find m and s for the binomial probability distribution, and use the normal distribution to approximate the probability P(4 x 6). Note that this value is a good approximation to the exact value of P(4 x 6) even though np 5. b. P(x 7) 6.41 Suppose the random variable x has a binomial distribution corresponding to n 20 and p.30. Use Table 1 of Appendix I to calculate these probabilities: a. P(x 5) 6.42 Refer to Exercise 6.41. Use the normal approximation to calculate P(x 5) and P(x 7). Compare with the exact values obtained from Table 1 in Appendix I. 6.43 Consider a binomial experiment with n 20 and p.4. Calculate P(x 10) using each of these methods: a. Table 1 in Appendix I b. The normal approximation to the binomial probabil- ity distribution 6.44 Find the normal approximation to P(355 x 360) for a binomial probability distribution with n 400 and p.9. APPLICATIONS 6.45 In-Home Movies How often do you watch movies at home? A USA Today Snapshot found that about 7 in 10 adults say they watch movies at home at least once a week.5 Suppose a random sample of How Often We Do At
-Home Movies Once a week or more 71% Every few weeks 11% Once a month 6% Every few months 4% Less often 8% n 50 adults are polled and asked if they had watched a movie at home this week. Let us assume that p.7 is, in fact, correct. What are the probabilities for the following events? a. Fewer than 30 individuals watched a movie at home this week? b. More than 42 individuals watched a movie at home this week? c. Fewer than 10 individuals did not watch a movie at home this week? 6.46 Genetic Defects Data collected over a long period of time show that a particular genetic defect occurs in 1 of every 1000 children. The records of a medical clinic show x 60 children with the defect in a total of 50,000 examined. If the 50,000 children were a random sample from the population of children represented by past records, what is the probability of observing a value of x equal to 60 or more? Would you say that the observation of x 60 children with genetic defects represents a rare event? 6.47 No Shows Airlines and hotels often grant reservations in excess of capacity to minimize losses due to no-shows. Suppose the records of a hotel show that, on the average, 10% of their prospective guests will not claim their reservation. If the hotel accepts 215 reservations and there are only 200 rooms in the hotel, what is the probability that all guests who arrive to claim a room will receive one? 6.48 Lung Cancer Compilation of large masses of data on lung cancer shows that approximately 1 of every 40 adults acquires the disease. Workers in a certain occupation are known to work in an air-polluted environment that may cause an increased rate of lung cancer. A random sample of n 400 workers shows 19 with identiﬁable cases of lung cancer. Do the data provide sufficient evidence to indicate a higher rate of lung cancer for these workers than for the national average? 6.49 Tall or Short? Is a tall president better than a short one? Do Americans tend to vote for the taller of the two candidates in a presidential selection? In 33 of our presidential elections between 1856 and 2006, 17 of the winners were taller than their opponents.1 Assume that Americans are not biased by a candidate’s height and that the winner is just as likely to be taller or shorter than his opponent. Is the observed number of taller winners in the U.S. presidential elections unusual? 6.4 THE NOR
MAL APPROXIMATION TO THE BINOMIAL PROBABILITY DISTRIBUTION (OPTIONAL) ❍ 245 a. Find the approximate probability of ﬁnding 17 or more of the 33 pairs in which the taller candidate wins. b. Based on your answer to part a, can you conclude that Americans might consider a candidate’s height when casting their ballot? 6.50 The Rh Factor In a certain population, 15% of the people have Rh-negative blood. A blood bank serving this population receives 92 blood donors on a particular day. a. What is the probability that 10 or fewer are Rh-negative? b. What is the probability that 15 to 20 (inclusive) of the donors are Rh-negative? c. What is the probability that more than 80 of the donors are Rh-positive? 6.51 Pepsi’s Market Share Two of the biggest soft drink rivals, Pepsi and Coke, are very concerned about their market share. The following pie chart, which appeared on the company website (http:// www.pepsico.com) in November, 2006, claims that Pepsi-Cola’s share of the U.S. beverage market is 26%.6 Assume that this proportion will be close to the probability that a person selected at random indicates a preference for a Pepsi product when choosing a soft drink. U.S. Liquid Refreshment Beverage Market Share % Volume in Measured Channels Private Label 14% Other 20% Coca-Cola 24% PepsiCo 26% Cadbury Schweppes 10% Nestle 6% PepsiCo has the leading share of the liquid refreshment beverage market. A test group of 500 consumers is randomly selected. Use the normal curve to approximate the following binomial probabilities: a. Exactly 150 consumers prefer a Pepsi product. b. Between 120 and 150 consumers (inclusive) prefer a Pepsi product. c. Fewer than 150 consumers prefer a Pepsi product. d. Would it be unusual to ﬁnd that 232 of the 500 consumers preferred a Pepsi product? If this were to occur, what conclusions would you draw? 6.52 Ready, Set, Relax! The typical American family spends lots of time driving to and from various activities, and lots of time in the drive-thru lines at fast-food restaurants. There is a rising amount of evidence suggesting that we are beginning to burn out! In fact, in a study conducted for the Center for a New American Dream, Time magazine reports that 60%
of Americans felt pressure to work too much, and 80% wished for more family time.7 Assume that these percentages are correct for all Americans, and that a random sample of 25 Americans is selected. a. Use Table 1 in Appendix I to ﬁnd the prob- ability that more than 20 felt pressure to work too much. b. Use the normal approximation to the binomial distribution to aproximate the probability in part a. Compare your answer with the exact value from part a. c. Use Table 1 in Appendix I to ﬁnd the probability that between 15 and 20 (inclusive) wished for more family time. d. Use the normal approximation to the binomial distribution to approximate the probability in part c. Compare your answer with the exact value from part c. 6.53 We said, “Relax!” The article in Time magazine7 (Exercise 6.52) also reported that 80% of men and 62% of women put in more than 40 hours a week on the job. Assume that these percentages are correct for all Americans, and that a random sample of 50 working women is selected. a. What is the average number of women who put in more than 40 hours a week on the job? b. What is the standard deviation for the number of women who put in more than 40 hours a week on the job? c. Suppose that in our sample of 50 working women, there are 25 who work more than 40 hours a week. Would you consider this to be an unusual occurrence? Explain. 246 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION CHAPTER REVIEW Key Concepts and Formulas I. Continuous Probability Distributions III. The Standard Normal Distribution 1. Continuous random variables 2. Probability distributions or probability density functions a. Curves are smooth. b. Area under the curve equals 1. c. The area under the curve between a and b represents the probability that x falls between a and b. d. P(x a) 0 for continuous random vari- ables. 1. The standard normal random variable z has mean 0 and standard deviation 1. 2. Any normal random variable x can be transformed to a standard normal random variable using m z x s 3. Convert necessary values of x to z. 4. Use Table 3 in Appendix I to compute standard normal probabilities. 5. Several important z-values have right-tail areas II. The Normal Probability Distribution as follows: 1. Symmetric
about its mean m 2. Shape determined by its standard deviation s Right-Tail Area z-Value.005 2.58.01 2.33.025 1.96.05.10 1.645 1.28 Normal Probabilities When the random variable of interest has a normal probability distribution, you can generate either of these probabilities: • Cumulative probabilities—P(x k)—for a given value of k • Inverse cumulative probabilities—the value of k such that the area to its left under the normal probability distribution is equal to a You must specify which normal distribution you are using and the necessary parameters: the mean m and the standard deviation s. As in Chapter 5, you have the option of specifying only one single value of k (or a) or several values of k (or a), which should be stored in a column (say, C1) of the MINITAB worksheet. Suppose that the average birth weights of babies born at hospitals owned by a major health maintenance organization (HMO) are approximately normal with mean 6.75 pounds and standard deviation.54 pound. What proportion of babies born at these hospitals weigh between 6 and 7 pounds? To use MINITAB to ﬁnd P(6 x 7), name column C1 as “x” and enter the critical values x 6 and x 7 into this column. Use Calc Probability Distributions Normal to generate the Dialog box, as shown in Figure 6.23. Type the values for m and s in the appropriate boxes (the default values generate probabilities for the standard normal z distribution), and select C1 for the input column. (If you do not type a column number for optional storage, MINITAB will display the results in the Session window.) Make sure that the radio button marked MY MINITAB ❍ 247 FI GUR E 6. 23 ● FI GUR E 6. 24 ● “Cumulative probability” is selected. The cumulative distribution function for x 6 and x 7 appears in the Session window when you click OK (see Figure 6.24). To ﬁnd P(6 x 7), remember that the cumulative probability is the area to the left of the given value of x. Hence, P(6 x 7) P(x 7) P(x 6).678305.082433.595872 You can check this calculation using Table 3 in Appendix I if you wish! To inverse calculate cumulative probabilities, use Calc→Pro
bability Distributions→Normal again. Make sure that the radio button marked “Inverse cumulative probability” is selected, and that the mean and standard deviation are entered into the proper boxes. Then enter the appropriate values of a in C1, or if you have only a single value, enter the value in the Input constant box. For example, to ﬁnd the 95th percentile of the birth weights, you look for a value k such that only 5% of the values of x exceed this value (and 95% are less than or equal to k). If you enter the probability.95 into the Input constant box and select the option marked “Inverse cumulative probability,” the 95th percentile will appear in the Session window, as in Figure 6.25. That is, 95% of all babies born at these hospitals weigh 7.63822 pounds or less. Would you consider a baby who weighs 9 pounds to be unusually large? 248 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION F IG URE 6. 25 ● Supplementary Exercises 6.54 Using Table 3 in Appendix I, calculate the area under the standard normal curve to the left of the following: a. z 1.2 c. z 1.46 b. z.9 d. z.42 6.55 Find the following probabilities for the standard normal random variable: a. P(.3 z 1.56) b. P(.2 z.2) 6.56 a. Find the probability that z is greater than.75. b. Find the probability that z is less than 1.35. 6.57 Find z0 such that P(z z0).5. 6.58 Find the probability that z lies between z 1.48 and z 1.48. 6.59 Find z0 such that P(z0 z z0).5. What percentiles do z0 and z0 represent? 6.60 Drill Bits The life span of oil-drilling bits depends on the types of rock and soil that the drill encounters, but it is estimated that the mean length of life is 75 hours. Suppose an oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 75 hours and a standard deviation equal to 12 hours. a. What proportion of the company’s drill bits will fail before 60 hours of use? b. What proportion will last at least 60 hours? c.
What proportion will have to be replaced after more than 90 hours of use? 6.61 Faculty Ages The inﬂux of new ideas into a college or university, introduced primarily by new young faculty, is becoming a matter of concern because of the increasing ages of faculty members; that is, the distribution of faculty ages is shifting upward due most likely to a shortage of vacant positions and an oversupply of PhDs. Thus, faculty members are reluctant to move and give up a secure position. If the retirement age at most universities is 65, would you expect the distribution of faculty ages to be normal? Explain. 6.62 Bearing Diameters A machine operation produces bearings whose diameters are normally distributed, with mean and standard deviation equal to.498 and.002, respectively. If speciﬁcations require that the bearing diameter equal.500 inch.004 inch, what fraction of the production will be unacceptable? 6.63 Used Cars A used-car dealership has found that the length of time before a major repair is required on the cars it sells is normally distributed with a mean equal to 10 months and a standard deviation of 3 months. If the dealer wants only 5% of the cars to fail before the end of the guarantee period, for how many months should the cars be guaranteed? 6.64 Restaurant Sales The daily sales total (excepting Saturday) at a small restaurant has a probability distribution that is approximately normal, with a mean m equal to $1230 per day and a standard deviation s equal to $120. a. What is the probability that the sales will exceed $1400 for a given day? b. The restaurant must have at least $1000 in sales per day to break even. What is the probability that on a given day the restaurant will not break even? 6.65 Washers The life span of a type of automatic washer is approximately normally distributed with mean and standard deviation equal to 10.5 and 3.0 years, respectively. If this type of washer is guaranteed for a period of 5 years, what fraction will need to be repaired and/or replaced? 6.66 Garage Door Openers Most users of automatic garage door openers activate their openers at distances that are normally distributed with a mean of 30 feet and a standard deviation of 11 feet. To minimize interference with other remote-controlled devices, the manufacturer is required to limit the operating distance to 50 feet. What percentage of the time will users attempt to operate the opener outside its operating limit? 6.
67 How Long Is the Test? The average length of time required to complete a college achievement test was found to equal 70 minutes with a standard deviation of 12 minutes. When should the test be terminated if you wish to allow sufficient time for 90% of the students to complete the test? (Assume that the time required to complete the test is normally distributed.) 6.68 Servicing Automobiles The length of time required for the periodic maintenance of an automobile will usually have a probability distribution that is mound-shaped and, because some long service times will occur occasionally, is skewed to the right. The length of time required to run a 5000-mile check and to service an automobile has a mean equal to 1.4 hours and a standard deviation of.7 hour. Suppose that the service department plans to service 50 automobiles per 8-hour day and that, in order to do so, it must spend no more than an average of 1.6 hours per automobile. What proportion of all days will the service department have to work overtime? 6.69 TV Viewers An advertising agency has stated that 20% of all television viewers watch a particular program. In a random sample of 1000 viewers, x 184 viewers were watching the program. Do these data present sufficient evidence to contradict the advertiser’s claim? SUPPLEMENTARY EXERCISES ❍ 249 forecasts “showed that the average estimate missed the mark by 15%.” a. Suppose the distribution of these forecast errors has a mean of 15% and a standard deviation of 10%. Is it likely that the distribution of forecast errors is approximately normal? b. Suppose the probability is.5 that a corporate executive’s forecast error exceeds 15%. If you were to sample the forecasts of 100 corporate executives, what is the probability that more than 60 would be in error by more than 15%? 6.71 Filling Soda Cups A soft drink machine can be regulated to discharge an average of m ounces per cup. If the ounces of ﬁll are normally distributed, with standard deviation equal to.3 ounce, give the setting for m so that 8-ounce cups will overﬂow only 1% of the time. 6.72 Light Bulbs A manufacturing plant uses 3000 electric light bulbs whose life spans are normally distributed, with mean and standard deviation equal to 500 and 50 hours, respectively. In order to minimize the number of bulbs that burn out during operating hours, all the bulbs are replaced after a given period of operation. How often should the
bulbs be replaced if we wish no more than 1% of the bulbs to burn out between replacement periods? 6.73 The Freshman Class The admissions office of a small college is asked to accept deposits from a number of qualiﬁed prospective freshmen so that, with probability about.95, the size of the freshman class will be less than or equal to 120. Suppose the applicants constitute a random sample from a population of applicants, 80% of whom would actually enter the freshman class if accepted. a. How many deposits should the admissions coun- selor accept? b. If applicants in the number determined in part a are accepted, what is the probability that the freshman class size will be less than 105? 6.74 No Shows An airline ﬁnds that 5% of the persons making reservations on a certain ﬂight will not show up for the ﬂight. If the airline sells 160 tickets for a ﬂight that has only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to ﬂy? 6.70 Forecasting Earnings A researcher notes that senior corporation executives are not very accurate forecasters of their own annual earnings. He states that his studies of a large number of company executive 6.75 Long Distance It is known that 30% of all calls coming into a telephone exchange are long-distance calls. If 200 calls come into the exchange, what is the probability that at least 50 will be long-distance calls? 250 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION 6.76 Plant Genetics In Exercise 5.75, a cross between two peony plants—one with red petals and one with streaky petals—produced offspring plants with red petals 75% of the time. Suppose that 100 seeds from this cross were collected and germinated, and x, the number of plants with red petals, was recorded. a. What is the exact probability distribution for x? b. Is it appropriate to approximate the distribution in part a using the normal distribution? Explain. c. Use an appropriate method to ﬁnd the approximate probability that between 70 and 80 (inclusive) offspring plants have red ﬂowers. d. What is the probability that 53 or fewer offspring plants had red ﬂowers? Is this an unusual occurrence? e. If you actually observed 53 of 100 offspring plants with red ﬂowers, and if you were certain that
the genetic ratio 3:1 was correct, what other explanation could you give for this unusual occurrence? 6.77 Suppliers A or B? A purchaser of electric relays buys from two suppliers, A and B. Supplier A supplies two of every three relays used by the company. If 75 relays are selected at random from those in use by the company, ﬁnd the probability that at most 48 of these relays come from supplier A. Assume that the company uses a large number of relays. 6.78 Snacking and TV Is television dangerous to your diet? Psychologists believe that excessive eating may be associated with emotional states (being upset or bored) and environmental cues (watching television, reading, and so on). To test this theory, suppose you randomly selected 60 overweight persons and matched them by weight and gender in pairs. For a period of 2 weeks, one of each pair is required to spend evenings reading novels of interest to him or her. The other member of each pair spends each evening watching television. The calorie count for all snack and drink intake for the evenings is recorded for each person, and you record x 19, the number of pairs for which the television watchers’ calorie intake exceeded the intake of the readers. If there is no difference in the effects of television and reading on calorie intake, the probability p that the calorie intake of one member of a pair exceeds that of the other member is.5. Do these data provide sufficient evidence to indicate a difference between the effects of television watching and reading on calorie intake? (HINT: Calculate the z-score for the observed value, x 19.) 6.79 Gestation Times The Biology Data Book reports that the gestation time for human babies averages 278 days with a standard deviation of 12 days.8 Suppose that these gestation times are normally distributed. a. Find the upper and lower quartiles for the gestation times. b. Would it be unusual to deliver a baby after only 6 months of gestation? Explain. 6.80 Tax Audits In Exercise 6.28, we suggested that the IRS assign auditing rates per state by randomly selecting 50 auditing percentages from a normal distribution with a mean equal to 1.55% and a standard deviation of.45%. a. What is the probability that a particular state would have more than 2% of its tax returns audited? b. What is the expected value of x, the number of states that will have more than 2% of their income tax returns aud
ited? c. Is it likely that as many as 15 of the 50 states will have more than 2% of their income tax returns audited? 6.81 Your Favorite Sport There is a difference in sports preferences between men and women, according to a recent survey. Among the 10 most popular sports, men include competition-type sports—pool and billiards, basketball, and softball—whereas women include aerobics, running, hiking, and calisthenics. However, the top recreational activity for men was still the relaxing sport of ﬁshing, with 41% of those surveyed indicating that they had ﬁshed during the year. Suppose 180 randomly selected men are asked whether they had ﬁshed in the past year. a. What is the probability that fewer than 50 had ﬁshed? b. What is the probability that between 50 and 75 had ﬁshed? c. If the 180 men selected for the interview were selected by the marketing department of a sportinggoods company based on information obtained from their mailing lists, what would you conclude about the reliability of their survey results? 6.82 Introvert or Extrovert? A psychological introvert–extrovert test produced scores that had a normal distribution with a mean and standard deviation of 75 and 12, respectively. If we wish to designate the highest 15% as extroverts, what would be the proper score to choose as the cutoff point? 6.83 Curving the Grades Students very often ask their professors whether they will be “curving the grades.” The traditional interpretation of “curving grades” required that the grades have a normal distribution, and that the grades will be assigned in these proportions: Letter Grade A B C D F Proportion of Students 10% 20% 40% 20% 10% a. If the average “C” grade is centered at the average grade for all students, and if we assume that the grades are normally distributed, how many standard deviations on either side of the mean will constitute the “C” grades? b. How many deviations on either side of the mean will be the cutoff points for the “B” and “D” grades? 6.84 Curving the Grades, continued Refer to Exercise 6.83. For ease of calculation, round the Exercises Use one of the three applets (Normal Distribution Probabilities, Normal Probabilities and z-Scores, or Normal App
roximation to Binomial Probabilities) described in this chapter to solve the following exercises. 6.86 Calculate the area under the standard normal curve to the left of these values: b. z 2.34 a. z.90 c. z 5.4 6.87 Calculate the area under the standard normal curve between these values: a. z 2.0 and z 2.0 b. z 2.3 and 1.5 6.88 Find the following probabilities for the standard normal random variable z: a. P(1.96 z 1.96) c. P(z 1.96) b. P(z 1.96) 6.89 a. Find a z0 such that P(z z0).9750. b. Find a z0 such that P(z z0).3594. 6.90 a. Find a z0 such that P(z0 z z0).95. b. Find a z0 such that P(z0 z z0).98. 6.91 A normal random variable x has mean m 5 and s 2. Find the following probabilities of these x-values: a. 1.2 x 10 b. x 7.5 c. x 0 MYAPPLET EXERCISES ❍ 251 number of standard deviations for “C” grades to.5 standard deviations, and for “B” and “D” grades to 1.5 standard deviations. Suppose that the distribution of grades for a large class of students has an average of 78 with a standard deviation of 11. Find the appropriate cutoff points for the grades A, B, C, D, and F. 6.85 Normal Temperatures In Exercise 1.67, Allen Shoemaker derived a distribution of human body temperatures, which has a distinct mound-shape.9 Suppose we assume that the temperatures of healthy humans is approximately normal with a mean of 98.6 degrees and a standard deviation of 0.8 degrees. a. If a healthy person is selected at random, what is the probability that the person has a temperature above 99.0 degrees? b. What is the 95th percentile for the body tempera- tures of healthy humans? 6.92 Let x be a binomial random variable with n 36 and p.54. Use the normal approximation to ﬁnd: a. P(x 25) c. P(x 30) b. P(15 x 20)
6.93 Cellphone Etiquette A Snapshot in USA Today indicates that 51% of Americans say the average person is not very considerate of others when talking on a cellphone.10 Suppose that 100 Americans are randomly selected. How Polite Are Cellphone Users? Somewhat 44% Not very 51% Extremely 4% a. Use the Calculating Binomial Probabilities applet from Chapter 5 to ﬁnd the exact probability that 60 252 ❍ CHAPTER 6 THE NORMAL PROBABILITY DISTRIBUTION or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone. b. Use the Normal Approximation to Binomial Probabilities to approximate the probability in part a. Compare your answers. 6.94 Stamps Philatelists (stamp collectors) often buy stamps at or near retail prices, but, when they sell, the price is considerably lower. For example, it may be reasonable to assume that (depending on the mix of a collection, condition, demand, economic conditions, etc.) a collection will sell at x% of the retail price, where x is normally distributed with a mean equal to 45% and a standard deviation of 4.5%. If a philatelist has a collection to sell that has a retail value of $30,000, what is the probability that the philatelist receives these amounts for the collection? a. More than $15,000 c. Less than $12,000 b. Less than $15,000 6.95 Test Scores The scores on a national achievement test were approximately normally distributed, with a mean of 540 and a standard deviation of 110. a. If you achieved a score of 680, how far, in standard deviations, did your score depart from the mean? b. What percentage of those who took the examina- tion scored higher than you? 6.96 Faculty Salaries Although faculty salaries at colleges and universities in the United States continue to rise, they do not always keep pace with the cost of living nor with salaries in the private sector. In 2005, the National Center for Educational Statistics indicated that the average salary for Assistant Professors at public four-year colleges was $50,581.11 Suppose that these salaries are normally distributed with a standard deviation of $4000. a. What proportion of assistant professors at public 4-year colleges will have salaries less than $45,000? b. What proportion of these professors will have salaries between $45,000 and $55,000? 6
.97 Transplanting Cells Briggs and King developed the technique of nuclear transplantation, in which the nucleus of a cell from one of the later stages of the development of an embryo is transplanted into a zygote (a single-cell fertilized egg) to see whether the nucleus can support normal development. If the probability that a single transplant from the early gastrula stage will be successful is.65, what is the probability that more than 70 transplants out of 100 will be successful? CASE STUDY The Long and Short of It If you were the boss, would height play a role in your selection of a successor for your job? In his Fortune column, Daniel Seligman discussed his ideas concerning height as a factor in Deng Xiaoping’s choice of Hu Yaobang as his replacement as Chairman of the Chinese Communist Party.12 As Seligman notes, the facts surrounding the case arouse suspicions when examined in the light of statistics. Deng, it seemed, was only 5 feet tall, a height that is short even in China. Therefore, the choice of Hu Yaobang, who was also 5 feet tall, raised (or lowered) some eyebrows because, as Seligman notes, “the odds against a ‘height-blind’ decision producing a chairman as short as Deng are about 40 to 1.” In other words, if we had the relative frequency distribution of the heights of all Chinese men, only 1 in 41 (i.e., 2.4%) would be 5 feet tall or shorter. To calculate these odds, Seligman notes that the Chinese equivalent of the U.S. Health Service does not exist and hence that health statistics on the current population of China are difficult to acquire. He says, however, that “it is generally held that a boy’s length at birth represents 28.6% of his ﬁnal height” and that, in prerevolutionary China, the average length of a Chinese boy at birth was 18.9 inches. From this, Seligman deduces that the mean height of mature Chinese men is.9 8 1 66.08 inches, or 5 feet 6.08 inches 8 6.2 CASE STUDY ❍ 253 He then assumes that the distribution of the heights of men in China follows a normal distribution (“as it does in the U.S.”) with a mean of 66 inches and a standard deviation equal to 2.7 inches,
“a ﬁgure that looks about right for that mean.” 1. Using Seligman’s assumptions, calculate the probability that a single adult Chinese man, chosen at random, will be less than or equal to 5 feet tall, or equivalently, 60 inches tall. 2. Do the results in part 1 agree with Seligman’s odds? 3. Comment on the validity of Seligman’s assumptions. Are there any basic ﬂaws in his reasoning? 4. Based on the results of parts 1 and 3, do you think that Deng Xiaoping took height into account in selecting his successor? Sampling Distributions GENERAL OBJECTIVES In the past several chapters, we studied populations and the parameters that describe them. These populations were either discrete or continuous, and we used probability as a tool for determining how likely certain sample outcomes might be. In this chapter, our focus changes as we begin to study samples and the statistics that describe them. These sample statistics are used to make inferences about the corresponding population parameters. This chapter involves sampling and sampling distributions, which describe the behavior of sample statistics in repeated sampling. CHAPTER INDEX ● The Central Limit Theorem (7.4) ● Random samples (7.2) ● The sampling distribution of the sample mean, x (7.5) ● The sampling distribution of the sample proportion, pˆ (7.6) ● Sampling plans and experimental designs (7.2) ● Statistical process control: x and p charts (7.7) ● Statistics and sampling distributions (7.3) How Do I Calculate Probabilities for the Sample Mean x? How Do I Calculate Probabilities for the Sample Proportion ˆp? 7 © PictureNet/CORBIS Sampling the Roulette at Monte Carlo How would you like to try your hand at gambling without the risk of losing? You could do it by simulating the gambling process, making imaginary bets, and observing the results. This technique, called a Monte Carlo procedure, is the topic of the case study at the end of this chapter. 254 7.2 SAMPLING PLANS AND EXPERIMENTAL DESIGNS ❍ 255 INTRODUCTION 7.1 Parameter ⇔ Population Statistic ⇔ Sample In the previous three chapters, you have learned a lot about probability distributions, such as the binomial and normal distributions. The shape of the normal distribution is determined by its mean m and its standard
deviation s, whereas the shape of the binomial distribution is determined by p. These numerical descriptive measures— called parameters—are needed to calculate the probability of observing sample results. In practical situations, you may be able to decide which type of probability distribution to use as a model, but the values of the parameters that specify its exact form are unknown. Here are two examples: • A pollster is sure that the responses to his “agree/disagree” questions will follow a binomial distribution, but p, the proportion of those who “agree” in the population, is unknown. • An agronomist believes that the yield per acre of a variety of wheat is approximately normally distributed, but the mean m and standard deviation s of the yields are unknown. In these cases, you must rely on the sample to learn about these parameters. The proportion of those who “agree” in the pollster’s sample provides information about the actual value of p. The mean and standard deviation of the agronomist’s sample approximate the actual values of m and s. If you want the sample to provide reliable information about the population, however, you must select your sample in a certain way! SAMPLING PLANS AND EXPERIMENTAL DESIGNS 7.2 The way a sample is selected is called the sampling plan or experimental design and determines the quantity of information in the sample. Knowing the sampling plan used in a particular situation will often allow you to measure the reliability or goodness of your inference. Simple random sampling is a commonly used sampling plan in which every sample of size n has the same chance of being selected. For example, suppose you want to select a sample of size n 2 from a population containing N 4 objects. If the four objects are identiﬁed by the symbols x1, x2, x3, and x4, there are six distinct pairs that could be selected, as listed in Table 7.1. If the sample of n 2 observations is selected so that each of these six samples has the same chance of selection, given by 1/6, then the resulting sample is called a simple random sample, or just a random sample. TABLE 7.1 ● Ways of Selecting a Sample of Size 2 from 4 Objects Sample Observations in Sample 1 2 3 4 5 6 x1, x2 x1, x3 x1, x4 x2, x3 x2, x4 x3, x4 256 �
� CHAPTER 7 SAMPLING DISTRIBUTIONS EXAMPLE 7.1 TABLE 7.2 Definition If a sample of n elements is selected from a population of N elements using a sampling plan in which each of the possible samples has the same chance of selection, then the sampling is said to be random and the resulting sample is a simple random sample. Perfect random sampling is difficult to achieve in practice. If the size of the population N is small, you might write each of N numbers on a poker chip, mix the chips, and select a sample of n chips. The numbers that you select correspond to the n measurements that appear in the sample. Since this method is not always very practical, a simpler and more reliable method uses random numbers—digits generated so that the values 0 to 9 occur randomly and with equal frequency. These numbers can be generated by computer or may even be available on your scientiﬁc calculator. Alternatively, Table 10 in Appendix I is a table of random numbers that you can use to select a random sample. A computer database at a downtown law ﬁrm contains ﬁles for N 1000 clients. The ﬁrm wants to select n 5 ﬁles for review. Select a simple random sample of 5 ﬁles from this database. Solution You must ﬁrst label each ﬁle with a number from 1 to 1000. Perhaps the ﬁles are stored alphabetically, and the computer has already assigned a number to each. Then generate a sequence of 10 three-digit random numbers. If you are using Table 10 of Appendix I, select a random starting point and use a portion of the table similar to the one shown in Table 7.2. The random starting point ensures that you will not use the same sequence over and over again. The ﬁrst three digits of Table 7.2 indicate the number of the ﬁrst ﬁle to be reviewed. The random number 001 corresponds to ﬁle #1, and the last ﬁle, #1000, corresponds to the random number 000. Using Table 7.2, you would choose the ﬁve ﬁles numbered 155, 450, 32, 882, and 350 for review. Alternately, you might choose to read across the lines, and choose ﬁles 155, 350, 989, 450 and 369 for review. ● Portion of a Table of Random Numbers 989
24 28630 50856 21121 35026 36933 78812 26053 15574 45045 03225 88292 The situation described in Example 7.1 is called an observational study because the data already existed before you decided to observe or describe their characteristics. Most sample surveys, in which information is gathered with a questionnaire, fall into this category. Computer databases make it possible to assign identiﬁcation numbers to each element even when the population is large and to select a simple random sample. You must be careful when conducting a sample survey, however, to watch for these frequently occurring problems: • Nonresponse: You have carefully selected your random sample and sent out your questionnaires, but only 50% of those surveyed return their questionnaires. Are the responses you received still representative of the entire population, or are they biased because only those people who were particularly opinionated about the subject chose to respond? EXAMPLE 7.2 7.2 SAMPLING PLANS AND EXPERIMENTAL DESIGNS ❍ 257 • Undercoverage: You have selected your random sample using telephone records as a database. Does the database you used systematically exclude certain segments of the population—perhaps those who do not have telephones? • Wording bias: Your questionnaire may have questions that are too complicated or tend to confuse the reader. Possibly the questions are sensitive in nature—for example, “Have you ever used drugs?” or “Have you ever cheated on your income tax?”—and the respondents will not answer truthfully. Methods have been devised to solve some of these problems, but only if you know that they exist. If your survey is biased by any of these problems, then your conclusions will not be very reliable, even though you did select a random sample! Some research involves experimentation, in which an experimental condition or treatment is imposed on the experimental units. Selecting a simple random sample is more difficult in this situation. A research chemist is testing a new method for measuring the amount of titanium (Ti) in ore samples. She chooses 10 ore samples of the same weight for her experiment. Five of the samples will be measured using a standard method, and the other 5 using the new method. Use random numbers to assign the 10 ore samples to the new and standard groups. Do these data represent a simple random sample from the population? Solution There are really two populations in this experiment. They consist of titanium measurements, using either the new or standard method, for all possible ore samples of
this weight. These populations do not exist in fact; they are hypothetical populations, envisioned in the mind of the researcher. Thus, it is impossible to select a simple random sample using the methods of Example 7.1. Instead, the researcher selects what she believes are 10 representative ore samples and hopes that these samples will behave as if they had been randomly selected from the two populations. The researcher can, however, randomly select the ﬁve samples to be measured with each method. Number the samples from 1 to 10. The ﬁve samples selected for the new method may correspond to 5 one-digit random numbers. Use this sequence of random digits generated on a scientiﬁc calculator: 948247817184610 Since you cannot select the same ore sample twice, you must skip any digit that has already been chosen. Ore samples 9, 4, 8, 2, and 7 will be measured using the new method. The other samples—1, 3, 5, 6, and 10—will be measured using the standard method. In addition to simple random sampling, there are other sampling plans that involve randomization and therefore provide a probabilistic basis for inference making. Three such plans are based on stratiﬁed, cluster, and systematic sampling. When the population consists of two or more subpopulations, called strata, a sampling plan that ensures that each subpopulation is represented in the sample is called a stratiﬁed random sample. Definition Stratiﬁed random sampling involves selecting a simple random sample from each of a given number of subpopulations, or strata. 258 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS Citizens’ opinions about the construction of a performing arts center could be collected using a stratiﬁed random sample with city voting wards as strata. National polls usually involve some form of stratiﬁed random sampling with states as strata. Another form of random sampling is used when the available sampling units are groups of elements, called clusters. For example, a household is a cluster of individuals living together. A city block or a neighborhood might be a convenient sampling unit and might be considered a cluster for a given sampling plan. Definition A cluster sample is a simple random sample of clusters from the available clusters in the population. When a particular cluster is included in the sample, a census of every element in the cluster is taken. Sometimes the population to be sampled is ordered, such as an
alphabetized list of people with driver’s licenses, a list of utility users arranged by service addresses, or a list of customers by account numbers. In these and other situations, one element is chosen at random from the ﬁrst k elements, and then every kth element thereafter is included in the sample. Definition A 1-in-k systematic random sample involves the random selection of one of the ﬁrst k elements in an ordered population, and then the systematic selection of every kth element thereafter. Not all sampling plans, however, involve random selection. You have probably heard of the nonrandom telephone polls in which those people who wish to express support for a question call one “900 number” and those opposed call a second “900 number.” Each person must pay for his or her call. It is obvious that those people who call do not represent the population at large. This type of sampling plan is one form of a convenience sample—a sample that can be easily and simply obtained without random selection. Advertising for subjects who will be paid a fee for participating in an experiment produces a convenience sample. Judgment sampling allows the sampler to decide who will or will not be included in the sample. Quota sampling, in which the makeup of the sample must reﬂect the makeup of the population on some preselected characteristic, often has a nonrandom component in the selection process. Remember that nonrandom samples can be described but cannot be used for making inferences! All sampling plans used for making inferences must involve randomization! 7.2 EXERCISES BASIC TECHNIQUES 7.1 A population consists of N 500 experimental units. Use a random number table to select a random sample of n 20 experimental units. (HINT: Since you need to use three-digit numbers, you can assign 2 three-digit numbers to each of the sampling units in the manner shown in the table.) What is the probability that each experimental unit is selected for inclusion in the sample? Experimental Units Random Numbers 1 2 3 4... 499 500 001, 501 002, 502 003, 503 004, 504... 499, 999 500, 000 7.2 A political analyst wishes to select a sample of n 20 people from a population of 2000. Use the random number table to identify the people to be included in the sample. 7.3 A population contains 50,000 voters. Use the random number table to identify the voters to be
included in a random sample of n 15. 7.4 A small city contains 20,000 voters. Use the random number table to identify the voters to be included in a random sample of n 15. 7.5 Every 10th Person A random sample of public opinion in a small town was obtained by selecting every 10th person who passed by the busiest corner in the downtown area. Will this sample have the characteristics of a random sample selected from the town’s citizens? Explain. 7.6 Parks and Recreation A questionnaire was mailed to 1000 registered municipal voters selected at random. Only 500 questionnaires were returned, and of the 500 returned, 360 respondents were strongly opposed to a surcharge proposed to support the city Parks and Recreation Department. Are you willing to accept the 72% ﬁgure as a valid estimate of the percentage in the city who are opposed to the surcharge? Why or why not? 7.7 DMV Lists In many states, lists of possible jurors are assembled from voter registration lists and Department of Motor Vehicles records of licensed drivers and car owners. In what ways might this list not cover certain sectors of the population adequately? 7.8 Sex and Violence One question on a survey questionnaire is phrased as follows: “Don’t you agree that there is too much sex and violence during prime TV viewing hours?” Comment on possible problems with the responses to this question. Suggest a better way to pose the question. APPLICATIONS 7.9 Cancer in Rats The Press Enterprise identiﬁed a byproduct of chlorination called MX that has been linked to cancer in rats.1 A scientist wants to conduct a validation study using 25 rats in the experimental group, each to receive a ﬁxed dose of MX, and 25 rats in a control group that will receive no MX. Determine a randomization scheme to assign the 50 individual rats to the two groups. 7.10 Racial Bias? Does the race of an interviewer matter? This question was investigated by Chris Gilberg 7.2 SAMPLING PLANS AND EXPERIMENTAL DESIGNS ❍ 259 and colleagues and reported in an issue of Chance magazine.2 The interviewer asked, “Do you feel that affirmative action should be used as an occupation selection criteria?” with possible answers of yes or no. a. What problems might you expect with responses to this question when asked by interviewers of different ethnic origins? b. When people were interviewed by an African- American, the response was
about 70% in favor of affirmative action, approximately 35% when interviewed by an Asian, and approximately 25% when interviewed by a Caucasian. Do these results support your answer in part a? 7.11 Native American Youth In the American Journal of Human Biology, Chery Smith and Stefanie Fila reported on a study of urban Native American youth.3 The study objective was to determine the appropriateness of an assessment tool to identify dietary measurements for use in this population. The subjects were Native American youth attending an after-school program in Minneapolis, MN. All 61 children between the ages of 9 and 13 who satisﬁed the requirements of the study objectives were included in the experiment. a. Describe the sampling plan used to select study par- ticipants. b. What chance mechanism was used to select this sample of 61 Native American 9- to 13-year-old individuals? c. Can valid inferences be made using the results of this study? Why or why not? d. If you had to devise an alternative sampling plan, what would you change? 7.12 Blood Thinner A study of an experimental blood thinner was conducted to determine whether it works better than the simple aspirin tablet in warding off heart attacks and strokes.4 The study, reported in the Press Enterprise, involved 19,185 people who had suffered heart attacks, strokes, or pain from clogged arteries. Each person was randomly assigned to take either aspirin or the experimental drug for 1 to 3 years. Assume that each person was equally likely to be assigned one of the two medications. a. Devise a randomization plan to assign the medica- tions to the patients. b. Will there be an equal number of patients in each treatment group? Explain. 7.13 Going to the Moon Two different Gallup Polls were conducted for CNN/USA Today, both of which involved people’s feelings about the U.S. space 260 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS program.5 Here is a question from each poll, along with the responses of the sampled Americans: Space Exploration CNN/USA Today/Gallup Poll. Dec. 5–7, 2003. Nationwide: "Would you favor or oppose a new U.S. space program that would send astronauts to the moon?" Form A (N 510, MoE 5) Favor % 53 Oppose % 45 No Opinion % 2 12/03 "Would you favor or oppose the U.S. government spending billions of dollars to
send astronauts to the moon?" Form B (N 494, MoE 5) Favor % 31 Oppose % 67 No Opinion % 2 12/03 a. Read the two poll questions. Which of the two wordings is more unbiased? Explain. b. Look at the responses for the two different polls. How would you explain the large differences in the percentages either favoring or opposing the new program? 7.14 Ask America A nationwide policy survey titled “Ask America” was sent by the National Republican Congressional Committee to voters in the Fortyfourth Congressional District, asking for opinions on a variety of political issues.6 Here are some questions from the survey: • • In recent years has the federal government grown more or less intrusive in your personal and business affairs? Is President Bush right in trying to rein in the size and scope of the federal government against the wishes of the big government Democrats? • Do you believe the death penalty is a deterrrent to crime? • Do you agree that the obstructionist Democrats should not be allowed to gain control of the U.S. Congress in the upcoming elections? Comment on the effect of wording bias on the responses gathered using this survey. STATISTICS AND SAMPLING DISTRIBUTIONS 7.3 When you select a random sample from a population, the numerical descriptive measures you calculate from the sample are called statistics. These statistics vary or change for each different random sample you select; that is, they are random variables. The probability distributions for statistics are called sampling distributions because, in repeated sampling, they provide this information: • What values of the statistic can occur • How often each value occurs Definition The sampling distribution of a statistic is the probability distribution for the possible values of the statistic that results when random samples of size n are repeatedly drawn from the population. There are three ways to ﬁnd the sampling distribution of a statistic: 1. Derive the distribution mathematically using the laws of probability. 2. Use a simulation to approximate the distribution. That is, draw a large number of samples of size n, calculating the value of the statistic for each sample, and tabulate the results in a relative frequency histogram. When the number of EXAMPLE 7.3 7.3 STATISTICS AND SAMPLING DISTRIBUTIONS ❍ 261 samples is large, the histogram will be very close to the theoretical sampling distribution. 3. Use statistical theorems to derive exact or approximate sampling distributions. The next example demonstrates how to derive the sampling distributions of two statistics
for a very small population. A population consists of N 5 numbers: 3, 6, 9, 12, 15. If a random sample of size n 3 is selected without replacement, ﬁnd the sampling distributions for the sample mean x and the sample median m. Solution You are sampling from the population shown in Figure 7.1. It contains ﬁve distinct numbers and each is equally likely, with probability p(x) 1/5. You can easily ﬁnd the population mean and median as m 3 6 9 12 15 5 9 and M 9 FI GUR E 7. 1 Probability histogram for the N 5 population values in Example 7.3 ● p(x).4.2 3 6 9 12 15 x Sampling distributions can be either discrete or continuous. There are 10 possible random samples of size n 3 and each is equally likely, with probability 1/10. These samples, along with the calculated values of x and m for each, are listed in Table 7.3. You will notice that some values of x are more likely than others because they occur in more than one sample. For example, 3 2.3.2 and P(m 6) P(x 8) 0 1 1 0 The values in Table 7.3 are tabulated, and the sampling distributions for x and m are shown in Table 7.4 and Figure 7.2. Since the population of N 5 values is symmetric about the value x 9, both the population mean and the median equal 9. It would seem reasonable, therefore, to consider using either x or m as a possible estimator of M m 9. Which estimator would you choose? From Table 7.3, you see that, in using m as an estimator, you would be in error by 9 6 3 with probability.3 or by 9 12 3 with probability.3. That is, the error in estimation using m would be 3 with probability.6. In using x, however, an error of 3 would occur with probability only.2. On these grounds alone, you may wish to use x as an estimator in preference to m. 262 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS Values of x and m for Simple Random Sampling TABLE 7.3 TABLE 7.4 ● when n 3 and N 5 x Sample Values Sample 1 2 3 4 5 6 7 8 9 10 3, 6, 9 3, 6, 12 3, 6, 15 3, 9,
12 3, 9, 15 3, 12, 15 6, 9, 12 6, 9, 15 6, 12, 15 9, 12, 15 6 7 8 8 9 10 9 10 11 12 m 6 6 6 9 9 12 9 9 12 12 Sampling Distributions for (a) the Sample Mean ● and (b) the Sample Median m (a) (b) p(x) x 6 9 12 6 7 8 9 10 11 12.1.1.2.2.2.1.1 F IG URE 7. 2 Probability histograms for the sampling distributions of the sample mean, x, and the sample median, m, in Example 7.3 ● p(x).4.3.2.1 p (m).3.4.3 p(m).4.3.2.1 6 7 8 9 10 11 12 x 6 7 8 9 10 11 12 m Almost every statistic has a mean and a standard deviation (or standard error) describing its center and spread. It was not too difficult to derive these sampling distributions in Example 7.3 because the number of elements in the population was very small. When this is not the case, you may need to use one of these methods: • Use a simulation to approximate the sampling distribution empirically. • Rely on statistical theorems and theoretical results. One important statistical theorem that describes the sampling distribution of statistics that are sums or averages is presented in the next section. 7.4 THE CENTRAL LIMIT THEOREM ❍ 263 THE CENTRAL LIMIT THEOREM 7.4 The Central Limit Theorem states that, under rather general conditions, sums and means of random samples of measurements drawn from a population tend to have an approximately normal distribution. Suppose you toss a balanced die n 1 time. The random variable x is the number observed on the upper face. This familiar random variable can take six values, each with probability 1/6, and its probability distribution is shown in Figure 7.3. The shape of the distribution is ﬂat or uniform and symmetric about the mean m 3.5, with a standard deviation s 1.71. (See Section 4.8 and Exercise 4.84.) ● FI GUR E 7. 3 Probability distribution for x, the number appearing on a single toss of a die 1/ Now, take a sample of size n 2 from this population; that is, toss two dice and record the sum of the numbers on the
two upper faces, Sxi x1 x2. Table 7.5 shows the 36 possible outcomes, each with probability 1/36. The sums are tabulated, and each of the possible sums is divided by n 2 to obtain an average. The result is the sampling distribution of x Sxi/n, shown in Figure 7.4. You should notice the dramatic difference in the shape of the sampling distribution. It is now roughly moundshaped but still symmetric about the mean m 3.5. TABLE 7.5 ● Sums of the Upper Faces of Two Dice Second Die First Die 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 264 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS F IG URE 7. 4 Sampling distribution of x for n 2 dice ●.15.10 ) r a b x ( p.05 0 1 2 3 4 5 x-bar 6 Average of Two Dice Using MINITAB, we generated the sampling distributions of x when n 3 and n 4. For n 3, the sampling distribution in Figure 7.5 clearly shows the mound shape of the normal probability distribution, still centered at m 3.5. Notice also that the spread of the distribution is slowly decreasing as the sample size n increases. Figure 7.6 dramatically shows that the distribution of x is approximately normally distributed based on a sample as small as n 4. This phenomenon is the result of an important statistical theorem called the Central Limit Theorem (CLT). F IG URE 7. 5 MINITAB sampling distribution of x for n 3 dice ● FIGURE 7.6 MINITAB sampling distribution of x for n 4 dice ●.15.10.05 0.15.10.05 ) -bar 6 Average of Three Dice 0 1 2 3 4 5 Average of Four Dice x-bar 6 7.4 THE CENTRAL LIMIT THEOREM ❍ 265 Central Limit Theorem If random samples of n observations are drawn from a nonnormal population with ﬁnite mean m and standard deviation s, then, when n is large, the sampling distribution of the sample mean x is approximately normally distributed, with mean m and standard deviation s n The approximation becomes more accurate as n becomes large. Regardless of its shape, the sampling distribution of x always has a mean identical to the mean of the sampled population and a standard deviation equal to the population standard deviation s divided by n. Consequently, the spread of the distribution of sample means is considerably less than the spread
of the sampled population. The Central Limit Theorem can be restated to apply to the sum of the sample measurements Sxi, which, as n becomes large, also has an approximately normal distribution with mean nm and standard deviation sn. The Java applet called The Central Limit Theorem can be used to perform a simulation for the sampling distributions of the average of one, two, three or four dice. Figure 7.7 shows the applet after the pair of dice (n 2) has been tossed 2500 times. This is not as hard as it seems, since you need only press the button 25 times. The simulation shows the possible values for x Sxi/10 and also shows the mean and standard deviation for these 2500 measurements. The mean, 3.5, is exactly equal to m 3.5. What is the standard deviation for these 2500 measurements? Is it close to the theoretical value, s/n 1.21? You will use this applet again for the MyApplet Exercises at the end of the chapter. 1.71 2 FI GUR E 7. 7 Central Limit Theorem applet ● 266 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS The sampling distribution of x always has a mean m and standard deviation s/n. The CLT helps describe its shape. The important contribution of the Central Limit Theorem is in statistical inference. Many estimators that are used to make inferences about population parameters are sums or averages of the sample measurements. When the sample size is sufficiently large, you can expect these estimators to have sampling distributions that are approximately normal. You can then use the normal distribution to describe the behavior of these estimators in repeated sampling and evaluate the probability of observing certain sample results. As in Chapter 6, these probabilities are calculated using the standard normal random variable an e M or t a im t s E z d n io iat d r da n a t S ev As you reread the Central Limit Theorem, you may notice that the approximation is valid as long as the sample size n is “large”—but how large is “large”? Unfortunately, there is no clear answer to this question. The appropriate value of n depends on the shape of the population from which you sample as well as on how you want to use the approximation. However, these guidelines will help: HOW DO I DECIDE WHEN THE SAMPLE SIZE IS LARGE ENOUGH? • If the sampled population is normal, then the sampling
distribution of x will also be normal, no matter what sample size you choose. This result can be proven theoretically, but it should not be too difficult for you to accept without proof. • When the sampled population is approximately symmetric, the sampling distribution of x becomes approximately normal for relatively small values of n. Remember how rapidly the “ﬂat” distribution in the dice example became mound-shaped (n 3). • When the sampled population is skewed, the sample size n must be larger, with n at least 30 before the sampling distribution of x becomes approximately normal. These guidelines suggest that, for many populations, the sampling distribution of x will be approximately normal for moderate sample sizes; an exception to this rule occurs in sampling a binomial population when either p or q (1 p) is very small. As speciﬁc applications of the Central Limit Theorem arise, we will give you the appropriate sample size n. 7.5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN If the population mean m is unknown, you might choose several statistics as an estimator; the sample mean x and the sample median m are two that readily come to mind. Which should you use? Consider these criteria in choosing the estimator for m: 7.5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN ❍ 267 Is it easy or hard to calculate? • • Does it produce estimates that are consistently too high or too low? • Is it more or less variable than other possible estimators? The sampling distributions for x and m with n 3 for the small population in Example 7.3 showed that, in terms of these criteria, the sample mean performed better than the sample median as an estimator of m. In many situations, the sample mean x has desirable properties as an estimator that are not shared by other competing estimators; therefore, it is more widely used. THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN, –x • • • If a random sample of n measurements is selected from a population with mean m and standard deviation s, the sampling distribution of the sample mean x will have mean m and standard deviation† s n If the population has a normal distribution, the sampling distribution of x will be exactly normally distributed, regardless of the sample size, n. If the population distribution is nonnormal, the sampling distribution of x will be approximately normally distributed for large samples (by the Central Limit Theorem). Standard Error Definition The standard deviation of
a statistic used as an estimator of a population parameter is also called the standard error of the estimator (abbreviated SE) because it refers to the precision of the estimator. Therefore, the standard deviation of x—given by s/n—is referred to as the standard error of the mean (abbreviated as SE(x) or just SE). †When repeated samples of size n are randomly selected from a ﬁnite population with N elements whose mean is m and whose variance is s 2, the standard deviation of x is s N n n 1 N where s 2 is the population variance. When N is large relative to the sample size n, (N n)(N 1) is approximately equal to 1, and the standard deviation of x is s n 268 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS How Do I Calculate Probabilities for the Sample Mean –x? If you know that the sampling distribution of x is normal or approximately normal, you can describe the behavior of the sample mean x by calculating the probability of observing certain values of x in repeated sampling. 1. Find m and calculate SE (x) s/n. 2. Write down the event of interest in terms of x, and locate the appropriate area on the normal curve. 3. Convert the necessary values of x to z-values using x z s/ m n 4. Use Table 3 in Appendix I to calculate the probability. Exercise Reps (Fill in the Blanks) A. You take a random sample of size n 36 from a distribution with mean m 75 and s 12. The sampling distribution of x will be approximately with a mean of and a standard deviation (or standard error) of. B. To ﬁnd the probability that the sample mean exceeds 80, write down the event of interest. When x 80, x z s/ m n Find the probability: P(x ) P(z ) 1 C. To ﬁnd the probability that the sample mean is between 70 and 72, write down the event of interest. When x 70 and x 72, x z s/ m n Find the probability: x and z s/ m n P( x ) P( z ) Progress Report • Still having trouble? Try again using the Exercise Reps at the end of this section. • No problems? You can skip the Exercise Reps at the end of this section! Answers are located on the perforated card at the back of this book. 7.
5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN ❍ 269 EXAMPLE 7.4 The duration of Alzheimer’s disease from the onset of symptoms until death ranges from 3 to 20 years; the average is 8 years with a standard deviation of 4 years. The administrator of a large medical center randomly selects the medical records of 30 deceased Alzheimer’s patients from the medical center’s database, and records the average duration. Find the approximate probabilities for these events: If x is normal, x is normal for any n. If x is not normal, x is approximately normal for large n. 1. The average duration is less than 7 years. 2. The average duration exceeds 7 years. 3. The average duration lies within 1 year of the population mean m 8. Solution Since the administrator has selected a random sample from the database at this medical center, he can draw conclusions about only past, present, or future patients with Alzheimer’s disease at this medical center. If, on the other hand, this medical center can be considered representative of other medical centers in the country, it may be possible to draw more far-reaching conclusions. What can you say about the shape of the sampled population? It is not symmetric, because the mean m 8 does not lie halfway between the maximum and minimum values. Since the mean is closer to the minimum value, the distribution is skewed to the right, with a few patients living a long time after the onset of the disease. Regardless of the shape of the population distribution, however, the sampling distribution of x has a mean m 8 and standard deviation s/n 4/30.73. In addition, because the sample size is n 30, the Central Limit Theorem ensures the approximate normality of the sampling distribution of x. 1. The probability that x is less than 7 is given by the shaded area in Figure 7.8. To ﬁnd this area, you need to calculate the value of z corresponding to x 7: m x 8 1.37 7 z s n / 73. From Table 3 in Appendix I, you can ﬁnd the cumulative area corresponding to z 1.37 and P(x 7) P(z 1.37).0853 FI GUR E 7. 8 The probability that x is less than 7 for Example 7.4 ● f(x) P(x < 7) 7 –1.37 µ = 8 0 x z [NOTE: You must use s/
n (not s) in the formula for z because you are ﬁnding an area under the sampling distribution for x, not under the probability distribution for x.] 270 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS Remember that for continuous random variables, there is no probability assigned to a single point. Therefore, P(x 7) P(x 7). 2. The event that x exceeds 7 is the complement of the event that x is less than 7. Thus, the probability that x exceeds 7 is P(x 7) 1 P(x 7) 1.0853.9147 3. The probability that x lies within 1 year of m 8 is the shaded area in Fig- ure 7.9. The z-value corresponding to x 7 is z 1.37, from part 1, and the z-value for x 9 is m x 8 1.37 9 z n s / 73. The probability of interest is P(7 x 9) P(1.37 z 1.37).9147.0853.8294 F IG URE 7. 9 The probability that x lies within 1 year of m 8 for Example 7.4 ● f(x) P(7 < x < 9) 7 –1.37 µ = 8 0 9 1.37 x z Example 7.4 can be solved using the Normal Probabilities for Means applet. If you enter the values for x, s, m, and n (press “Enter” to record each change) and adjust the dropdown list at the bottom of the applet, you can calculate the area to the right or left of z0, the area in two tails, or the area between z0 and z0. Conversely, if you need to ﬁnd the value of x that cuts off a certain area under the curve, enter the area in the box marked “prob:” at the bottom of the applet, and the applet will provide the value of x. The applet in Figure 7.10 is set to calculate P(7 x 9).829, correct to three decimal places. You will use this applet for the MyApplet Exercises at the end of the chapter. FI GUR E 7. 10 Normal Probabilities for Means applet ● 7.5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN ❍ 271 EXAMPLE 7.5 To avoid difficulties with the
Federal Trade Commission or state and local consumer protection agencies, a beverage bottler must make reasonably certain that 12-ounce bottles actually contain 12 ounces of beverage. To determine whether a bottling machine is working satisfactorily, one bottler randomly samples 10 bottles per hour and measures the amount of beverage in each bottle. The mean x of the 10 ﬁll measurements is used to decide whether to readjust the amount of beverage delivered per bottle by the ﬁlling machine. If records show that the amount of ﬁll per bottle is normally distributed, with a standard deviation of.2 ounce, and if the bottling machine is set to produce a mean ﬁll per bottle of 12.1 ounces, what is the approximate probability that the sample mean x of the 10 test bottles is less than 12 ounces? Solution The mean of the sampling distribution of the sample mean x is identical to the mean of the population of bottle ﬁlls—namely, m 12.1 ounces—and the standard error of x is s.2 SE.063 n 10 (NOTE: s is the standard deviation of the population of bottle ﬁlls, and n is the number of bottles in the sample.) Since the amount of ﬁll is normally distributed, x is also normally distributed, as shown in Figure 7.11. To ﬁnd the probability that x is less than 12 ounces, express the value x 12 in units of standard deviations: m x 2.1 1.59 1 12 z n s / 3 06. Then P(x 12) P(z 1.59).0559.056 272 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS ● f(x) F IG URE 7. 11 Probability distribution of x, the mean of the n 10 bottle ﬁlls, for Example 7.5 12 –1.59 µ = 12.1 0 x z Thus, if the machine is set to deliver an average ﬁll of 12.1 ounces, the mean ﬁll x of a sample of 10 bottles will be less than 12 ounces with a probability equal to.056. When this danger signal occurs (x is less than 12), the bottler takes a larger sample to recheck the setting of the ﬁlling machine. 7.5 EXERCISES EXERCISE REPS (FILL IN THE BLANKS) These exercises refer back to the
MyPersonal Trainer section on page 268. 7.15 You take a random sample of size n 49 from a distribution with mean m 53 and s 21. The sampling distribution of x will be approximately mean of and a standard deviation (or standard error) of with a. 7.16 Refer to Exercise 7.15. To ﬁnd the probability that the sample mean is greater than 55, write down the event of interest. When x 55, x z s m n / Find the probability: P(x ) P(z ) 1 7.17 You take a random sample of size n 40 from a distribution with mean m 100 and s 20. The sampling distribution of x will be approximately with a mean of and a standard deviation (or standard error) of. 7.18 Refer to Exercise 7.17. To ﬁnd the probability that the sample mean is between 105 and 110, write down the event of interest. When x 105 and x 110, x z s m n / x and z s m n / Find the probability: P( x ) P( z ) BASIC TECHNIQUES 7.19 Random samples of size n were selected from populations with the means and variances given here. Find the mean and standard deviation of the sampling distribution of the sample mean in each case: a. n 36, m 10, s 2 9 b. n 100, m 5, s 2 4 c. n 8, m 120, s 2 1 7.20 Refer to Exercise 7.19. a. If the sampled populations are normal, what is the sampling distribution of x for parts a, b, and c? b. According to the Central Limit Theorem, if the sampled populations are not normal, what can be said about the sampling distribution of x for parts a, b, and c? 7.21 A population consists of N = 5 numbers: 1, 3, 5, 6, and 7. It can be shown that the mean EX0721 and standard deviation for this population are m = 4.4 and s = 2.15, respectively. a. Construct a probability histogram for this population. b. Use the random number table, Table 10 in Appendix I, to select a random sample of size n = 10 with replacement from the population. Calculate the sample mean, x. Repeat this procedure, calculating the sample mean x for your second sample. (HINT: Assign the random digits 0 and 1 to the measurement x = 1; assign digits 2
and 3 to the measurement x = 3, and so on.) c. To simulate the sampling distribution of x, we have selected 50 more samples of size n = 10 with replacement, and have calculated the corresponding sample means. Construct a relative frequency histogram for these 50 values of x. What is the shape of this distribution? 4.8 3.0 4.6 5.0 4.4 4.2 5.9 4.1 4.6 4.2 4.2 5.7 3.4 4.1 4.2 4.5 4.2 4.9 5.1 5.2 4.3 4.4 4.1 3.4 5.4 4.3 4.8 4.0 5.9 4.8 5.0 5.0 3.7 5.0 3.6 4.0 5.1 4.3 4.3 5.0 3.3 4.8 4.3 4.5 4.5 4.7 4.2 4.5 3.9 4.9 7.22 Refer to Exercise 7.21. a. Use the data entry method in your calculator to ﬁnd the mean and standard deviation of the 50 values of x given in Exercise 7.21, part c. b. Compare the values calculated in part a to the theoretical mean m and the theoretical standard deviation s/n for the sampling distribution of x. How 7.5 THE SAMPLING DISTRIBUTION OF THE SAMPLE MEAN ❍ 273 close do the values calculated from the 50 measurements come to the theoretical values? c. n 4 f. n 25 b. n 2 e. n 16 7.23 A random sample of n observations is selected from a population with standard deviation s 1. Calculate the standard error of the mean (SE) for these values of n: a. n 1 d. n 9 g. n 100 7.24 Refer to Exercise 7.23. Plot the standard error of the mean (SE) versus the sample size n and connect the points with a smooth curve. What is the effect of increasing the sample size on the standard error? 7.25 Suppose a random sample of n 25 observations is selected from a population that is normally distributed with mean equal to 106 and standard deviation equal to 12. a. Give the mean and the standard deviation of the sampling distribution of the sample mean x. b. Find the probability that x exceeds 110. c. Find the probability that the sample mean deviates
from the population mean m 106 by no more than 4. APPLICATIONS 7.26 Faculty Salaries Suppose that college faculty with the rank of professor at two-year institutions earn an average of $64,571 per year7 with a standard deviation of $4,000. In an attempt to verify this salary level, a random sample of 60 professors was selected from a personnel database for all two-year institutions in the United States. a. Describe the sampling distribution of the sample mean x. b. Within what limits would you expect the sample average to lie, with probability.95? c. Calculate the probability that the sample mean x is greater than $66,000? d. If your random sample actually produced a sample mean of $66,000, would you consider this unusual? What conclusion might you draw? 7.27 Measurement Error When research chemists perform experiments, they may obtain slightly different results on different replications, even when the experiment is performed identically each time. These 274 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS differences are due to a phenomenon called “measurement error.” a. List some variables in a chemical experiment that might cause some small changes in the ﬁnal response measurement. b. If you want to make sure that your measurement error is small, you can replicate the experiment and take the sample average of all the measurements. To decrease the amount of variability in your average measurement, should you use a large or a small number of replications? Explain. 7.28 Tomatoes Explain why the weight of a package of one dozen tomatoes should be approximately normally distributed if the dozen tomatoes represent a random sample. 7.29 Bacteria in Water Use the Central Limit Theorem to explain why a Poisson random variable— say, the number of a particular type of bacteria in a cubic foot of water—has a distribution that can be approximated by a normal distribution when the mean m is large. (HINT: One cubic foot of water contains 1728 cubic inches of water.) 7.30 Paper Strength A manufacturer of paper used for packaging requires a minimum strength of 20 pounds per square inch. To check on the quality of the paper, a random sample of 10 pieces of paper is selected each hour from the previous hour’s production and a strength measurement is recorded for each. The standard deviation s of the strength measurements, computed by pooling the sum of squares of deviations of many samples, is known to equal 2 pounds per square inch
, and the strength measurements are normally distributed. a. What is the approximate sampling distribution of the sample mean of n 10 test pieces of paper? b. If the mean of the population of strength measurements is 21 pounds per square inch, what is the approximate probability that, for a random sample of n 10 test pieces of paper, x 20? c. What value would you select for the mean paper strength m in order that P(x 20) be equal to.001? 7.31 Potassium Levels The normal daily human potassium requirement is in the range of 2000 to 6000 milligrams (mg), with larger amounts required during hot summer weather. The amount of potassium in food varies, depending on the food. For example, there are approximately 7 mg in a cola drink, 46 mg in a beer, 630 mg in a banana, 300 mg in a carrot, and 440 mg in a glass of orange juice. Suppose the distribution of potassium in a banana is normally distributed, with mean equal to 630 mg and standard deviation equal to 40 mg per banana. You eat n 3 bananas per day, and T is the total number of milligrams of potassium you receive from them. a. Find the mean and standard deviation of T. b. Find the probability that your total daily intake of potassium from the three bananas will exceed 2000 mg. (HINT: Note that T is the sum of three random variables, x1, x2, and x3, where x1 is the amount of potassium in banana number 1, etc.) 7.32 Deli Sales The total daily sales, x, in the deli section of a local market is the sum of the sales generated by a ﬁxed number of customers who make purchases on a given day. a. What kind of probability distribution do you expect the total daily sales to have? Explain. b. For this particular market, the average sale per customer in the deli section is $8.50 with s $2.50. If 30 customers make deli purchases on a given day, give the mean and standard deviation of the probability distribution of the total daily sales, x. 7.33 Normal Temperatures In Exercise 1.67, Allen Shoemaker derived a distribution of human body temperatures with a distinct mound shape.8 Suppose we assume that the temperatures of healthy humans is approximately normal with a mean of 98.6 degrees and a standard deviation of 0.8 degrees. a. If 130 healthy people are selected at random, what is the probability that
the average temperature for these people is 98.25 degrees or lower? b. Would you consider an average temperature of 98.25 degrees to be an unlikely occurrence, given that the true average temperature of healthy people is 98.6 degrees? Explain. 7.34 Sports and Achilles Tendon Injuries Some sports that involve a signiﬁcant amount of running, jumping, or hopping put participants at risk for Achilles tendinopathy (AT), an inﬂammation and thickening of the Achilles tendon. A study in The American Journal of Sports Medicine looked at the diameter (in mm) of the affected and nonaffected tendons for patients who participated in these types of sports activities.9 Suppose that the Achilles tendon diameters in the general population have a mean of 5.97 millimeters (mm) with a standard deviation of 1.95 mm. a. What is the probability that a randomly selected sample of 31 patients would produce an average diameter of 6.5 mm or less for the nonaffected tendon? b. When the diameters of the affected tendon were measured for a sample of 31 patients, the average diameter was 9.80. If the average tendon 7.6 THE SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION ❍ 275 diameter in the population of patients with AT is no different than the average diameter of the nonaffected tendons (5.97 mm), what is the probability of observing an average diameter of 9.80 or higher? c. What conclusions might you draw from the results of part b? THE SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION 7.6 Q: How do you know if it’s binomial or not? A: Look to see if the measurement taken on a single experimental unit in the sample is a “success/failure” type. If so, it’s probably binomial. ● FI GUR E 7. 12 Sampling distribution of the binomial random variable x and the sample proportion ˆp There are many practical examples of the binomial random variable x. One common application involves consumer preference or opinion polls, in which we use a random sample of n people to estimate the proportion p of people in the population who have a speciﬁed characteristic. If x of the sampled people have this characteristic, then the sample proportion x ˆp n can be used to estimate the population proportion p (Figure 7.12).† The
binomial random variable x has a probability distribution p(x), described in Chapter 5, with mean np and standard deviation npq. Since ˆp is simply the value of x, the sampling distribution of ˆp is identical to x, expressed as a proportion ˆp the probability distribution of x, except that it has a new scale along the horizontal axis. n 0.3 0.2 0.1 0.0 0 0 1 2 3 4 1/5 2/5 3/5 4/5 x ˆp 5 1 Because of this change of scale, the mean and standard deviation of ˆp are also rescaled, so that the mean of the sampling distribution of ˆp is p, and its standard error is SE( ˆp) p q where q 1 p n Finally, just as we can approximate the probability distribution of x with a normal distribution when the sample size n is large, we can do the same with the sampling distribution of ˆp. †A “hat” placed over the symbol of a population parameter denotes a statistic used to estimate the population parameter. For example, the symbol ˆp denotes the sample proportion. 276 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS PROPERTIES OF THE SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION, ˆp • If a random sample of n observations is selected from a binomial population with parameter p, then the sampling distribution of the sample proportion x ˆp n will have a mean p and a standard deviation SE( ˆp) p q where q 1 p n EXAMPLE 7.6 • When the sample size n is large, the sampling distribution of ˆp can be approximated by a normal distribution. The approximation will be adequate if np 5 and nq 5. In a survey, 500 mothers and fathers were asked about the importance of sports for boys and girls. Of the parents interviewed, 60% agreed that the genders are equal and should have equal opportunities to participate in sports. Describe the sampling distribution of the sample proportion ˆp of parents who agree that the genders are equal and should have equal opportunities. Solution You can assume that the 500 parents represent a random sample of the parents of all boys and girls in the United States and that the true proportion in the population is equal to some unknown value that you can call p. The sampling distribution of
ˆp can be approximated by a normal distribution,† with mean equal to p (see Figure 7.13) and standard error q SE( ˆp) p n ● ˆ f(p) F IG URE 7. 13 The sampling distribution for ˆp based on a sample of n 500 parents for Example 7.6 p 2SE.044 ˆp †Checking the conditions that allow the normal approximation to the distribution of ˆp, you can see that n 500 is adequate for values of p near.60 because nˆp 300 and n ˆq 200 are both greater than 5. 7.6 THE SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION ❍ 277 You can see from Figure 7.13 that the sampling distribution of ˆp is centered over its mean p. Even though you do not know the exact value of p (the sample proportion ˆp.60 may be larger or smaller than p), an approximate value for the standard deviation of the sampling distribution can be found using the sample proportion ˆp.60 to approximate the unknown value of p. Thus, SE p qˆ q ˆp n n (.60.40).022 ( ) 0 0 5 Therefore, approximately 95% of the time, ˆp will fall within 2SE.044 of the (unknown) value of p. How Do I Calculate Probabilities for the Sample Proportion ˆp? 1. Find the necessary values of n and p. 2. Check whether the normal approximation to the binomial distribution is appro- priate (np 5 and nq 5). 3. Write down the event of interest in terms of ˆp, and locate the appropriate area on the normal curve. 4. Convert the necessary values of ˆp to z-values using z ˆp p q p n 5. Use Table 3 in Appendix I to calculate the probability. Exercise Reps (Fill in the Blanks) A. You take a random sample of size n 36 from a binomial distribution with mean p.4. The sampling distribution of ˆp will be approximately with a mean of. B. To ﬁnd the probability that the sample proportion exceeds.5, write down the and a standard deviation (or standard error) of event of interest. When ˆp.5, z ˆp p q
p n Find the probability: P( ˆp ) P(z ) 1 (continued) 278 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS C. To ﬁnd the probability that the sample proportion is between.5 and.6, write down the event of interest. When ˆp.5 and ˆp.6, z ˆp p q p n and z ˆp p q p n Find the probability: P( ˆp ) P( z ) Progress Report • Still having trouble? Try again using the Exercise Reps at the end of this section. • No problems? You can skip the Exercise Reps at the end of this section! Answers are located on the perforated card at the back of this book. EXAMPLE 7.7 Refer to Example 7.6. Suppose the proportion p of parents in the population is actually equal to.55. What is the probability of observing a sample proportion as large as or larger than the observed value ˆp.60? Solution Figure 7.14 shows the sampling distribution of ˆp when p.55, with the observed value ˆp.60 located on the horizontal axis. The probability of observing a sample proportion ˆp equal to or larger than.60 is approximated by the shaded area in the upper tail of this normal distribution with p.55 and SE p q (.55.45).0222 ( ) 0 0 5 n F IG URE 7. 14 The sampling distribution of ˆp for n 500 and p.55 for Example 7.7 ● f(ˆp) P(ˆp ≥.60) p =.55.60 ˆp To ﬁnd this shaded area, ﬁrst calculate the z-value corresponding to ˆp.60: p ˆp.55 2.25.60 z n 222.0 p q/ 7.6 THE SAMPLING DISTRIBUTION OF THE SAMPLE PROPORTION ❍ 279 Using Table 3 in Appendix I, you ﬁnd P( ˆp.60) P(z 2.25) 1.9878.0122 That is, if you were to select a random sample of n 500 observations from a population with proportion p equal to.55, the probability that the sample proportion ˆp would be as large as or larger
than.60 is only.0122. When the normal distribution was used in Chapter 6 to approximate the binomial probabilities associated with x, a correction of.5 was applied to improve the approximation. The equivalent correction here is (.5/n). For example, for ˆp.60 the value of z with the correction is z1 (.60.001).55.45) (.55 ( ) 0 0 5 2.20 with P( ˆp.60).0139. To two-decimal-place accuracy, this value agrees with the earlier result. When n is large, the effect of using the correction is generally negligible. You should solve problems in this and the remaining chapters without the correction factor unless you are speciﬁcally instructed to use it. 7.6 EXERCISES EXERCISE REPS (FILL IN THE BLANKS) These exercises refer back to the MyPersonal Trainer section on page 277. 7.35 You take a random sample of size n 50 from a binomial distribution with a mean of p.7. The sampling distribution of ˆp will be approximately mean of and a standard deviation (or standard error) of. with a 7.36 To ﬁnd the probability that the sample proportion is less than.8, write down the event of interest. When ˆp.8, ˆp p q p z n Find the probability: P( ˆp ) P(z ) BASIC TECHNIQUES 7.37 Random samples of size n were selected from binomial populations with population parameters p given here. Find the mean and the standard deviation of the sampling distribution of the sample proportion ˆp in each case: a. n 100, p.3 b. n 400, p.1 c. n 250, p.6 7.38 Is it appropriate to use the normal distribution to approximate the sampling distribution of ˆp in the following circumstances? a. n 50, p.05 b. n 75, p.1 c. n 250, p.99 280 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS 7.39 Random samples of size n 75 were selected from a binomial population with p.4. Use the normal distribution to approximate the following probabilities: a. P( ˆp.43) b. P(.35 ˆp.43) 7.40 Random samples of size n 500
were selected from a binomial population with p.1. a. Is it appropriate to use the normal distribution to approximate the sampling distribution of ˆp? Check to make sure the necessary conditions are met. Using the results of part a, ﬁnd these probabilities: b. ˆp.12 c. ˆp.10 d. ˆp lies within.02 of p 7.41 Calculate SE( ˆp) for n 100 and these values of p: a. p.01 d. p.50 g. p.99 h. Plot SE( ˆp) versus p on graph paper and sketch a b. p.10 e. p.70 c. p.30 f. p.90 smooth curve through the points. For what value of p is the standard deviation of the sampling distribution of ˆp a maximum? What happens to the standard error when p is near 0 or near 1.0? 7.42 a. Is the normal approximation to the sampling distribution of ˆp appropriate when n 400 and p.8? b. Use the results of part a to ﬁnd the probability that ˆp is greater than.83. c. Use the results of part a to ﬁnd the probability that ˆp lies between.76 and.84. APPLICATIONS 7.43 Losing Weight News reports tell us that the average American is overweight. Many of us have tried to trim down to our weight when we ﬁnished high school or college. And, in fact, only 19% of adults say they do not suffer from weight-loss woes.10 Suppose that the 19% ﬁgure is correct, and that a random sample of n 100 adults is selected. a. Does the distribution of ˆp, the sample proportion of adults who do not suffer from excess weight, have an approximate normal distribution? If so, what is its mean and standard deviation? b. What is the probability that the sample proportion ˆp exceeds.25? c. What is the probability that ˆp lies within the interval.25 to.30? d. What might you conclude about p if the sample proportion exceeded.30? 7.44 Prescription Costs The cost of brand-name prescriptions are set to provide support for research and development of these drugs, which may cover as many as 20 years. Nonetheless, a majority of
Americans say that costs of prescription drugs (66%), hospital costs (64%), and doctors visits (55%) are unreasonably high.11 Suppose that you take a random sample of n 1000 adults. Let ˆp be the proportion of adults who say that prescription drugs are unreasonably high. a. What is the exact distribution of ˆp? How can you approximate the distribution of ˆp? b. What is the probability that ˆp exceeds.68? c. What is the probability that ˆp lies between.64 and.68? d. Would a sample percentage of 70% contradict the reported value of 66%? 7.45 M&M’S According to the M&M’S website, the average percentage of brown M&M’S candies in a package of milk chocolate M&M’S is 13%.12 (This percentage varies, however, among the different types of packaged M&M’S.) Suppose you randomly select a package of milk chocolate M&M’S that contains 55 candies and determine the proportion of brown candies in the package. a. What is the approximate distribution of the sample proportion of brown candies in a package that contains 55 candies? b. What is the probability that the sample percentage of brown candies is less than 20%? c. What is the probability that the sample percentage exceeds 35%? d. Within what range would you expect the sample proportion to lie about 95% of the time? 7.46 The “Cheeseburger Bill” In the spring of 2004, the U.S. Congress considered a bill that would prevent Americans from suing fast-food giants like McDonald’s for making them overweight. Although the fast-food industry may not be to blame, a study by Children’s Hospital in Boston reports that about twothirds of adult Americans and about 15% of children and adolescents are overweight.13 A random sample of 100 children is selected. a. What is the probability that the sample percentage of overweight children exceeds 25%? 7.7 A SAMPLING APPLICATION: STATISTICAL PROCESS CONTROL (OPTIONAL) ❍ 281 b. What is the probability that the sample percentage of overweight children is less than 12%? c. Would it be unusual to ﬁnd that 30% of the sam- pled children were overweight? Explain. 7.47 Oh, Nuts! Are you
a chocolate “purist,” or do you like other ingredients in your chocolate? American Demographics reports that almost 75% of consumers like traditional ingredients such as nuts or caramel in their chocolate. They are less enthusiastic about the taste of mint or coffee that provide more distinctive ﬂavors.14 A random sample of 200 consumers is selected and the number who like nuts or caramel in their chocolate is recorded. a. What is the approximate sampling distribution for the sample proportion ˆp? What are the mean and standard deviation for this distribution? b. What is the probability that the sample percentage is greater than 80%? c. Within what limits would you expect the sample proportion to lie about 95% of the time? A SAMPLING APPLICATION: STATISTICAL PROCESS CONTROL (OPTIONAL) 7.7 Statistical process control (SPC) methodology was developed to monitor, control, and improve products and services. Steel bearings must conform to size and hardness speciﬁcations, industrial chemicals must have a low prespeciﬁed level of impurities, and accounting ﬁrms must minimize and ultimately eliminate incorrect bookkeeping entries. It is often said that statistical process control consists of 10% statistics, 90% engineering and common sense. We can statistically monitor a process mean and tell when the mean falls outside preassigned limits, but we cannot tell why it is out of control. Answering this last question requires knowledge of the process and problemsolving ability—the other 90%! Product quality is usually monitored using statistical control charts. Measurements on a process variable to be monitored change over time. The cause of a change in the variable is said to be assignable if it can be found and corrected. Other variation— small haphazard changes due to alteration in the production environment—that is not controllable is regarded as random variation. If the variation in a process variable is solely random, the process is said to be in control. The ﬁrst objective in statistical process control is to eliminate assignable causes of variation in the process variable and then get the process in control. The next step is to reduce variation and get the measurements on the process variable within speciﬁcation limits, the limits within which the measurements on usable items or services must fall. Once a process is in control and is producing a satisfactory product, the process variables are monitored with control charts. Samples of n items are drawn from the process at speci
ﬁed intervals of time, and a sample statistic is computed. These statistics are plotted on the control chart, so that the process can be checked for shifts in the process variable that might indicate control problems. A Control Chart for the Process Mean: The ¯x Chart Assume that n items are randomly selected from the production process at equal intervals and that measurements are recorded on the process variable. If the process is in control, the sample means should vary about the population mean m in a random manner. Moreover, according to the Central Limit Theorem, the sampling distribution 282 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS of x should be approximately normal, so that almost all of the values of x fall into the interval (m 3 SE) m 3(s/n). Although the exact values of m and s are unknown, you can obtain accurate estimates by using the sample measurements. Every control chart has a centerline and control limits. The centerline for the x chart is the estimate of m, the grand average of all the sample statistics calculated from the measurements on the process variable. The upper and lower control limits are placed three standard deviations above and below the centerline. If you monitor the process mean based on k samples of size n taken at regular intervals, the centerline is x, the average of the sample means, and the control limits are at x 3(s/n), with s estimated by s, the standard deviation of the nk measurements. A statistical process control monitoring system samples the inside diameters of n 4 bearings each hour. Table 7.6 provides the data for k 25 hourly samples. Construct an x chart for monitoring the process mean. Solution The sample mean was calculated for each of the k 25 samples. For example, the mean for sample 1 is x.992 1.007 1.016.991 4 1.0015 EXAMPLE 7.8 TABLE 7.6 ● n 4 Bearings per Sample 25 Hourly Samples of Bearing Diameters, Sample Sample Measurements Sample Mean 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25.992 1.015.988.996 1.015 1.000.989.994 1.018.997 1.020 1.007 1.016.982 1.001.992 1.020.993.978.984.990 1.015.983 1.011.987 1.007.984.993 1.020 1.006.
982 1.009 1.010 1.016 1.005.986.986 1.002.995 1.000 1.008.988.987 1.006 1.009 1.012.983.990 1.012.987 1.016.976 1.011 1.004 1.002 1.005 1.019 1.009.990.989 1.002.981 1.010 1.011.983 1.001 1.015 1.006 1.002.983 1.010 1.003.997.991 1.007.991 1.000.981.999 1.001.989.994.990 1.011 1.001.989.995.999.987 1.002.996.986 1.001.982.986 1.007.989 1.002 1.008.995 1.00150.99375.99325 1.00475 1.00600.99400 1.00275 1.00075 1.00875.99800.99925.99225 1.00675.99375.99650.99925 1.00225.99675.99200.99050 1.00475.99750.99300 1.00550.99400 The sample means are shown in the last column of Table 7.6. The centerline is located at the average of the sample means, or 675.9987 9 x 24. 5 2 7.7 A SAMPLING APPLICATION: STATISTICAL PROCESS CONTROL (OPTIONAL) ❍ 283 The calculated value of s, the sample standard deviation of all nk 4(25) 100 observations, is s.011458, and the estimated standard error of the mean of n 4 observations is s 58 14.01.005729 4 n The upper and lower control limits are found as s.9987 3(.005729) 1.015887 UCL x 3 n and s.9987 3(.005729).981513 LCL x 3 n Figure 7.15 shows a MINITAB printout of the x chart constructed from the data. If you assume that the samples used to construct the x chart were collected when the process was in control, the chart can now be used to detect changes in the process mean. Sample means are plotted periodically, and if a sample mean falls outside the control limits,
a warning should be conveyed. The process should be checked to locate the cause of the unusually large or small mean. FI GUR E 7. 15 MINITAB x chart for Example 7..02 1.01 1.00 0.99 0.98 Xbar Chart of Diameter UCL 1.01589 X 0.9987 LCL 0.98151 1 3 5 7 9 11 13 15 17 19 21 23 25 Sample A Control Chart for the Proportion Defective: The p Chart Sometimes the observation made on an item is simply whether or not it meets speciﬁcations; thus, it is judged to be defective or nondefective. If the fraction defective produced by the process is p, then x, the number of defectives in a sample of n items, has a binomial distribution. To monitor a process for defective items, samples of size n are selected at periodic intervals and the sample proportion ˆp is calculated. When the process is in control, ˆp should fall into the interval p 3SE, where p is the proportion of defectives in the population (or the process fraction defective) with standard error SE p q p(1 p) n n 284 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS The process fraction defective is unknown but can be estimated by the average of the k sample proportions: S ˆpi p k and the standard error is estimated by SE p(1 p) n The centerline for the p chart is located at p, and the upper and lower control limits are UCL p 3p(1 p) n and LCL p 3p(1 p) n EXAMPLE 7.9 TABLE 7.7 A manufacturer of ballpoint pens randomly samples 400 pens per day and tests each to see whether the ink ﬂow is acceptable. The proportions of pens judged defective each day over a 40-day period are listed in Table 7.7. Construct a control chart for the proportion ˆp defective in samples of n 400 pens selected from the process. ● Proportions of Defectives in Samples of n 400 Pens Proportion Proportion Proportion Proportion Day Day Day Day 1 2 3 4 5 6 7 8 9 10.0200.0125.0225.0100.0150.0200.0275.0175.0200.0250 11 12 13 14 15 16 17 18 19 20.0100.0175.0250.0175
.0275.0200.0225.0100.0175.0200 21 22 23 24 25 26 27 28 29 30.0300.0200.0125.0175.0225.0150.0200.0250.0150.0175 31 32 33 34 35 36 37 38 39 40.0225.0175.0225.0100.0125.0300.0200.0150.0150.0225 Solution The estimate of the process proportion defective is the average of the k 40 sample proportions in Table 7.7. Therefore, the centerline of the control chart is located at S ˆpi p k.0200.0125.0225 40 00.019 6.7 0 4 An estimate of SE, the standard error of the sample proportions, is p) (.019 (0 p(1 ) 0 4 n.981).00683 and 3SE (3)(.00683).0205. Therefore, the upper and lower control limits for the p chart are located at UCL p 3SE.0190.0205.0395 and LCL p 3SE.0190.0205.0015 7.7 A SAMPLING APPLICATION: STATISTICAL PROCESS CONTROL (OPTIONAL) ❍ 285 Or, since p cannot be negative, LCL 0. The p control chart is shown in Figure 7.16. Note that all 40 sample proportions fall within the control limits. If a sample proportion collected at some time in the future falls outside the control limits, the manufacturer should be concerned about an increase in the defective rate. He should take steps to look for the possible causes of this increase. FI GUR E 7. 16 MINITAB p chart for Example 7..04 0.03 0.02 0.01 0.00 P Chart of Defects UCL 0.03948 p 0.019 LCL 0 1 5 9 13 17 25 29 33 37 21 Day Other commonly used control charts are the R chart, which is used to monitor variation in the process variable by using the sample range, and the c chart, which is used to monitor the number of defects per item. 7.7 EXERCISES BASIC TECHNIQUES 7.48 The sample means were calculated for 30 samples of size n 10 for a process that was judged to be in control. The means of the 30 x-values and the standard deviation of the combined 300 measurements were x 20.74 and
s.87, respectively. a. Use the data to determine the upper and lower con- trol limits for an x chart. b. What is the purpose of an x chart? c. Construct an x chart for the process and explain how it can be used. 7.49 The sample means were calculated for 40 samples of size n 5 for a process that was judged to be in control. The means of the 40 values and the standard deviation of the combined 200 measurements were x 155.9 and s 4.3, respectively. a. Use the data to determine the upper and lower con- trol limits for an x chart. b. Construct an x chart for the process and explain how it can be used. 7.50 Explain the difference between an x chart and a p chart. 7.51 Samples of n 100 items were selected hourly over a 100-hour period, and the sample proportion of defectives was calculated each hour. The mean of the 100 sample proportions was.035. a. Use the data to ﬁnd the upper and lower control limits for a p chart. b. Construct a p chart for the process and explain how it can be used. 7.52 Samples of n 200 items were selected hourly over a 100-hour period, and the sample proportion of defectives was calculated each hour. The mean of the 100 sample proportions was.041. a. Use the data to ﬁnd the upper and lower control limits for a p chart. 286 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS b. Construct a p chart for the process and explain how it can be used. APPLICATIONS 7.53 Black Jack A gambling casino records and plots the mean daily gain or loss from ﬁve blackjack tables on an x chart. The overall mean of the sample means and the standard deviation of the combined data over 40 weeks were x $10,752 and s $1605, respectively. a. Construct an x chart for the mean daily gain per blackjack table. b. How can this x chart be of value to the manager of the casino? 7.54 Brass Rivets A producer of brass rivets randomly samples 400 rivets each hour and calculates the proportion of defectives in the sample. The mean sample proportion calculated from 200 samples was equal to.021. Construct a control chart for the proportion of defectives in samples of 400 rivets. Explain how the control chart can be of value to a manager. EX0755 7
.55 Lumber Specs The manager of a building-supplies company randomly samples incoming lumber to see whether it meets quality speciﬁcations. From each shipment, 100 pieces of 2 4 lumber are inspected and judged according to whether they are ﬁrst (acceptable) or second (defective) grade. The proportions of second-grade 2 4s recorded for 30 shipments were as follows:.14.21.14.21.15.20.19.23.18.22.22.19.19.26.23.20.22.13.18.12.22.23.19.21.25.15.20.17.21.26 Construct a control chart for the proportion of secondgrade 2 4s in samples of 100 pieces of lumber. Explain how the control chart can be of use to the manager of the building-supplies company. 7.56 Coal Burning Power Plant A coal-burning power plant tests and measures three specimens of coal each day to monitor the percentage of ash in the coal. The overall mean of 30 daily sample means and the combined standard deviation of all the data were x 7.24 and s.07, respectively. Construct an x chart for the process and explain how it can be of value to the manager of the power plant. EX0757 7.57 Nuclear Power Plant The data in the table are measures of the radiation in air particulates at a nuclear power plant. Four measurements were recorded at weekly intervals over a 26-week period. Use the data to construct an x chart and plot the 26 values of x. Explain how the chart can be used. Week Radiation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.031.025.029.035.022.030.019.027.034.017.022.016.015.029.031.014.019.024.029.032.041.034.021.029.016.020.032.026.029.037.024.029.019.028.032.016.020.018.017.028.029.016.019.024.027.030.042.036.022.029.017.021.030.025.031.034.022.030.018.028.033.018.020.017.018.029.030.016.021.024.028.031
.038.036.024.030.017.020.031.025.030.035.023.030.019.028.033.018.021.017.017.029.031.017.020.025.028.030.039.035.022.029.016.022 7.58 Baseball Bats A hardwoods manufacturing plant has several different production lines to make baseball bats of different weights. One such production line is designed to produce bats weighing 32 ounces. During a period of time when the production process was known to be in statistical control, the average bat weight was found to be 31.7 ounces. The observed data were gathered from 50 samples, each consisting of 5 measurements. The standard deviation of all samples was found to be s.2064 ounces. Construct an x-chart to monitor the 32-ounce bat production process. 7.59 More Baseball Bats Refer to Exercise 7.58 and suppose that during a day when the state of the 32-ounce bat production process was unknown, the following measurements were obtained at hourly intervals. x Hour Hour x 1 2 3 31.6 32.5 33.4 4 5 6 33.1 31.6 31.8 Each measurement represents a statistic computed from a sample of ﬁve bat weights selected from the production process during a certain hour. Use the control chart constructed in Exercise 7.58 to monitor the process. CHAPTER REVIEW ❍ 287 CHAPTER REVIEW Key Concepts and Formulas I. Sampling Plans and Experimental Designs 1. Simple random sampling a. Each possible sample of size n is equally likely to occur. b. Use a computer or a table of random num- bers. 3. Probabilities involving the sample mean can be calculated by standardizing the value of x using z: m x z n s / IV. Sampling Distribution of the Sample c. Problems are nonresponse, undercoverage, Proportion and wording bias. 2. Other sampling plans involving randomization a. Stratiﬁed random sampling b. Cluster sampling c. Systematic 1-in-k sampling 3. Nonrandom sampling a. Convenience sampling b. Judgment sampling c. Quota sampling II. Statistics and Sampling Distributions 1. Sampling distributions describe the possible values of a statistic and how often they occur in repeated sampling. 2. Sampling distributions can be derived mathematically, approximated empirically, or found using statistical theorems. 3. The Central Limit The
orem states that sums and averages of measurements from a nonnormal population with ﬁnite mean m and standard deviation s have approximately normal distributions for large samples of size n. III. Sampling Distribution of the Sample Mean 1. When samples of size n are randomly drawn from a normal population with mean m and variance s 2, the sample mean x has a normal distribution with mean m and standard deviation s/n. 2. When samples of size n are randomly drawn from a nonnormal population with mean m and variance s 2, the Central Limit Theorem ensures that the sample mean x will have an approximately normal distribution with mean m and standard deviation s/n when n is large (n 30). 1. When samples of size n are drawn from a binomial population with parameter p, the sample proportion ˆp will have an approximately normal distribution with mean p and standard deviation pq/n as long as np 5 and nq 5. 2. Probabilities involving the sample proportion can be calculated by standardizing the value ˆp using z: z ˆp p q p n V. Statistical Process Control 1. To monitor a quantitative process, use an x chart. Select k samples of size n and calculate the overall mean x and the standard deviation s of all nk measurements. Create upper and lower control limits as s x 3 n If a sample mean exceeds these limits, the process is out of control. 2. To monitor a binomial process, use a p chart. Select k samples of size n and calculate the average of the sample proportions as S ˆpi p k Create upper and lower control limits as p 3p(1 p) n If a sample proportion exceeds these limits, the process is out of control. 288 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS The Central Limit Theorem at Work MINITAB provides a perfect tool for exploring the way the Central Limit Theorem works in practice. Remember that, according to the Central Limit Theorem, if random samples of size n are drawn from a nonnormal population with mean m and standard deviation s, then when n is large, the sampling distribution of the sample mean x will be approximately normal with the same mean m and with standard error s/n. Let’s try sampling from a nonnormal population with the help of MINITAB. In a new MINITAB worksheet, generate 100 samples of size n 30 from a nonnormal distribution called the exponential distribution. Use Calc Random
Data Exponential. Type 100 for the number of rows of data, and store the results in C1–C30 (see Figure 7.17). Leave the mean at the default of 1.0, the threshold at 0.0, and click OK. The data are generated and stored in the worksheet. Use Graph Histogram Simple to look at the distribution of some of the data—say, C1 (as in Figure 7.18). Notice that the distribution is not mound-shaped; it is highly skewed to the right. F IG URE 7. 17 ● For the exponential distribution that we have used, the mean and standard deviation are m 1 and s 1, respectively. Check the descriptive statistics for one of the columns (use Stat Basic Statistics Display Descriptive Statistics), and you will ﬁnd that the 100 observations have a sample mean and standard deviation that are both close to but not exactly equal to 1. Now, generate 100 values of x based on samples of size n 30 by creating a column of means for the 100 rows. Use Calc Row Statistics, and select Mean. To average the entries in all 30 columns, select or type C1–C30 in the Input variables box, and store the results in C31 (see Figure 7.19). You can now look at the distribution of the sample means using Graph Histogram Simple, selecting C31 and clicking OK. The distribution of the 100 sample means generated for our example is shown in Figure 7.20. FI GUR E 7. 18 ● MY MINITAB ❍ 289 F IGU RE 7.1 9 ● 290 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS F IG URE 7. 20 ● Notice the distinct mound shape of the distribution in Figure 7.20 compared to the original distribution in Figure 7.18. Also, if you check the descriptive statistics for C31, you will ﬁnd that the mean and standard deviation of our 100 sample means are not too different from the theoretical values, m 1 and s/n 1/30.18. (For our data, the sample mean is.9645 and the standard deviation is.1875.) Since we had only 100 samples, our results are not exactly equal to the theoretical values. If we had generated an inﬁnite number of samples, we would have gotten an exact match. This is the Central Limit Theorem at work! Supplementary Exercises 7.60 A ﬁnite population
consists of four elements: 6, 1, 3, 2. a. How many different samples of size n 2 can be selected from this population if you sample without replacement? (Sampling is said to be without replacement if an element cannot be selected twice for the same sample.) b. List the possible samples of size n 2. c. Compute the sample mean for each of the samples given in part b. d. Find the sampling distribution of x. Use a probability histogram to graph the sampling distribution of x. e. If all four population values are equally likely, calculate the value of the population mean m. Do any of the samples listed in part b produce a value of x exactly equal to m? 7.61 Refer to Exercise 7.60. Find the sampling distribution for x if random samples of size n 3 are selected without replacement. Graph the sampling distribution of x. 7.62 Lead Pipes Studies indicate that drinking water supplied by some old lead-lined city piping systems may contain harmful levels of lead. An important study of the Boston water supply system showed that the distribution of lead content readings for individual water specimens had a mean and standard deviation of approximately.033 milligrams per liter (mg/l) and.10 mg/l, respectively.15 a. Explain why you believe this distribution is or is not normally distributed. b. Because the researchers were concerned about the shape of the distribution in part a, they calculated the average daily lead levels at 40 different locations on each of 23 randomly selected days. What can you say about the shape of the distribution of the average daily lead levels from which the sample of 23 days was taken? c. What are the mean and standard deviation of the distribution of average lead levels in part b? 7.63 Biomass The total amount of vegetation held by the earth’s forests is important to both ecologists and politicians because green plants absorb carbon dioxide. An underestimate of the earth’s vegetative mass, or biomass, means that much of the carbon dioxide emitted by human activities (primarily fossilburning fuels) will not be absorbed, and a climatealtering buildup of carbon dioxide will occur. Studies16 indicate that the biomass for tropical woodlands, thought to be about 35 kilograms per square meter (kg/m2), may in fact be too high and that tropical biomass values vary regionally—from about 5 to 55 kg/m2. Suppose you measure the tropical biomass in 400 randomly selected square-meter plots. a. Approximate
s, the standard deviation of the biomass measurements. b. What is the probability that your sample average is within two units of the true average tropical biomass? c. If your sample average is x 31.75, what would you conclude about the overestimation that concerns the scientists? 7.64 Hard Hats The safety requirements for hard hats worn by construction workers and others, established by the American National Standards Institute (ANSI), specify that each of three hats pass the following test. A hat is mounted on an aluminum head form. An 8-pound steel ball is dropped on the hat from a height of 5 feet, and the resulting force is measured at the bottom of the head form. The force exerted on the head form by each of the three hats must be less than 1000 pounds, and the average of the three must be less than 850 pounds. (The relationship between this test and actual human head damage is unknown.) Suppose SUPPLEMENTARY EXERCISES ❍ 291 the exerted force is normally distributed, and hence a sample mean of three force measurements is normally distributed. If a random sample of three hats is selected from a shipment with a mean equal to 900 and s 100, what is the probability that the sample mean will satisfy the ANSI standard? 7.65 Imagery and Memory A research psychologist is planning an experiment to determine whether the use of imagery—picturing a word in your mind— affects people’s ability to memorize. He wants to use two groups of subjects: a group that memorizes a set of 20 words using the imagery technique, and a control group that does not use imagery. a. Use a randomization technique to divide a group of 20 subjects into two groups of equal size. b. How can the researcher randomly select the group of 20 subjects? c. Suppose the researcher offers to pay subjects $50 each to participate in the experiment and uses the ﬁrst 20 students who apply. Would this group behave as if it were a simple random sample of size n 20? 7.66 Legal Abortions The results of a Newsweek poll concerning views on abortion given in the table that follows show that there is no consensus on this issue among Americans.17 Newsweek Poll conducted by Princeton Survey Research Associates International. Oct. 26–27, 2006. N 1002 adults nationwide. MoE 3 (for all adults). “Which side of the political debate on the abortion issue do you sympathize with more: the right-to-life movement that believes abortion is the taking of
human life and should be outlawed; OR, the pro-choice movement that believes a woman has the right to choose what happens to her body, including deciding to have an abortion?” (Options rotated) Right-to-Life % Pro-Choice % Neither % Unsure % ALL adults Republicans Democrats Independents 39 62 25 35 53 31 69 57 3 4 2 4 5 3 4 4 a. Is this an observational study or a planned experi- ment? b. Is there the possibility of problems in responses arising because of the somewhat sensitive nature of the subject? What kinds of biases might occur? 7.67 Sprouting Radishes A biology experiment was designed to determine whether sprouting radish seeds inhibit the germination of lettuce seeds.18 Three 292 ❍ CHAPTER 7 SAMPLING DISTRIBUTIONS 10-centimeter petri dishes were used. The ﬁrst contained 26 lettuce seeds, the second contained 26 radish seeds, and the third contained 13 lettuce seeds and 13 radish seeds. a. Assume that the experimenter had a package of 50 radish seeds and another of 50 lettuce seeds. Devise a plan for randomly assigning the radish and lettuce seeds to the three treatment groups. b. What assumptions must the experimenter make about the packages of 50 seeds in order to assure randomness in the experiment? 7.68 9/11 A study of about n 1000 individuals in the United States during September 21–22, 2001, revealed that 43% of the respondents indicated that they were less willing to ﬂy following the events of September 11, 2001.19 a. Is this an observational study or a designed experi- ment? b. What problems might or could have occurred because of the sensitive nature of the subject? What kinds of biases might have occurred? 7.69 Telephone Service Suppose a telephone company executive wishes to select a random sample of n 20 (a small number is used to simplify the exercise) out of 7000 customers for a survey of customer attitudes concerning service. If the customers are numbered for identiﬁcation purposes, indicate the customers whom you will include in your sample. Use the random number table and explain how you selected your sample. 7.70 Rh Positive The proportion of individuals with an Rh-positive blood type is 85%. You have a random sample of n 500 individuals. a. What are the mean and standard deviation of ˆp, the sample proportion with Rh-positive blood type? b. Is the distribution

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