text stringlengths 235 3.08k |
|---|
Oct. 2). Teen drivers: Get the facts. Retrieved from http://www.cdc.gov/ Motorvehiclesafety/Teen_Drivers/teendrivers_factsheet.html Daily Mail. (2011, June 9). One born every minute: the maternity unit where mothers are THREE to a bed. Retrieved http://www.dailymail.co.uk/news/article-2001422/Busiest-maternity-ward-planet-averages-60-babies-dayfrom mothersbed.html Department of Aviation at the Hartsfield-Jackson Atlanta International Airport. (2013). ATL fact sheet. Retrieved from http://www.atlanta-airport.com/Airport/ATL/ATL_FactSheet.aspx Lenhart, A. (2012). Teens, smartphones & testing: Texting volume is up while the frequency of voice calling is down. About one in four teens say they own smartphones. Pew Internet. Retrieved from http://www.pewinternet.org/~/media/Files/ Reports/2012/PIP_Teens_Smartphones_and_Texting.pdf Ministry of Health, Labour, and Welfare. (n.d.). Children and childrearing. Retrieved from http://www.mhlw.go.jp/english/ policy/children/children-childrearing/index.html Pew Internet. (2013). How Americans use text messaging. Retrieved from http://pewinternet.org/Reports/2011/Cell-PhoneTexting-2011/Main-Report.aspx South Carolina Department of Mental Health. (2006). Eating disorder statistics. Retrieved from http://www.state.sc.us/dmh/ anorexia/statistics.htm The Guardian. (2011, June 8). Giving birth in Manila: The maternity ward at the Dr Jose Fabella Memorial Hospital in Manila, in the Philippines, where there is an average of 60 births a day. Retrieved from http://www.theguardian.com/world/gallery/2011/jun/08/philippines-health#/?picture=375471900&index=2 the busiest Vanderkam, 8). http://management.fortune.cnn.com/2012/10/08/stop-checking-your-email-now/ (2012, Oct. checking email, Stop your L. now. CNNMoney. Retrieved from World Earthquakes. http://www.worldearthquakes |
.com/index.php?option=ethq_prediction earthquakes: Live (2012). World earthquake news and highlights. Retrieved from SOLUTIONS 1 x P(x) 0 1 2 3 4 5 6.12.18.30.15.10.10.05 Table 4.38 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 317 3.10 +.05 =.15 5 1 7.35 +.40 +.10 =.85 9 1(.15) + 2(.35) + 3(.40) + 4(.10) =.15 +.70 + 1.20 +.40 = 2.45 11 x P(x) 0 1 2 3.03.04.08.85 Table 4.39 13 Let X = the number of events Javier volunteers for each month. 15 x P(x) 0 1 2 3 4 5.05.05.10.20.25.35 Table 4.40 17 1 –.05 =.95 19.2 + 1.2 + 2.4 + 1.6 = 5.4 21 The values of P(x) do not sum to one. 23 Let X = the number of years a physics major will spend doing postgraduate research. 25 1 –.35 –.20 –.15 –.10 –.05 =.15 27 1(.35) + 2(.20) + 3(.15) + 4(.15) + 5(.10) + 6(.05) =.35 +.40 +.45 +.60 +.50 +.30 = 2.6 years 29 X is the number of years a student studies ballet with the teacher. 31.10 +.05 +.10 =.25 33 The sum of the probabilities sum to one because it is a probability distribution. 35 −2 ⎛ ⎝ 40 52 ⎞ ⎠ + 30 ⎛ ⎝ 12 52 ⎞ ⎠ = − 1.54 + 6.92 = 5.38 37 X = the number that reply yes 39 0, 1, 2, 3, 4, 5, 6, 7, 8 41 5.7 318 43.4151 Chapter 4 | Discrete Random Variables 45 X = the number of freshmen selected from the study until one replied yes to |
the law that was passed. 47 1,2,… 49 1.4 51 X = the number of business majors in the sample. 53 2, 3, 4, 5, 6, 7, 8, 9 55 6.26 57 0, 1, 2, 3, 4, … 59.0485 61.0214 63 X = the number of United States teens who die from motor vehicle injuries per day. 65 0, 1, 2, 3, 4,... 67 no 71 The variable of interest is X, or the gain or loss, in dollars. The face cards jack, queen, and king. There are (3)(4) = 12 face cards and 52 – 12 = 40 cards that are not face cards. We first need to construct the probability distribution for X. We use the card and coin events to determine the probability for each outcome, but we use the monetary value of X to determine the expected value. Card Event X net gain/loss P(X) Face Card and Heads Face Card and Tails 6 2 (Not Face Card) and (H or T) –2 Table 4.41 ⎛ ⎝ 12 52 ⎞ ⎛ ⎝ ⎠ 1 2 ⎞ ⎠ = ⎛ ⎝ 6 52 ⎞ ⎠ ⎛ ⎝ 12 52 ⎞ ⎛ ⎝ ⎠ 1 2 ⎞ ⎠ = ⎛ ⎝ 6 52 ⎞ ⎠ ⎛ ⎝ 40 52 ⎞ ⎠(1) = ⎛ ⎝ 40 52 ⎞ ⎠ ⎛ • Expected value = (6) ⎝ 6 52 ⎛ ⎞ ⎠ + (2) ⎝ 6 52 ⎛ ⎞ ⎠ + ( − 2) ⎝ 40 52 ⎞ ⎠ = – 32 52 • Expected value = –$0.62, rounded to the nearest cent • If you play this game repeatedly, over a long string of games, you would expect to lose 62 cents per game, on average. • You should not play this game to win money because the expected value indicates an expected average loss. 73 a..1 b. 1.6 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 |
Chapter 4 | Discrete Random Variables 319 75 a. Software Company x 5,000,000 1,000,000 –1,000,000 Table 4.42 P(x).10.30.60 Hardware Company x 3,000,000 1,000,000 –1,000,00 Table 4.43 P(x).20.40.40 Biotech Firm x 6,000,000 0 P(x).10.70 –1,000,000.20 Table 4.44 b. $200,000; $600,000; $400,000 c. d. e. third investment because it has the lowest probability of loss first investment because it has the highest probability of loss second investment 77 4.85 years 79 b 81 Let X = the amount of money to be won on a ticket. The following table shows the PDF for X: x 0 5 P(x).969 250 10,000 =.025 Table 4.45 320 Chapter 4 | Discrete Random Variables x P(x) 25 100 50 10,000 10 10,000 =.005 =.001 Table 4.45 Calculate the expected value of X. 0(.969) + 5(.025) + 25(.005) + 100(.001) =.35 A fair price for a ticket is $0.35. Any price over $0.35 will enable the lottery to raise money. 83 X = the number of patients calling in claiming to have the flu, who actually have the flu. X = 0, 1, 2,...25 85.0165 87 a. X = the number of DVDs a Video to Go customer rents b. c. d..12.11.77 89 d. 4.43 91 c 93 • X = number of questions answered correctly • X ~ B ⎛ ⎝32, 1 3 ⎞ ⎠ • We are interested in MORE THAN 75 percent of 32 questions correct. 75 percent of 32 is 24. We want to find P(x > 24). The event more than 24 is the complement of less than or equal to 24. • Using your calculator's distribution menu: 1 – binomcdf ⎛ ⎝32, 1 3, 24 ⎞ ⎠ • P(x > 24) = 0 • The probability of getting more than 75 percent of the 32 questions correct when randomly guessing is very small and practically zero. 95 |
a. X = the number of college and universities that offer online offerings. b. 0, 1, 2, …, 13 c. X ~ B(13, 0.96) d. 12.48 e..0135 f. P(x = 12) =.3186 P(x = 13) = 0.5882 More likely to get 13. 97 a. X = the number of fencers who do not use the foil as their main weapon b. 0, 1, 2, 3,... 25 c. X ~ B(25,.40) d. 10 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 321 e..0442 f. The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising. 99 a. X = the number of audits in a 20-year period b. 0, 1, 2, …, 20 c. X ~ B(20,.02) d. e. f..4.6676.0071 101 1. X = the number of matches 2. 0, 1, 2, 3 3. X ~ B ⎛ ⎝3, 1 6 ⎞ ⎠ 4. 5. In dollars: −1, 1, 2, 3 1 2 6. Multiply each Y value by the corresponding X probability from the PDF table. The answer is −.0787. You lose about eight cents, on average, per game. 7. The house has the advantage. 103 a. X ~ B(15,.281) Figure 4.10 b. i. Mean = μ = np = 15(.281) = 4.215 ii. Standard Deviation = σ = npq = 15(.281)(.719) = 1.7409 c. P(x > 5) = 1 – P(x ≤ 5) = 1 – binomcdf(15,.281, 5) = 1 – 0.7754 =.2246 P(x = 3) = binompdf(15,.281, 3) =.1927 P(x = 4) = binompdf(15,.281, 4) =.2259 It is more likely that four people are literate than three people are. 322 105 Chapter 4 | Discrete Random Variables a |
. X = the number of adults in America who are surveyed until one says he or she will watch the Super Bowl. b. X ~ G(.40) c. 2.5 d. e..0187.2304 107 a. X = the number of pages that advertise footwear b. X takes on the values 0, 1, 2,..., 20 c. X ~ B(20, 29 192 ) d. 3.02 e. no f..9997 g. X = the number of pages we must survey until we find one that advertises footwear. X ~ G( 29 192 ) h..3881 i. 6.6207 pages 109 0, 1, 2, and 3 111 a. X ~ G(.25) b. i. mean = μ = 1 p = 1 0.25 = 4 ii. standard deviation = σ = 1 − p p2 = 1 −.25.252 ≈ 3.4641 c. P(x = 10) = geometpdf(.25, 10) =.0188 d. P(x = 20) = geometpdf(.25, 20) =.0011 e. P(x ≤ 5) = geometcdf(.25, 5) =.7627 113 a. X = the number of pages that advertise footwear b. 0, 1, 2, 3,..., 20 c. X ~ H(29, 163, 20), r = 29, b = 163, n = 20 d. 3.03 e. 1.5197 115 a. X = the number of Patriots picked b. 0, 1, 2, 3, 4 c. X ~ H(4, 8, 9) d. without replacement 117 a. X ~ P(5.5); μ = 5.5; σ = 5.5 ≈ 2.3452 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 4 | Discrete Random Variables 323 b. P(x ≤ 6) = poissoncdf(5.5, 6) ≈.6860 c. There is a 15.7 percent probability that the law staff will receive more calls than they can handle. d. P(x > 8) = 1 – P(x ≤ 8) = 1 – poissoncdf(5.5, 8) ≈ 1 – |
.8944 =.1056 119 Let X = the number of defective bulbs in a string. Using the Poisson distribution: • μ = np = 100(.03) = 3 • X ~ P(3) • P(x ≤ 4) = poissoncdf(3, 4) ≈.8153 Using the binomial distribution • X ~ B(100,.03) • P(x ≤ 4) = binomcdf(100,.03, 4) ≈.8179 The Poisson approximation is very good—the difference between the probabilities is only.0026. 121 a. X = the number of children for a Spanish woman b. 0, 1, 2, 3,... c. X ~ P(1.47) d. e. f..2299.5679.4321 123 a. X = the number of fortune cookies that have an extra fortune b. 0, 1, 2, 3,... 144 c. X ~ B(144,.03) or P(4.32) d. 4.32 e. f..0124 or.0133.6300 or.6264 g. As n gets larger, the probabilities get closer together. 125 a. X = the number of people audited in one year b. 0, 1, 2,..., 100 c. X ~ P(2) d. 2 e. f..1353.3233 127 a. X = the number of shell pieces in one cake b. 0, 1, 2, 3,... c. X ~ P(1.5) d. 1.5 e. f..2231.0001 g. yes 324 129 d 130 Chapter 4 | Discrete Random Variables a. You can use randInt (0,1,5) to generate five trials of the experiment. Count the number of 1’s generated to determine the number of successes. b. Student answers may vary. c. Student answers may vary. d. The theoretical mean is (5)(.5) = 2.5. The theoretical standard deviation is (5)(.5)(0.5) = 1.25. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 325 5 | CONTINUOUS RANDOM VARIABLES Figure 5.1 The heights of these radish plants are continuous random |
variables. (credit: Rev Stan) Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: • Recognize and understand continuous probability density functions in general • Recognize the uniform probability distribution and apply it appropriately • Recognize the exponential probability distribution and apply it appropriately Continuous random variables have many applications. Baseball batting averages, IQ scores, the length of time a longdistance telephone call lasts, the amount of money a person carries, the length of time a computer chip lasts, and SAT scores are just a few. The field of reliability depends on a variety of continuous random variables. 326 NOTE Chapter 5 | Continuous Random Variables The values of discrete and continuous random variables can be ambiguous. For example, if X is equal to the number of miles (to the nearest mile) you drive to work, then X is a discrete random variable. You count the miles. If X is the distance you drive to work, then you measure values of X and X is a continuous random variable. For a second example, if X is equal to the number of books in a backpack, then X is a discrete random variable. If X is the weight of a book, then X is a continuous random variable because weights are measured. How the random variable is defined is very important. Properties of Continuous Probability Distributions The graph of a continuous probability distribution is a curve. Probability is represented by the area under the curve. The curve is called the probability density function (abbreviated as pdf). We use the symbol f(x) to represent the curve. f(x) is the function that corresponds to the graph; we use the density function f(x) to draw the graph of the probability distribution. Area under the curve is given by a different function called the cumulative distribution function (abbreviated as cdf). The cumulative distribution function is used to evaluate probability as area. • The outcomes are measured, not counted. • The entire area under the curve and above the x-axis is equal to one. • Probability is found for intervals of x values rather than for individual x values. • P(c < x < d) is the probability that the random variable X is in the interval between the values c and d. P(c < x < d) is the area under the curve, above the x-axis, to the right of c and the left of d. • P(x = c) = 0 The probability that x takes on any single |
individual value is zero. The area below the curve, above the x-axis, and between x = c and x = c has no width, and therefore no area (area = 0). Since the probability is equal to the area, the probability is also zero. • P(c < x < d) is the same as P(c ≤ x ≤ d) because probability is equal to area. We will find the area that represents probability by using geometry, formulas, technology, or probability tables. In general, calculus is needed to find the area under the curve for many probability density functions. When we use formulas to find the area in this textbook, we are using formulas that were found by using the techniques of integral calculus. However, because most students taking this course have not studied calculus, we will not be using calculus in this textbook. There are many continuous probability distributions. When probability is modeled by use of a continuous probability distribution, the distribution used is selected to model and fit the particular situation in the best way. In this chapter and the next, we will study the uniform distribution, the exponential distribution, and the normal distribution. The following graphs illustrate these distributions: Figure 5.2 The graph shows a uniform distribution with the area between x = 3 and x = 6 shaded to represent the probability that the value of the random variable X is in the interval between three and six. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 327 Figure 5.3 The graph shows an exponential distribution with the area between x = 2 and x = 4 shaded to represent the probability that the value of the random variable X is in the interval between two and four. Figure 5.4 The graph shows the standard normal distribution with the area between x = 1 and x = 2 shaded to represent the probability that the value of the random variable X is in the interval between one and two. 5.1 | Continuous Probability Functions We begin by defining a continuous probability density function. We use the function notation f(x). Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function f(x) so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one, the maximum area is also one. For continuous probability distributions, PROBABILITY = AREA. Example 5.1 |
Consider the function f(x) = 1 20 for 0 ≤ x ≤ 20. x = a real number. The graph of f(x) = 1 20 is a horizontal line. However, since 0 ≤ x ≤ 20, f(x) is restricted to the portion between x = 0 and x = 20, inclusive. 328 Chapter 5 | Continuous Random Variables Figure 5.5 f(x) = 1 20 for 0 ≤ x ≤ 20. The graph of f(x) = 1 20 is a horizontal line segment when 0 ≤ x ≤ 20. The area between f(x) = 1 20 = 1 20. where 0 ≤ x ≤ 20 and the x-axis is the area of a rectangle with base = 20 and height Suppose we want to find the area between f(x) = 1 20 and the x-axis where 0 < x < 2. AREA = 20 ⎛ ⎝ 1 20 ⎞ ⎠ = 1 Figure 5.6 ⎛ AREA = (2 – 0) ⎝ ⎞ ⎠ = 0.1 1 20 (2 – 0) = 2 = base of a rectangle REMINDER area of a rectangle = (base)(height) The area corresponds to a probability. The probability that x is between zero and two is 0.1, which can be written This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 329 mathematically as P(0 < x < 2) = P(x < 2) = 0.1. Suppose we want to find the area between f(x) = 1 20 and the x-axis where 4 < x < 15. Figure 5.7 ⎛ AREA = (15 – 4) ⎝ ⎞ ⎠ = 0.55 1 20 (15 – 4) = 11 = the base of a rectangle The area corresponds to the probability P(4 < x < 15) = 0.55. Suppose we want to find P(x = 15). On an x-y graph, x = 15 is a vertical line. A vertical line has no width (or zero ⎞ width). Therefore, P(x = 15) = (base)(height) = (0) ⎛ ⎠ ⎝ = 0 1 20 Figure 5.8 P(X <= x), which can also be written |
as P(X < x) for continuous distributions, is called the cumulative distribution function or CDF. Notice the less than or equal to symbol. We can also use the CDF to calculate P(X > x). The CDF gives area to the left and P(X > x) gives area to the right. We calculate P(X > x) for continuous distributions as follows: P(X > x) = 1 – P (X < x). 330 Chapter 5 | Continuous Random Variables Figure 5.9 Label the graph with f(x) and x. Scale the x and y axes with the maximum x and y values. f(x) = 1 20, 0 ≤ x ≤ 20. To calculate the probability that x is between two values, look at the following graph. Shade the region between x = 2.3 and x = 12.7. Then calculate the shaded area of a rectangle. Figure 5.10 ⎛ P(2.3 < x < 12.7) = (base)(height) = (12.7 − 2.3) ⎝ ⎞ ⎠ = 0.52 1 20 5.1 Consider the function f(x) = 1 8 for 0 ≤ x ≤ 8. Draw the graph of f(x) and find P(2.5 < x < 7.5). 5.2 | The Uniform Distribution The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data are inclusive or exclusive of endpoints. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 331 Example 5.2 The data in Table 5.1 are 55 smiling times, in seconds, of an eight-week-old baby. 10.4 19.6 18.8 13.9 17.8 16.8 21.6 17.9 12.5 11.1 4.9 12.8 14.8 22.8 20.0 15.9 16.3 13.4 17.1 14.5 19.0 22.8 1.3 5.8 8.9 0.7 6.9 9.4 8.9 2.6 9.4 11.9 10.9 7.3 5.9 5.8 7.6 21.7 11. |
8 3.4 10.0 3.3 6.7 3.7 2.1 7.8 17.9 19.2 9.8 4.5 6.3 10.7 11.6 13.8 18.6 Table 5.1 The sample mean = 11.49 and the sample standard deviation = 6.23. We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution. Let X = length, in seconds, of an eight-week-old baby's smile. The notation for the uniform distribution is X ~ U(a, b) where a = the lowest value of x and b = the highest value of x. The probability density function is f(x) = For this example, X ~ U(0, 23) and f(x) = 1 b − a for a ≤ x ≤ b. 1 23 − 0 for 0 ≤ X ≤ 23. Formulas for the theoretical mean and standard deviation are For this problem, the theoretical mean and standard deviation are μ = a + b 2 and σ = (b − a)2 12 μ = 0 + 23 2 = 11.50 seconds and σ = (23 − 0)2 12 = 6.64 seconds. Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example. 332 Chapter 5 | Continuous Random Variables 5.2 The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of a and b. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation. 1 12 4 10 4 14 11 7 11 4 13 2 4 6 3 10 0 12 6 9 10 5 13 4 10 14 12 11 6 10 11 0 11 13 2 Table 5.2 Example 5.3 a. Refer to Example 5.2. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds? Solution 5.3 P(2 < x < 18) = (base)(height) = (18 – 2) ⎛ ⎝ � |
�� ⎠ 1 23 = 16 23 Figure 5.11 b. Find the 90th percentile for an eight-week-old baby's smiling time. Solution 5.3 b. Ninety percent of the smiling times fall below the 90th percentile, k, so P(x < k) = 0.90. ⎛ ⎛ ⎠ ⎝base⎞ P(x < k) = 0.90 ⎝height⎞ ⎠ = 0.90 ⎞ ⎛ ⎠ = 0.90 (k − 0) ⎝ 1 23 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 333 k = (23)(0.90) = 20.7 Figure 5.12 c. Find the probability that a random eight-week-old baby smiles more than 12 seconds knowing that the baby smiles more than eight seconds. Solution 5.3 c. This probability question is a conditional. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds. Find P(x > 12|x > 8) There are two ways to do the problem. For the first way, use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds. for 8 < x < 23 Write a new f(x): f(x) = 1 23 − 8 = 1 15 for 8 < x < 23. P(x > 12|x > 8) = (23 − 12) ⎛ ⎝ ⎞ ⎠ 1 15 = 11 15 Figure 5.13 For the second way, use the conditional formula from Probability Topics with the original distribution. P(A|B) = P(A AND B) P(B) For this problem, A is (x > 12) and B is (x > 8). 334 Chapter 5 | Continuous Random Variables So, P(x > 12|x > 8) = (x > 12 AND x > 8) P(x > 8) = P(x > 12) P(x > 8) = 11 23 15 23 = 11 15 Figure 5.14 5.3 A distribution is given as X ~ U |
(0, 20). What is P(2 < x < 18)? Find the 90th percentile. Example 5.4 The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive. a. What is the probability that a person waits fewer than 12.5 minutes? Solution 5.4 a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 15. X ~ U(0, 15). Write the probability density function. f (x) = 1 15 − 0 = 1 15 for 0 ≤ x ≤ 15. Find P (x < 12.5). Draw a graph. ⎛ P(x < k) = (base)(height) = (12.5 - 0) ⎝ ⎞ ⎠ = 0.8333 1 15 The probability a person waits fewer than 12.5 minutes is 0.8333. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 335 Figure 5.15 b. On the average, how long must a person wait? Find the mean, μ, and the standard deviation, σ. Solution 5.4 b. μ = a + b 2 = 15 + 0 2 σ = (b - a)2 12 = (15 - 0)2 12 = 7.5. On the average, a person must wait 7.5 minutes. = 4.3. The standard deviation is 4.3 minutes. c. Ninety percent of the time, the minutes a person must wait falls below what value? This question asks for the 90th percentile. Solution 5.4 c. Find the 90th percentile. Draw a graph. Let k = the 90th percentile. P(x < k) = (base)(height) = (k − 0)( 1 15 ) ⎛ 0.90 = (k) ⎝ ⎞ ⎠ 1 15 k = (0.90)(15) = 13.5 k is sometimes called a critical value. The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes. 336 Chapter 5 | Continuous Random Variables Figure 5.16 5.4 The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 |
hours and 521 hours inclusive. a. Find a and b and describe what they represent. b. Write the distribution. c. Find the mean and the standard deviation. d. What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours? e. What is the 65th percentile for the duration of games for a team for the 2011 season? Example 5.5 Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let X = the time, in minutes, it takes a nine-year-old child to eat a doughnut. Then X ~ U(0.5, 4). a. The probability that a randomly selected nine-year-old child eats a doughnut in at least two minutes is _______. Solution 5.5 a. 0.5714 b. Find the probability that a different nine-year-old child eats a doughnut in more than two minutes given that the child has already been eating the doughnut for more than 1.5 minutes. The second question has a conditional probability. You are asked to find the probability that a nine-year-old child eats a doughnut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see Example 5.3). You must reduce the sample space. First way: Since you know the child has already been eating the doughnut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes. Write a new f(x): Find P(x > 2|x > 1.5). Draw a graph. f (x) = 1 4 − 1.5 = 2 5 for 1.5 ≤ x ≤ 4. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 337 Figure 5.17 Solution 5.5 b. 4 5 P(x > 2|x > 1.5) = (base)(new height) = (4 – 2 The probability that a nine-year-old child eats a donut in more than two minutes given that the child has already been eating the doughnut for more than 1.5 minutes is 4 5. Second way: Draw the original graph for X ~ U |
(0.5, 4). Use the conditional formula P(x > 2|x > 1.5) = P(x > 2 AND x > 1.5) P(x > 1.5) = P(x > 2) P(x > 1.5) = 2 3.5 2.5 3.5 = 0.8 = 4 5 5.5 Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let X = the time, in minutes, it takes a student to finish a quiz. Then X ~ U(6, 15). Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes. Example 5.6 Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let x = the time needed to fix a furnace. Then x ~ U(1.5, 4). a. Find the probability that a randomly selected furnace repair requires more than two hours. b. Find the probability that a randomly selected furnace repair requires less than three hours. c. Find the 30th percentile of furnace repair times. d. The longest 25 percent of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25 percent of repair times.) What percentile does this represent? e. Find the mean and standard deviation 338 Chapter 5 | Continuous Random Variables Solution 5.6 a. To find f(x): f (x) = 1 4 − 1.5 = 1 2.5 so f(x) = 0.4 P(x > 2) = (base)(height) = (4 – 2)(0.4) = 0.8 Figure 5.18 Uniform distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time x is greater than two Solution 5.6 b. P(x < 3) = (base)(height) = (3 – 1.5)(0.4) = 0.6 The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between x = 1.5 and x = 3. Note that the shaded area starts at x = 1.5 rather |
than at x = 0. Because X ~ U(1.5, 4), x cannot be less than 1.5. Figure 5.19 Uniform distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time x is less than three Solution 5.6 c. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 339 Figure 5.20 Uniform distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30 percent of repair times. P (x < k) = 0.30 P(x < k) = (base)(height) = (k – 1.5)(0.4) 0.3 = (k – 1.5) (0.4); Solve to find k: 0.75 = k – 1.5, obtained by dividing both sides by 0.4 k = 2.25, obtained by adding 1.5 to both sides The 30th percentile of repair times is 2.25 hours. 30 percent of repair times are 2.5 hours or less. Solution 5.6 d. Figure 5.21 Uniform distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25 percent of repair times. P(x > k) = 0.25 P(x > k) = (base)(height) = (4 – k)(0.4) 0.25 = (4 – k)(0.4); Solve for k: 0.625 = 4 − k, obtained by dividing both sides by 0.4 −3.375 = −k, obtained by subtracting four from both sides: k = 3.375 The longest 25 percent of furnace repairs take at least 3.375 hours (3.375 hours or longer). Note: Since 25 percent of repair times are 3.375 hours or longer, that means that 75 percent of repair times are 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times. 340 Chapter 5 | Continuous Random Variables Solution 5.6 e. μ = a + b 2 and σ = (b − a)2 12 μ = 1.5 + 4 2 = 2.75 hours and σ = (4 – 1.5)2 12 = 0.7217 hours 5. |
6 The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let X = the time needed to change the oil on a car. a. Write the random variable X in words. X = __________________. b. Write the distribution. c. Graph the distribution. d. Find P (x > 19). e. Find the 50th percentile. 5.3 | The Exponential Distribution (Optional) The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long-distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution. Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money. Exponential distributions are commonly used in calculations of product reliability, or the length of time a product lasts. Example 5.7 Let X = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes. X is a continuous random variable since time is measured. It is given that μ = 4 minutes. To do any calculations, you must know m, the decay parameter. m = 1 μ. Therefore, m = 1 4 = 0.25. The standard deviation, σ, is the same as the mean. μ = σ The distribution notation is X ~ Exp(m). Therefore, X ~ Exp(0.25). The probability density function is f(x) = me-mx. The number e = 2.71828182846... It is a number that is used often in mathematics. Scientific calculators have the key "ex." If you enter one for x, the calculator will display the value e. The curve is f(x) = 0.25e–0.25x where x is at least zero and m = 0.25. For example, f(5) = 0.25e |
(−0.25)(5) = 0.072. The probability that the postal clerk spends five minutes with the This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 341 customers is 0.072. The graph is as follows: Figure 5.22 Notice the graph is a declining curve. When x = 0, f(x) = 0.25e(−0.25)(0) = (0.25)(1) = 0.25 = m. The maximum value on the y-axis is m. 5.7 The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution. Example 5.8 a. Using the information in Example 5.7, find the probability that a clerk spends four to five minutes with a randomly selected customer. Solution 5.8 a. Find P(4 < x < 5). The cumulative distribution function (CDF) gives the area to the left. P(x < x) = 1 – e – mx P(x < 5) = 1 – e ( – 0.25)(5) = 0.7135 and P(x < 4) = 1 – e ( – 0.25)(4) = 0.6321 342 Chapter 5 | Continuous Random Variables Figure 5.23 NOTE You can do these calculations easily on a calculator. The probability that a postal clerk spends four to five minutes with a randomly selected customer is P(4 < x < 5) = P(x < 5) – P(x < 4) = 0.7135 − 0.6321 = 0.0814. On the home screen, enter (1 – e^(–0.25*5))–(1–e^(–0.25*4)) or enter e^(–0.25*4) – e^(–0.25*5). b. Half of all customers are finished within how long? (Find the 50th percentile). Solution 5.8 b. Find the 50th percentile. Figure 5.24 P(x < k) = 0.50, k = 2.8 minutes (calculator or computer) Half of all customers are finished within 2.8 minutes. You can also do the calculation as |
follows: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 343 P(x < k) = 0.50 and P(x < k) = 1 – e – 0.25k Therefore, 0.50 = 1 − e−0.25k and e−0.25k = 1 − 0.50 = 0.5. Take natural logs: ln(e–0.25k) = ln(0.50). So, –0.25k = ln(0.50). Solve for k: k = ln(0.50) -0.25 following two notes. NOTE = 2.8 minutes. The calculator simplifies the calculation for percentile k. See the A formula for the percentile k is k = ln(1 − AreaToTheLe f t) −m where ln is the natural log. On the home screen, enter ln(1 – 0.50)/–0.25. Press the (–) for the negative. c. Which is larger, the mean or the median? Solution 5.8 c. From Part b, the median or 50th percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger. 5.8 The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than 10 days in advance. How many days do half of all travelers wait? Have each class member count the change he or she has in his or her pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean. Let X = the amount of money a student in your class has in his or her pocket or purse. The distribution for X is approximately exponential with mean, μ = _______ and m = _______. The standard deviation, σ = ________. Draw the appropriate exponential graph. You should label the x– and y–axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than $0.40 in his or her pocket or purse. (Shade P(x < |
0.40)). 344 Chapter 5 | Continuous Random Variables Example 5.9 On the average, a certain computer part lasts 10 years. The length of time the computer part lasts is exponentially distributed. a. What is the probability that a computer part lasts more than seven years? Solution 5.9 a. Let x = the amount of time (in years) a computer part lasts. Find P(x > 7). Draw the graph. μ = 10, so m = 1 μ = 1 10 = 0.1 P(x > 7) = 1 – P(x < 7). Since P(X < x) = 1 – e–mx then P(X > x) = 1 – (1 – e–mx) = e-mx P(x > 7) = e(–0.1)(7) = 0.4966. The probability that a computer part lasts more than seven years is 0.4966. On the home screen, enter e^(-.1*7). Figure 5.25 b. On the average, how long would five computer parts last if they are used one after another? Solution 5.9 b. On the average, one computer part lasts 10 years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years. c. Eighty percent of computer parts last at most how long? Solution 5.9 c. Find the 80th percentile. Draw the graph. Let k = the 80th percentile. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 345 Figure 5.26 Solve for k: Eighty percent of the computer parts last at most 16.1 years. k = ln(1 – 0.80) – 0.1 = 16.1years. On the home screen, enter ln(1 – 0.80) – 0.1. d. What is the probability that a computer part lasts between nine and 11 years? Solution 5.9 d. Find P(9 < x < 11). Draw the graph. Figure 5.27 P(9 < x < 11) = P(x < 11) – P(x < 9) = (1 – e(–0.1)(11)) – (1 – e(–0.1)(9)) = 0.6671 – |
0.5934 = 0.0737. The probability that a computer part lasts between nine and 11 years is 0.0737. 346 Chapter 5 | Continuous Random Variables On the home screen, enter e^(–0.1*9) – e^(–0.1*11). 5.9 On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day? Example 5.10 Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter 1 12. If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let X = the length of a phone call, in minutes. What is m, μ, and σ? The probability that you must wait more than five minutes is _______. Solution 5.10 • m = 1 12 • μ = 12 • σ = 12 P(x > 5) = 0.6592 5.10 Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter 1. Let X = the distance people are willing to commute in miles. What is m, μ, and σ? What 20 is the probability that a person is willing to commute more than 25 miles? Example 5.11 The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed. a. On average, how many minutes elapse between two successive arrivals? b. When the store first opens, how long on average does it take for three customers to arrive? c. After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 347 d. After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive. e. Seventy percent of the customers arrive within how many minutes of |
the previous customer? f. Is an exponential distribution reasonable for this situation? Solution 5.11 a. Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average. b. Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive. c. Let X = the time between arrivals, in minutes. By Part a, μ = 2, so m = 1 2 = 0.5. Therefore, X ~ Exp(0.5). The cumulative distribution function is P(X < x) = 1 – e(–0.5)(x). Therefore P(X < 1) = 1 – e(–0.5)(1) ≈ 0.3935. 1 - e^(–0.5) ≈ 0.3935 Figure 5.28 d. P(X > 5) = 1 – P(X < 5) = 1 – (1 – e(−0.5)(5)) = e–2.5 ≈ 0.0821. Figure 5.29 348 Chapter 5 | Continuous Random Variables 1 – (1 – e ^ (– 0.50)(5)) or e ^ (– 0.50)(5) e. We want to solve 0.70 = P(X < x) for x. Substituting in the cumulative distribution function gives 0.70 = 1 – e–0.5x, so that e−0.5x = 0.30. Converting this to logarithmic form gives –0.5x = ln(0.30), or x = ln(0.30) – 0.5 ≈ 2.41 minutes. Thus, 70 percent of customers arrive within 2.41 minutes of the previous customer. You are finding the 70th percentile k so you can use the formula k = ln(1 – Area _ To _ The _ Le f t _ O f _ k) ( – m) k = ln(1 – 0.70) ( – 0.5) ≈ 2.41 minutes Figure 5.30 f. This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are bus |
ier than others. 5.11 Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution. a. On average, how many seconds elapse between two successive cars? b. After a car passes by, how long on average will it take for another seven cars to pass by? c. Find the probability that after a car passes by, the next car will pass within the next 20 seconds. d. Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds. Memorylessness of the Exponential Distribution In Example 5.7 recall that the amount of time between customers is exponentially distributed with a mean of two minutes (X ~ Exp(0.5)). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 349 distribution, this is not the case—the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property. Specifically, the memoryless property says the following P (X > r + t | X > r) = P (X > t) for all r ≥ 0 and t ≥ 0 For example, if five minutes have elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using r = 5 and t = 1 in the foregoing equation. P (X > 5 + 1 | X > 5) = P (X > 1) = e ( – 0.5)(1) ≈ 0.6065. This is the same probability as that of waiting more than one minute for a customer to arrive after the previous arrival. The exponential distribution is often used to model the longevity of an electrical or a mechanical device. In Example 5.9, the lifetime of a certain computer part has the exponential distribution with a mean of ten years (X ~ Exp(0.1)). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any |
more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is P(X > 17|X > 10) = P(X > 7) = 0.4966. Example 5.12 Refer to Example 5.7 where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk? The decay parameter of X is m = 1 4 = 0.25, so X ~ Exp(0.25). The cumulative distribution function is P(X < x) = 1 – e–0.25x. We want to find P(X > 7|X > 4). The memoryless property says that P(X > 7|X > 4) = P (X > 3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk. This is P(X > 3) = 1 – P (X < 3) = 1 – (1 – e–0.25⋅3) = e–0.75 ≈ 0.4724. Figure 5.31 1–(1–e^(–0.25*3)) = e^(–0.25*3). 350 Chapter 5 | Continuous Random Variables 5.12 Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of more than 19 years. Relationship Between the Poisson and the Exponential Distribution There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of μ units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean λ = 1/μ. Recall from the chapter on Discrete Random Variables that if X has the Poisson distribution with mean λ, then P(X = k) = λk e−λ |
k! time between events follows the exponential distribution. (k! = k*(k–1*)(k–2)*(k–3)…3*2*1). Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of Suppose X has the Poisson distribution with mean λ. Compute P(X = k) by entering 2nd, VARS(DISTR), C: poissonpdf(λ, k). To compute P(X ≤ k), enter 2nd, VARS (DISTR), D:poissoncdf(λ, k). Example 5.13 At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution. a. Find the average time between two successive calls. b. Find the probability that after a call is received, the next call occurs in less than 10 seconds. c. Find the probability that exactly five calls occur within a minute. d. Find the probability that fewer than five calls occur within a minute. e. Find the probability that more than 40 calls occur in an eight-minute period. Solution 5.13 a. On average four calls occur per minute, so 15 seconds, or 15 60 calls on average. b. Let T = time elapsed between calls. From Part a, μ = 0.25, so m = = 0.25 minutes occur between successive 1 0.25 = 4. Thus, T ~ Exp(4). The cumulative distribution function is P(T < t) = 1 – e–4t. The probability that the next call occurs in less than 10 seconds (10 seconds = 1/6 minute) is ⎞ ⎛ ⎝ ⎠ ⎞ ⎠ ⎛ ⎝.4866. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | |
Continuous Random Variables 351 Figure 5.32 c. Let X = the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute. Therefore, X ~ Poisson(4), and so P(X = 5) = 45 e−4 5! poissonpdf(4, 5) = 0.1563 ≈ 0.1563. (5! = (5)(4)(3)(2)(1)) d. Keep in mind that X must be a whole number, so P(X < 5) = P(X ≤ 4). To compute this, we could take P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4). Using technology, we see that P(X ≤ 4) = 0.6288. poisssoncdf(4, 4) = 0.6288 e. Let Y = the number of calls that occur during an eight-minute period. Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight minute period. Hence, Y ~ Poisson(32). Therefore, P(Y > 40) = 1 – P(Y ≤ 40) = 1 – 0.9294 = 0.0706. 1 – poissoncdf(32, 40). = 0.0706 5.13 In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week. a. Calculate the probability that at most two accidents occur in any given week. b. What is the probability that there are at least two weeks between any two accidents? 352 Chapter 5 | Continuous Random Variables 5.4 | Continuous Distribution This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 353 5.1 Continuous Distribution Student Learning Outcomes • The student will compare and contrast empirical data from a random number generator with the uniform distribution. Collect the Data Use a random number generator to generate 50 values between zero and one (inclusive). List them in Table 5.3. Round the numbers to four decimal places or set the calculator MODE to four places. 1. Complete the table. __________ __________ __________ __________ |
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 5.3 2. Calculate the following: a. b. c. d. x¯ = _______ s = _______ first quartile = _______ third quartile = _______ e. median = _______ Organize the Data 1. Construct a histogram of the empirical data. Make eight bars. 354 Chapter 5 | Continuous Random Variables Figure 5.33 2. Construct a histogram of the empirical data. Make five bars. Figure 5.34 Describe the Data 1. In two to three complete sentences, describe the shape of each graph. (Keep it simple. Does the graph go straight across, does it have a V shape, does it have a hump in the middle or at either end (and so on). One way to help you determine a shape is to draw a smooth curve roughly through the top of the bars.) 2. Describe how changing the number of bars might change the shape. Theoretical Distribution 1. In words, X = _____________________________________. 2. The theoretical distribution of X is X ~ U(0,1). 3. In theory, based upon the distribution X ~ U(0,1), complete the following. a. μ = ______ b. σ = ______ c. d. first quartile = ______ third quartile = ______ e. median = __________ 4. Are the empirical values (the data) in the section titled Collect the Data close to the corresponding theoretical values? Why or why not? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 355 Plot the Data 1. Construct a box plot of the data. Be sure to use a ruler to scale accurately and draw straight edges. 2. Do you notice any potential outliers? If |
so, which values are they? Either way, justify your answer numerically. (Recall that any data that are less than Q1 – 1.5(IQR) or more than Q3 + 1.5(IQR) are potential outliers. IQR means interquartile range.) Compare the Data 1. For each of the following parts, use a complete sentence to comment on how the value obtained from the data compares to the theoretical value you expected from the distribution in the section titled Theoretical Distribution: a. minimum value: _______ b. first quartile: _______ c. median: _______ d. third quartile: _______ e. maximum value: _______ f. width of IQR: _______ g. overall shape: _______ 2. Based on your comments in the section titled Collect the Data, how does the box plot fit or not fit what you would expect of the distribution in the section titled Theoretical Distribution? Discussion Question 1. Suppose that the number of values generated was 500, not 50. How would that affect what you would expect the empirical data to be and the shape of its graph to look like? 356 Chapter 5 | Continuous Random Variables KEY TERMS conditional probability the likelihood that an event will occur given that another event has already occurred decay parameter The decay parameter describes the rate at which probabilities decay to zero for increasing values of x. It is the value m in the probability density function f(x) = me(–mx) of an exponential random variable. It is also equal to m = 1 μ, where μ is the mean of the random variable. exponential distribution a continuous random variable (RV) that appears when we are interested in the intervals of time between some random events, for example, the length of time between emergency arrivals at a hospital; the notation is X ~ Exp(m). The mean is μ = 1 m. The probability density function is f(x) = me−mx, x ≥ 0 and and the standard deviation is σ = 1 m the cumulative distribution function is P(X ≤ x) = 1 − e−mx. memoryless property for an exponential random variable X, the statement that knowledge of what has occurred in the past has no effect on future probabilities This means that the probability that X exceeds x + k, given that it has exceeded x, is the same as the probability that X would exceed k if we had no knowledge about it. In symbols we say that P(X > x |
+ k|X > x) = P(X > k). Poisson distribution a distribution function that gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event; if there is a known average of λ events occurring per unit time, and these events are independent of each other, then the number of events X occurring in one unit of time has the Poisson distribution. The probability of k events occurring in one unit time is equal to P(X = k) = λk e−λ k!. uniform distribution a continuous random variable (RV) that has equally likely outcomes over the domain, a < x < b. Notation—X ~ U(a,b). The mean is μ = a + b 2 and the standard deviation is σ = (b − a)2 12. The probability density function is f(x) = 1 b − a for a < x < b or a ≤ x ≤ b. The cumulative distribution is P(X ≤ x) = x − a b − a. CHAPTER REVIEW 5.1 Continuous Probability Functions The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to P(a < x < b). The cumulative distribution function (cdf) gives the probability as an area. If X is a continuous random variable, the probability density function (pdf), f(x), is used to draw the graph of the probability distribution. The total area under the graph of f(x) is one. The area under the graph of f(x) and between values a and b gives the probability P(a < x < b). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 357 Figure 5.35 The cumulative distribution function (cdf) of X is defined by P (X ≤ x). It is a function of x that gives the probability that the random variable is less than or equal to x. 5.2 The Uniform Distribution If X has a uniform distribution where a < x < b or a ≤ x ≤ b, then X takes on values between a and b (may |
include a and b). All values x are equally likely. We write X ∼ U(a, b). The mean of X is μ = a + b. The standard deviation of X is 2 σ = (b − a)2 12 of X is P(X ≤ x is continuous.. The probability density function of X is f (x) = 1 for a ≤ x ≤ b. The cumulative distribution function b − a Figure 5.36 The probability P(c < X < d) may be found by computing the area under f(x), between c and d. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height. 5.3 The Exponential Distribution (Optional) If X has an exponential distribution with mean μ, then the decay parameter is m = 1 μ, and we write X ~ Exp(m) where x ≥ 0 and m > 0. The probability density function of X is f(x) = me-mx (or equivalently f (x) = 1 μe distribution function of X is P(X ≤ x) = 1 – e–mx. − x / μ. The cumulative The exponential distribution has the memoryless property, which says that future probabilities do not depend on any past information. Mathematically, it says that P(X > x + k|X > x) = P(X > k). If T represents the waiting time between events, and if T ~ Exp(λ), then the number of events X per unit time follows the Poisson distribution with mean λ. The probability density function of X is P(X = k) = λk e−k k!. This may be computed 358 Chapter 5 | Continuous Random Variables using a TI-83, 83+, 84, 84+ calculator with the command poissonpdf(λ, k). The cumulative distribution function P(X ≤ k) may be computed using the TI-83, 83+,84, 84+ calculator with the command poissoncdf(λ, k). FORMULA REVIEW 5.1 Continuous Probability Functions Probability density function (pdf) f(x): • f(x) ≥ 0 • The total area under the curve f(x) is one. Cumulative distribution function (cdf): P(X ≤ x) 5.2 The Uniform Distribution • cdf: P(X ≤ x) = x − a b − a • mean µ = a + b 2 |
• standard deviation σ = (b − a)2 12 • P(c < X < d) = (d – c real number between a and b (in some instances, X can take on the values a and b). a = smallest X, b = largest X 5.3 The Exponential Distribution (Optional) Exponential: X ~ Exp(m) where m = the decay parameter X ~ U(a, b) The mean is μ = a + b 2. The standard deviation is σ = (b – a)2 12. • pdf: f(x) = me(–mx) where x ≥ 0 and m > 0 • cdf: P(X ≤ x) = 1 – e(–mx) • mean µ = 1 m • standard deviation σ = µ Probability density function: f (x) = 1 b − a for • percentile k: k = ln(1 − AreaToTheLe f tO f k) ( − m) • Additionally ◦ P(X > x) = e(–mx) ◦ P(a < X < b) = e(–ma) – e(–mb) • Memoryless property: P(X > x + k|X > x) = P (X > k) • Poisson probability: P(X = k) = λk e−k k! with mean λ • k! = k*(k−1)*(k−2)*(k−3)*…3*2*1 a ≤ X ≤ b Area to the left of x: P(X < x) = (x – a) ⎛ ⎝ 1 b − a ⎞ ⎠ Area to the right of x: P(X > x) = (b – x) ⎛ ⎝ 1 b − a ⎞ ⎠ Area between c and d: P(c < x < d) = (base)(height) = (d – c) ⎛ ⎝ ⎞ ⎠ 1 b − a Uniform: X ~ U(a, b) where a < x < b • pdf: f (x) = 1 b − a for a ≤ x ≤ b PRACTICE This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 359 5.1 |
Continuous Probability Functions 1. Which type of distribution does the graph illustrate? Figure 5.37 2. Which type of distribution does the graph illustrate? Figure 5.38 3. Which type of distribution does the graph illustrate? Figure 5.39 360 Chapter 5 | Continuous Random Variables 4. What does the shaded area represent? P(___< x < ___) Figure 5.40 5. What does the shaded area represent? P(___< x < ___) Figure 5.41 6. For a continuous probablity distribution, 0 ≤ x ≤ 15. What is P(x > 15)? 7. What is the area under f(x) if the function is a continuous probability density function? 8. For a continuous probability distribution, 0 ≤ x ≤ 10. What is P(x = 7)? 9. A continuous probability function is restricted to the portion between x = 0 and 7. What is P(x = 10)? 10. f(x) for a continuous probability function is 1 5, and the function is restricted to 0 ≤ x ≤ 5. What is P(x < 0)? 11. f(x), a continuous probability function, is equal to 1 12 12)?, and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 361 12. Find the probability that x falls in the shaded area. Figure 5.42 13. Find the probability that x falls in the shaded area. Figure 5.43 14. Find the probability that x falls in the shaded area. Figure 5.44 15. f(x), a continuous probability function, is equal to 1 3 ⎛ ⎝x > 3 and the function is restricted to 1 ≤ x ≤ 4. Describe P 2 ⎞ ⎠. 5.2 The Uniform Distribution Use the following information to answer the next 10 questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes: 362 Chapter 5 | Continuous Random Variables 1.5 2.4 3.6 2.6 1.6 2.4 2.0 3.5 2.5 1.8 2.4 2.5 3.5 4.0 2.6 1.6 2.2 1.8 3.8 2.5 1 |
.5 2.8 1.8 4.5 1.9 1.9 3.1 1.6 Table 5.4 The sample mean = 2.50 and the sample standard deviation = 0.8302. The distribution can be written as X ~ U(1.5, 4.5). 16. What type of distribution is this? 17. In this distribution, outcomes are equally likely. What does this mean? 18. What is the height of f(x) for the continuous probability distribution? 19. What are the constraints for the values of x? 20. Graph P(2 < x < 3). 21. What is P(2 < x < 3)? 22. What is P(x < 3.5| x < 4)? 23. What is P(x = 1.5)? 24. What is the 90th percentile of square footage for homes? 25. Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet. Use the following information to answer the next eight exercises. A distribution is given as X ~ U(0, 12). 26. What is a? What does it represent? 27. What is b? What does it represent? 28. What is the probability density function? 29. What is the theoretical mean? 30. What is the theoretical standard deviation? 31. Draw the graph of the distribution for P(x > 9). 32. Find P(x > 9). 33. Find the 40th percentile. Use the following information to answer the next 12 exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years. 34. What is being measured here? 35. In words, define the random variable X. 36. Are the data discrete or continuous? 37. The interval of values for x is ______. 38. The distribution for X is ______. 39. Write the probability density function. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 363 40. Graph the probability distribution. a. Sketch the graph of the probability distribution. Figure 5.45 Identify the following values: b. i. Lowest value for x¯ : _______ ii. Highest value for x¯ : _______ iii. Height of the rectangle: |
_______ iv. Label for x-axis (words): _______ v. Label for y-axis (words): _______ 41. Find the average age of the cars in the lot. 42. Find the probability that a randomly chosen car in the lot was less than four years old. a. Sketch the graph, and shade the area of interest. Figure 5.46 b. Find the probability. P(x < 4) = _______ 364 Chapter 5 | Continuous Random Variables 43. Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old. a. Sketch the graph, shade the area of interest. Figure 5.47 b. Find the probability. P(x < 4|x < 7.5) = _______ 44. What has changed in the previous two problems that made the solutions different? 45. Find the third quartile of ages of cars in the lot. This means you will have to find the value such that 3 4, or 75 percent, of the cars are at most (less than or equal to) that age. a. Sketch the graph, and shade the area of interest. Figure 5.48 b. Find the value k such that P(x < k) = 0.75. c. The third quartile is _______ 5.3 The Exponential Distribution (Optional) Use the following information to answer the next 10 exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2) 46. What type of distribution is this? 47. Are outcomes equally likely in this distribution? Why or why not? 48. What is m? What does it represent? 49. What is the mean? 50. What is the standard deviation? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 365 51. State the probability density function. 52. Graph the distribution. 53. Find P(2 < x < 10). 54. Find P(x > 6). 55. Find the 70th percentile. Use the following information to answer the next eight exercises. A distribution is given as X ~ Exp(0.75). 56. What is m? 57. What is the probability density function? 58. What is the cumulative |
distribution function? 59. Draw the distribution. 60. Find P(x < 4). 61. Find the 30th percentile. 62. Find the median. 63. Which is larger, the mean or the median? Use the following information to answer the next eight exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14. 64. What is being measured here? 65. Are the data discrete or continuous? 66. In words, define the random variable X. 67. What is the decay rate (m)? 68. The distribution for X is ______. 69. Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. The question means that you need to find P(x < 5,730). a. Sketch the graph, and shade the area of interest. Figure 5.49 b. Find the probability. P(x < 5,730) = __________ 366 Chapter 5 | Continuous Random Variables 70. Find the percentage of carbon-14 lasting longer than 10,000 years. a. Sketch the graph, and shade the area of interest. Figure 5.50 b. Find the probability. P(x > 10,000) = ________ 71. Thirty percent of carbon-14 will decay within how many years? a. Sketch the graph, and shade the area of interest. Figure 5.51 b. Find the value k such that P(x < k) = 0.30. HOMEWORK 5.1 Continuous Probability Functions For each probability and percentile problem, draw the picture. 72. Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine percentage of nurses in America with an R.N. (registered nurse) degree. You ask nurses if they have an R.N. degree. The nurses answer yes or no. You then calculate the percentage of nurses with an R.N. degree. You give that percentage to your supervisor. a. What part of the experiment will yield discrete data? b. What part of the experiment will yield continuous data? 73. When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why? 5.2 The Uniform Distribution For each |
probability and percentile problem, draw the picture. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 367 74. Births are approximately uniformly distributed between the 52 weeks of the year. They can be said to follow a uniform distribution from one to 53 (spread of 52 weeks). f(x) = _________ a. X ~ _________ b. Graph the probability distribution. c. d. μ = _________ e. σ = _________ f. Find the probability that a person is born at the exact moment week 19 starts. That is, find P(x = 19) = _________ g. P(2 < x < 31) = _________ h. Find the probability that a person is born after week 40. i. P(12 < x|x < 28) = _________ j. Find the 70th percentile. k. Find the minimum for the upper quarter. 75. A random number generator picks a number from one to nine in a uniform manner. f(x) = _________ a. X ~ _________ b. Graph the probability distribution. c. d. μ = _________ e. σ = _________ f. P(3.5 < x < 7.25) = _________ g. P(x > 5.67) h. P(x > 5|x > 3) = _________ i. Find the 90th percentile. 76. According to a study by Dr. John McDougall of his live-in weight loss program, the people who follow his program lose between six and 15 pounds a month until they approach trim body weight. Let’s suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month. f(x) = _________ a. Define the random variable. X = _________ b. X ~ _________ c. Graph the probability distribution. d. e. μ = _________ f. σ = _________ g. Find the probability that the individual lost more than 10 pounds in a month. h. Suppose it is known that the individual lost more than 10 pounds in a month. Find the probability that he lost less than 12 pounds in the month. i. P(7 < x < 13|x > 9) = |
__________. State this result in a probability question, similarly to Parts g and h, draw the picture, and find the probability. 77. A subway train arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution. f(x) = _______ a. Define the random variable. X = _______ b. X ~ _______ c. Graph the probability distribution. d. e. μ = _______ f. σ = _______ g. Find the probability that the commuter waits less than one minute. h. Find the probability that the commuter waits between three and four minutes. i. Sixty percent of commuters wait more than how long for the train? State this result in a probability question, similarly to Parts g and h, draw the picture, and find the probability. 368 Chapter 5 | Continuous Random Variables 78. The age of a first grader on September 1 at Garden Elementary School is uniformly distributed from 5.8 to 6.8 years. We randomly select one first grader from the class. f(x) = _________ a. Define the random variable. X = _________ b. X ~ _________ c. Graph the probability distribution. d. e. μ = _________ f. σ = _________ g. Find the probability that she is over 6.5 years old. h. Find the probability that she is between four and six years old. i. Find the 70th percentile for the age of first graders on September 1 at Garden Elementary School. Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution. 79. What is the average waiting time (in minutes)? a. zero two b. three c. four d. 80. Find the 30th percentile for the waiting times (in minutes). a. two b. 2.4 c. 2.75 three d. 81. The probability of waiting more than seven minutes given a person has waited more than four minutes is? a. 0.125 b. 0.25 c. 0.5 d. 0.75 82. The time (in minutes) until the next bus departs a major bus depot follows a distribution with f(x) = 1 20 where x |
goes from 25 to 45 minutes. a. Define the random variable. X = ________ b. X ~ ________ c. Graph the probability distribution. d. The distribution is ______________ (name of distribution). It is _____________ (discrete or continuous). e. μ = ________ f. σ = ________ g. Find the probability that the time is at most 30 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. h. Find the probability that the time is between 30 and 40 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. i. P(25 < x < 55) = _________. State this result in a probability statement, similarly to Parts g and h, draw the picture, and find the probability. j. Find the 90th percentile. This means that 90 percent of the time, the time is less than _____ minutes. k. Find the 75th percentile. In a complete sentence, state what this means. (See Part j.) l. Find the probability that the time is more than 40 minutes given (or knowing that) it is at least 30 minutes. 83. Suppose that the value of a stock varies each day from $16 to $25 with a uniform distribution. a. Find the probability that the value of the stock is more than $19. b. Find the probability that the value of the stock between $19 and $22. c. Find the upper quartile — 25 percent of all days the stock is above what value? Draw the graph. d. Given that the stock is greater than $18, find the probability that the stock is more than $21. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 369 84. A fireworks show is designed so that the time between fireworks is between one and five seconds, and follows a uniform distribution. a. Find the average time between fireworks. b. Find the probability that the time between fireworks is greater than four seconds. 85. The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution. a. Find the probability that the truck driver goes more than 650 miles in a day. b. Find the probability that the truck driver goes between 400 and 650 miles in a day. c. |
At least how many miles does the truck driver travel on the 10 percent of days with the highest mileage? 5.3 The Exponential Distribution (Optional) 86. Suppose that the length of long-distance phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes. Is X continuous or discrete? a. Define the random variable. X = ________________. b. c. X ~ ________ d. μ = ________ e. σ = ________ f. Draw a graph of the probability distribution. Label the axes. g. Find the probability that a phone call lasts less than nine minutes. h. Find the probability that a phone call lasts more than nine minutes. i. Find the probability that a phone call lasts between seven and nine minutes. j. If 25 phone calls are made one after another, on average, what would you expect the total to be? Why? 87. Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.025. We are interested in the life of the battery. Is X continuous or discrete? a. Define the random variable. X = _________________________________. b. c. X ~ ________ d. On average, how long would you expect one car battery to last? e. On average, how long would you expect nine car batteries to last, if they are used one after another? f. Find the probability that a car battery lasts more than 36 months. g. Seventy percent of the batteries last at least how long? 88. The percent of persons (ages five and older) in each state who speak a language at home other than English is approximately exponentially distributed with a mean of 9.848. Suppose we randomly pick a state. Is X continuous or discrete? a. Define the random variable. X = _________________________________. b. c. X ~ ________ d. μ = ________ e. σ = ________ f. Draw a graph of the probability distribution. Label the axes. g. Find the probability that percentage is less than 12. h. Find the probability that percentage is between eight and 14. i. The percent of all individuals living in the United States who speak a language at home other than English is 13.8. i. Why is this number different from 9.848 percent? ii. What would make this number higher than 9.848 percent? 370 Chapter 5 | Continuous Random Variables 89 |
. The time (in years) after reaching age 60 that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Suppose we randomly pick one retired individual. We are interested in the time after age 60 to retirement. Is X continuous or discrete? a. Define the random variable. X = _________________________________. b. c. X ~ = ________ d. μ = ________ e. σ = ________ f. Draw a graph of the probability distribution. Label the axes. g. Find the probability that the person retired after age 70. h. Do more people retire before age 65 or after age 65? i. In a room of 1,000 people over age 80, how many do you expect will not have retired yet? 90. The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of $150. a. Define the random variable. X = _________________________________. b. X ~ = ________ c. μ = ________ d. σ = ________ e. Draw a graph of the probability distribution. Label the axes. f. Find the probability that a car required over $300 for maintenance during its first year. Use the following information to answer the next three exercises. The average lifetime of a certain new cell phone is three years. The manufacturer will replace any cell phone failing within two years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. 91. What is the decay rate? a. 0.3333 b. 0.5000 c. 2 d. 3 92. What is the probability that a phone will fail within two years of the date of purchase? a. 0.8647 b. 0.4866 c. 0.2212 d. 0.9997 93. What is the median lifetime of these phones (in years)? a. 0.1941 b. 1.3863 c. 2.0794 d. 5.5452 94. Let X ~ Exp(0.1). a. decay rate = ________ b. μ = ________ c. Graph the probability distribution function. d. On the graph, shade the area corresponding to P(x < 6), and find the probability. e. Sketch a new graph, shade the area corresponding to P(3 < x < 6), and find the probability. f. Sketch a new graph, shade the area corresponding to P(x < 7), and find the probability. |
g. Sketch a new graph, shade the area corresponding to the 40th percentile and find the value. h. Find the average value of x. 95. Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. a. Find the probability that a light bulb lasts less than one year. b. Find the probability that a light bulb lasts between six and 10 years. c. Seventy percent of all light bulbs last at least how long? d. A company decides to offer a warranty to give refunds to light bulbs whose lifetime is among the lowest two percent of all bulbs. To the nearest month, what should be the cutoff lifetime for the warranty to take place? If a light bulb has lasted seven years, what is the probability that it fails within the 8th year? e. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 371 96. At a 911 call center, calls come in at an average rate of one call every two minutes. Assume that the time that elapses from one call to the next has the exponential distribution. a. On average, how much time occurs between five consecutive calls? b. Find the probability that after a call is received, it takes more than three minutes for the next call to occur. c. Ninety-percent of all calls occur within how many minutes of the previous call? d. Suppose that two minutes have elapsed since the last call. Find the probability that the next call will occur within the next minute. e. Find the probability that fewer than 20 calls occur within an hour. 97. In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential. a. What is the probability that an entire season elapses with a single no-hitter? b. If an entire season elapses without any no-hitters, what is the probability that there are no no-hitters in the following season? c. What is the probability that there are more than three no-hitters in a single season? 98. During the years 1998–2012, a total of 29 earthquakes of magnitude greater than 6.5 occurred in Papua New Guinea. Assume that the time spent waiting between earthquakes |
is exponential. Assume that the current year is 2013 a. What is the probability that the next earthquake occurs within the next three months? b. Given that six months has passed without an earthquake in Papua New Guinea, what is the probability that the next three months will be free of earthquakes? c. What is the probability of zero earthquakes occurring in 2014? d. What is the probability that at least two earthquakes will occur in 2014? 99. According to the American Red Cross, about one out of nine people in the United States have type B blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of type B blood types that arrive roughly follows the Poisson distribution. a. If 100 people arrive, how many on average would be expected to have type B blood? b. What is the probability that more than 10 people out of these 100 have type B blood? c. What is the probability that more than 20 people arrive before a person with type B blood is found? 100. A website experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution. a. Find the probability that the duration between two successive visits to the website is more than 10 minutes. b. The top 25 percent of durations between visits are at least how long? c. Suppose that 20 minutes have passed since the last visit to the website. What is the probability that the next visit will occur within the next five minutes? d. Find the probability that fewer than seven visits occur within a one-hour period. 101. At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed. a. Find the probability that the time between two successive visits to the urgent care facility is less than two minutes. b. Find the probability that the time between two successive visits to the urgent care facility is more than 15 minutes. If 10 minutes have passed since the last arrival, what is the probability that the next person will arrive within the c. next five minutes? d. Find the probability that more than eight patients arrive during a half-hour period. REFERENCES 5.2 The Uniform Distribution McDougall, J. A. (1995). The McDougall program for maximum weight loss. New York: Plume 5.3 The Exponential Distribution (Optional) Baseball-Reference.com. (2013). No-hitter. |
Retrieved from http://www.baseball-reference.com/bullpen/No-hitter U.S. Census Bureau. (n.d.). Retrieved from https://www.census.gov/ 372 Chapter 5 | Continuous Random Variables World Earthquakes. (2013). Earthquake data for Papua New Guinea. Retrieved from http://www.world-earthquakes.com/ Zhou, Rick. (2013). Exponential distribution lecture slides. Retrieved from www.public.iastate.edu/~riczw/stat330s11/ lecture/lec13.pdf SOLUTIONS 1 Uniform distribution 3 Normal distribution 5 P(6 < x < 7) 7 one 9 zero 11 one 13 0.625 15 The probability is equal to the area from x = 3 2 to x = 4 above the x-axis and up to f(x) = 1 3. 17 It means that the value of x is just as likely to be any number between 1.5 and 4.5. 19 1.5 ≤ x ≤ 4.5 21 0.3333 23 zero 25 0.6 27 b is 12, and it represents the highest value of x. 29 six 31 Figure 5.52 33 4.8 35 X = The age (in years) of cars in the staff parking lot 37 0.5 to 9.5 39 f(x) = 1 9 41 μ = 5 where x is between 0.5 and 9.5, inclusive. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 373 43 a. Check student’s solution. b. 3.5 7 45 a. Check student's solution b. k = 7.25 c. 7.25 47 No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time. 49 five 51 f(x) = 0.2e–0.2x 53 0.5350 55 6.02 57 f(x) = 0.75e–0.75x 59 Figure 5.53 61 0.4756 63 The mean is larger. The mean is 1 m = 1 0.75 ≈ 1.33, which is greater than 0.9242. 65 continuous 67 m = 0.000121 69 a. |
Check student's solution b. P(x < 5,730) = 0.5001 71 a. Check student's solution b. k = 2947.73 73 Age is a measurement, regardless of the accuracy used. 75 a. X ~ U(1, 9) 374 Chapter 5 | Continuous Random Variables b. Check student’s solution c. f (x) = 1 8 where 1 ≤ x ≤ 9 d. five e. 2.3 f. g. h. 15 32 333 800 2 3 i. 8.2 77 a. X represents the length of time a commuter must wait for a train to arrive on the Red Line. b. X ~ U(0, 8) c. f (x) = 1 8 where ≤ x ≤ 8 d. four e. 2.31 f. g. 1 8 1 8 h. 3.2 79 d 81 b 83 a. The probability density function of X is 1 25 − 16 = 1 9. P(X > 19) = (25 – 19 Figure 5.54 b. P(19 < X < 22) = (22 – 19 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables 375 Figure 5.55 c. The area must be 0.25, and 0.25 = (width) ⎛ ⎝ ⎞ ⎠ 1 9, so width = (0.25)(9) = 2.25. Thus, the value is 25 – 2.25 = 22.75. d. This is a conditional probability question. P(x > 21| x > 18). You can do this two ways: ◦ Draw the graph where a is now 18 and b is still 25. The height is 1 (25 − 18) = 1 7 So, P(x > 21|x > 18) = (25 – 21) ⎛ ⎝ = 4/7. ⎞ ⎠ 1 7 ◦ Use the formula: P(x > 21|x > 18) = P(x > 21 AND x > 18) P(x > 18) = P(x > 21) P(x > 18) = (25 − 21) (25 − 18). = 4 7 85 a. P(X > 650) = 700 − 650 700 − 300 = 50 400 = 1 8 = 0. |
125. b. P(400 < X < 650) = 650 − 400 700 − 300 = 250 400 = 0.625 c. 0.10 = width 700 − 300 farthest 10 percent of days., so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the 87 a. X = the useful life of a particular car battery, measured in months. b. X is continuous. c. X ~ Exp(0.025) d. 40 months e. 360 months f. 0.4066 g. 14.27 89 a. X = the time (in years) after reaching age 60 that it takes an individual to retire b. X is continuous. c. X ~ Exp ⎛ ⎝ ⎞ ⎠ 1 5 d. five Chapter 5 | Continuous Random Variables 376 e. five f. Check student’s solution. g. 0.1353 h. before i. 18.3 91 a 93 c 95 Let T = the life time of a light bulb. The decay parameter is m = 1/8, and T ~ Exp(1/8). The cumulative distribution function is P(T < t) = 1 − e a. Therefore, P(T < 1.1175. b. We want to find P(6 < t < 10). To do this, P(6 < t < 10) – P(t < 6) * 10⎞ – 1 8 – 1 8 * 6⎞ = = ⎛ ⎜1 – e ⎝ ⎛ ⎜.7135 – 0.5276 = 0.1859 ⎠ Figure 5.56 c. We want to find 0.70 = P(T > t) = 1 – ⎛ ⎜. ⎞ ⎟ = e ⎠ Solving for t, e – t 8 = 0.70, so – t 8 = ln(0.70), and t = –8ln(0.70) ≈ 2.85 years Or use t = ln(area_to_the_right) ( – m) = ln(0.70) – 1 8 ≈ 2.85 years. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 5 | Continuous Random Variables |
377 Figure 5.57 d. We want to find 0.02 = P(T < t) = 1 – e – t 8. Solving for t, e – t 8 = 0.98, so – t 8 = ln(0.98), and t = –8ln(0.98) ≈ 0.1616 years, or roughly two months. The warranty should cover light bulbs that last less than 2 months. Or use ln(area_to_the_right) = 0.1616. ( – m) = ln(1 – 0.2) – 1 8 e. We must find P(T < 8|T > 7). Notice that by the rule of complement events, P(T < 8|T > 7) = 1 – P(T > 8|T > 7). By the memoryless property (P(X > r + t|X > r) = P(X > t)). So P(T > 8|T > 7) = P(T > 1) = 1 – ⎛ ⎜.8825 Therefore, P(T < 8|T > 7) = 1 – 0.8825 = 0.1175. 97 Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3. Therefore, (X = 0) = 30 e – 3 = e–3 ≈ 0.0498 0! You could let T = duration of time between no-hitters. Since the time is exponential and there are three no-hitters per season, then the time between no-hitters is 1 3 season. For the exponential, µ = 1 3. Therefore, m = 1 μ = 3 and T ~ Exp(3). a. The desired probability is P(T > 1) = 1 – P(T < 1) = 1 – (1 – e–3) = e–3 ≈ 0.0498. b. Let T = duration of time between no-hitters. We find P(T > 2|T > 1), and by the memoryless property this is simply P(T > 1), which we found to be 0.0498 in part a. c. Let X = the number of no-hitters is a season. Assume that X is Po |
isson with mean λ = 3. Then P(X > 3) = 1 – P(X ≤ 3) = 0.3528. 99 a. 100 9 = 11.11 b. P(X > 10) = 1 – P(X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532. 378 Chapter 5 | Continuous Random Variables c. The number of people with Type B blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean μ = 9 and m = 1 9. The cumulative distribution function of X is P(X < x) = 1 − e − x 9. Thus hus, P(X > 20) = 1 - P(X ≤ 20) = 1 − − 20 9 ⎛ ⎜1 − e ⎝ ⎞ ⎟ ≈ 0.1084. ⎠ NOTE We could also deduce that each person arriving has a 8 9 chance of not having type B blood. So the probability that none of the first 20 people arrive have type B blood is ⎛ ⎝ 8 9 than the exponential because the number of people between type B people is discrete instead of continuous.) ≈ 0.0948. (The geometric distribution is more appropriate ⎞ ⎠ 20 101 Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = 1 7. The cdf is P(T < t) = 1 − e t 7 a. P(T < 2.2485. b. P(T > 15) = 1 − P(T < 15) = 1 − c. P(T > 15|T > 10) = P(T > 5) = 1 − − 15 7 ⎛ ⎜1 − e ⎝ ⎞ ⎟ ≈ e ⎠ − 15 7 ≈ 0.1173. − 5 7 ⎛ ⎜.4895. d. Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of 30 7, X ~ Poisson ⎛ ⎝ ⎞ ⎠ 30 7. Find P(X > 8) = 1 – P(X ≤ 8) |
≈ 0.0311. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 379 6 | THE NORMAL DISTRIBUTION Figure 6.1 If you ask enough people about their shoe size, you will find that your graphed data is shaped like a bell curve and can be described as normally distributed. (credit: Ömer Ünlϋ) Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: • Recognize the normal probability distribution and apply it appropriately • Recognize the standard normal probability distribution and apply it appropriately • Compare normal probabilities by converting to the standard normal distribution The normal, a continuous distribution, is the most important of all the distributions. It is widely used and even more widely abused. Its graph is bell-shaped. You see the bell curve in almost all disciplines, including psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are normally distributed. Often, real-estate prices fit a normal distribution. The normal distribution is extremely important, but it cannot be applied to everything in the real world. In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with 380 them. Chapter 6 | The Normal Distribution The normal distribution has two parameters: —the mean (μ) and the standard deviation (σ). If X is a quantity to be measured that has a normal distribution with mean (μ) and standard deviation (σ), we designate this by writing Figure 6.2 The curve is symmetric about a vertical line drawn through the mean, μ. In theory, the mean is the same as the median, because the graph is symmetric about μ. With a normal distribution, the mean, median, and mode all lie at the same point. The normal distribution depends only on the mean and the standard deviation. The location of the mean simply indicates the location of the line of symmetry, in a normal distribution. Since the area under the curve must equal one, a change in the standard deviation, σ, causes a change in the shape of the curve; the curve becomes fatter or skinnier depending on σ. A change in μ causes the graph to shift to the left or right. The location of the mean simply indicates the location of the line of symmetry, in a normal |
distribution. This means there are an infinite number of normal probability distributions. One distribution of special interest is called the standard normal distribution. Your instructor will record the heights of both men and women in your class, separately. Draw histograms of your data. Then draw a smooth curve through each histogram. Is each curve somewhat bell-shaped? Do you think that if you had recorded 200 data values for men and 200 for women that the curves would look bell-shaped? Calculate the mean for each data set. Write the means on the x-axis of the appropriate graph below the peak. Shade the approximate area that represents the probability that one randomly chosen male is taller than 72 inches. Shade the approximate area that represents the probability that 1 randomly chosen female is shorter than 60 inches. If the total area under each curve is one, does either probability appear to be more than 0.5? 6.1 | The Standard Normal Distribution The standardized normal distribution is a type of normal distribution, with a mean of 0 and standard deviation of 1. It represents a distribution of standardized scores, called z-scores, as opposed to raw scores (the actual data values). A z-score indicates the number of standard deviation a score falls above or below the mean. Z-scores allow for comparison of scores, occurring in different data sets, with different means and standard deviations. It would not make sense to compare apples and oranges. Likewise, it does not make sense to compare scores from two different samples that have different means and standard deviations. Z-scores can be looked up in a Z-Table of Standard Normal Distribution, in order to find the area under the standard normal curve, between a score and the mean, between two scores, or above or below a score. The standard normal distribution allows us to interpret standardized scores and provides us with one table that we may use, in order to compute areas under the normal curve, for an infinite number of data sets, no matter what the mean or standard deviation. A z-score is calculated as z = sides of the equation by σ gives: (z)(σ) = x − μ. Adding μ to both sides of the equation gives μ + (z)(σ) = x.. The score itself can be found by using algebra and solving for x. Multiplying both x − μ σ This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 381 Suppose |
we have a data set with a mean of 5 and standard deviation of 2. We want to determine the number of standard deviations the score of 11 falls above the mean. We can find this answer (or z-score) by writing or we can solve for z. z = 11 − 5 2 = 3 5 + (z)(2) = 11, 2z = 6 z = 3 We have determined that the score of 11 falls 3 standard deviations above the mean of 5. With a standard normal distribution, we indicate the distribution by writing Z ~ N(0, 1) which shows the normal distribution has a mean of 0 and standard deviation of 1. This notation simply indicates that a standard normal distribution is being used. Z-Scores As described previously, if X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is z = x – μ σ. The z-score tells you how many standard deviations the value x is above, to the right of, or below, to the left of, the mean, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero. When determining the z-score for an x-value, for a normal distribution, with a given mean and standard deviation, the notation above for a normal distribution, will be given. Example 6.1 Suppose X ~ N(5, 6). This equation says that X is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then, z = x – μ σ = 17 – 5 6 = 2. This means that x = 17 is two standard deviations (2σ) above, or to the right, of the mean μ = 5. Notice that 5 + (2)(6) = 17. The pattern is μ + zσ = x. Now suppose x = 1. Then0.67, rounded to two decimal places. This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. This z-score shows that x = 1 is less than 1 standard deviation below the mean of 5. Therefore, the score doesn't fall very far below the mean. Summarizing, when z is positive, x is above or to |
the right of μ, and when z is negative, x is to the left of or below μ. Or, when z is positive, x is greater than μ, and when z is negative, x is less than μ. The absolute value of z indicates how far the score is from the mean, in either direction. 6.1 What is the z-score of x, when x = 1 and X ~ N(12, 3)? 382 Chapter 6 | The Normal Distribution Example 6.2 Some doctors believe that a person can lose five pounds, on average, in a month by reducing his or her fat intake and by consistently exercising. Suppose weight loss has a normal distribution. Let X = the amount of weight lost, in pounds, by a person in a month. Use a standard deviation of two pounds. X ~ N(5, 2). Fill in the blanks. a. Suppose a person lost 10 pounds in a month. The z-score when x = 10 pounds is z = 2.5 (verify). This z-score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). Solution 6.2 a. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five. b. Suppose a person gained three pounds, a negative weight loss. Then z = __________. This z-score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean. Solution 6.2 b. z = –4. This z-score tells you that x = –3 is four standard deviations to the left of the mean. c. Suppose the random variables X and Y have the following normal distributions: X ~ N(5, 6) and Y ~ N(2, 1). If x = 17, then z = 2. This was previously shown. If y = 4, what is z? Solution 6.2 c, where µ = 2 and σ = 1. The z-score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two of their own standard deviations to the right of their respective means. The z-score allows us to compare data that are scaled differently. To better understand the concept, suppose X ~ N |
(5, 6) represents weight gains for one group of people who are trying to gain weight in a six-week period and Y ~ N(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means. 6.2 Fill in the blanks. Jerome averages 16 points a game with a standard deviation of four points. X ~ N(16, 4). Suppose Jerome scores 10 points in a game. The z-score when x = 10 is –1.5. This score tells you that x = 10 is _____ standard deviations to the ______ (right or left) of the mean______ (What is the mean?). The Empirical Rule If X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule states the following: • About 68 percent of the x values lie between –1σ and +1σ of the mean µ (within one standard deviation of the mean). • About 95 percent of the x values lie between –2σ and +2σ of the mean µ (within two standard deviations of the mean). • About 99.7 percent of the x values lie between –3σ and +3σ of the mean µ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean. • The z-scores for +1σ and –1σ are +1 and –1, respectively. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 383 • The z-scores for +2σ and –2σ are +2 and –2, respectively. • The z-scores for +3σ and –3σ are +3 and –3, respectively. So, in other words, this is that about 68 percent of the values lie between z-scores of –1 and 1, about 95% of the values lie between z-scores of –2 and 2, and about 99.7 percent of the values lie between z-scores of -3 and 3. These facts can be checked, by looking up the mean to z area in a z-table for each positive z-score and |
multiplying by 2. The empirical rule is also known as the 68–95–99.7 rule. Figure 6.3 Example 6.3 The mean height of 15-to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15-to 18-year-old male from Chile in 2009–2010. Then X ~ N(170, 6.28). a. Suppose a 15-to 18-year-old male from Chile was 168 cm tall in 2009–2010. The z-score when x = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). Solution 6.3 a. –0.32, 0.32, left, 170 b. Suppose that the height of a 15-to 18-year-old male from Chile in 2009–2010 has a z-score of z = 1.27. What is the male’s height? The z-score (z = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean. Solution 6.3 b. 177.98 cm, 1.27, right 384 Chapter 6 | The Normal Distribution 6.3 Use the information in Example 6.3 to answer the following questions: a. Suppose a 15-to 18-year-old male from Chile was 176 cm tall from 2009–2010. The z-score when x = 176 cm is z = _______. This z-score tells you that x = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). b. Suppose that the height of a 15-to 18-year-old male from Chile in 2009–2010 has a z-score of z = –2. What is the male’s height? The z-score (z = –2) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean. Example 6.4 From 1984 to 1985, the mean height of 15-to 18-year-old males from Chile was 172.36 cm, and the standard deviation was |
6.34 cm. Let Y = the height of 15-to 18-year-old males from 1984–1985, and y = the height of one male from this group. Then Y ~ N(172.36, 6.34). The mean height of 15-to 18-year-old males from Chile in 2009–2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15-to 18-year-old male from Chile in 2009–2010, and x = the height of one male from this group. Then X ~ N(170, 6.28). Find the z-scores for x = 160.58 cm and y = 162.85 cm. Interpret each z-score. What can you say about x = 160.58 cm and y = 162.85 cm as they compare to their respective means and standard deviations? Solution 6.4 The z-score for x = 160.58 cm is z = –1.5. The z-score for y = 162.85 cm is z = –1.5. Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction. 6.4 In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean µ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then, X ~ N(496, 114). Find the z-scores for x1 = 325 and x2 = 366.21. Interpret each z-score. What can you say about x1 = 325 and x2 = 366.21, as they compare to their respective means and standard deviations? Example 6.5 Suppose x has a normal distribution with mean 50 and standard deviation 6. • About 68 percent of the x values lie within one standard deviation of the mean. Therefore, about 68 percent of the x values lie between –1σ = (–1)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively. • About 95 percent of the |
x values lie within two standard deviations of the mean. Therefore, about 95 percent of the x values lie between –2σ = (–2)(6) = –12 and 2σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 385 • About 99.7 percent of the x values lie within three standard deviations of the mean. Therefore, about 95 percent of the x values lie between –3σ = (–3)(6) = –18 and 3σ = (3)(6) = 18 of the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations from the mean 50. The z-scores are –3 and +3 for 32 and 68, respectively. 6.5 Suppose X has a normal distribution with mean 25 and standard deviation five. Between what values of x do 68 percent of the values lie? Example 6.6 From 1984–1985, the mean height of 15-to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15-to 18-year-old males in 1984–1985. Then Y ~ N(172.36, 6.34). a. About 68 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. b. About 95 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively. c. About 99.7 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. Solution 6.6 a. About 68 percent of the values lie between 166.02 cm and 178.7 cm. The z-scores are –1 and 1. b. About 95 percent of the values lie between 159.68 cm and 185.04 cm. The z-scores are –2 and 2. c. About 99.7 percent of the values lie between 1153.34 cm and 191.38 cm. The |
z-scores are –3 and 3. 6.6 The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ = 11 points. a. About 68 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. b. About 95 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. c. About 99.7 percent of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. 6.2 | Using the Normal Distribution The shaded area in the following graph indicates the area to the left of x. This area could represent the percentage of students scoring less than a particular grade on a final exam. This area is represented by the probability P(X < x). Normal tables, computers, and calculators are used to provide or calculate the probability P(X < x). 386 Chapter 6 | The Normal Distribution Figure 6.4 The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions. Suppose the graph above were to represent the percentage of students scoring less than 75 on a final exam, with this probability equal to 0.39. This would also indicate that the percentage of students scoring higher than 75 was equal to 1 minus 0.39 or 0.61. Calculations of Probabilities Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators. NOTE To calculate the probability, use the probability tables provided in Appendix H without the use of technology. The tables include instructions for how to use them. The probability is represented by the area under the normal curve. To find the probability, calculate the z-score and look up the z-score in the z-table under the z-column. Most z-tables show the area under the normal curve to the left of z. |
Others show the mean to z area. The method used will be indicated on the table. We will discuss the z-table that represents the area under the normal curve to the left of z. Once you have located the z-score, locate the corresponding area. This will be the area under the normal curve, to the left of the z-score. This area can be used to find the area to the right of the z-score, or by subtracting from 1 or the total area under the normal curve. These areas can also be used to determine the area between two z-scores. Example 6.7 If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772. 6.7 If the area to the left of x is 0.012, then what is the area to the right? Example 6.8 The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 387 a. Find the probability that a randomly selected student scored more than 65 on the exam. Solution 6.8 a. Let X = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5. Draw a graph. Calculate the z-score: z = x − μ σ = 65 − 63 5 = 2 5 =.40 The z-table shows that the area to the left of z is 0.6554. Subtracting this area from 1 gives 0.3446. Then, find P(x > 65). P(x > 65) = 0.3446 Figure 6.5 The probability that any student selected at random scores more than 65 is 0.3446. Go into 2nd DISTR. After pressing 2nd DISTR, press 2:normalcdf. The syntax for the instructions is as follows: normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 1099) by pressing 1, the EE key—a 2nd key—and then 99. Or, you can enter 10^99 instead. The number 1099 is way out in the right |
tail of the normal curve. We are calculating the area between 65 and 1099. In some instances, the lower number of the area might be –1E99 (= –1099). The number –1099 is way out in the left tail of the normal curve. We chose the exponent of 99 because this produces such a large number that we can reasonably expect all of the values under the curve to fall below it. This is an arbitrary value and one that works well, for our purpose. HISTORICAL NOTE The TI probability program calculates a z-score and then the probability from the z-score. Before technology, the z-score was looked up in a standard normal probability table, also known as a Z-table—the math involved to find probability is cumbersome. In this example, a standard normal table with area to the left of the z-score was used. You calculate the z-score and look up the area to the left. The probability is the area to the right. 388 Chapter 6 | The Normal Distribution Calculate the z-score *Press 2nd Distr *Press 3:invNorm( *Enter the area to the left of z followed by ) *Press ENTER. For this Example, the steps are 2nd Distr 3:invNorm(.6554) ENTER The answer is 0.3999, which rounds to 0.4. b. Find the probability that a randomly selected student scored less than 85. Solution 6.8 b. Draw a graph. Then find P(x < 85), and shade the graph. Using a computer or calculator, find P(x < 85) = 1. normalcdf(0,85,63,5) = 1 (rounds to one) The probability that one student scores less than 85 is approximately one, or 100 percent. c. Find the 90th percentile, —that is, find the score k that has 90 percent of the scores below k and 10 percent of the scores above k. Solution 6.8 c. Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile. This time, we are looking for a score that corresponds to a given area under the curve. Let k = the 90th percentile. The variable k is located on the x-axis. P(x < k) is the area to the left of k. The 90th percentile k separates the exam scores |
into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and 10 percent are the same or higher. The variable k is often called a critical value. We know the mean, standard deviation, and area under the normal curve. We need to find the z-score that corresponds to the area of 0.9 and then substitute it with the mean and standard deviation, into our z-score formula. The z-table shows a z-score of approximately 1.28, for an area under the normal curve to the left of z (larger portion) of approximately 0.9. Thus, we can write the following: Multiplying each side of the equation by 5 gives Adding 63 to both sides of the equation gives Thus, our score, k, is 69.4. 1.28 = x − 63 5 6.4 = x − 63 69.4 = x. k = 69.4 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 389 Figure 6.6 The 90th percentile is 69.4. This means that 90 percent of the test scores fall at or below 69.4 and 10 percent fall at or above. To get this answer on the calculator, follow this next step: invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation) For this problem, invNorm(0.90,63,5) = 69.4 d. Find the 70th percentile, —that is, find the score k such that 70 percent of scores are below k and 30 percent of the scores are above k. Solution 6.8 d. Find the 70th percentile. Draw a new graph and label it appropriately. k = 65.6 The 70th percentile is 65.6. This means that 70 percent of the test scores fall at or below 65.5 and 30 percent fall at or above. invNorm(0.70,63,5) = 65.6 6.8 The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a randomly selected golfer scored less than 65. Example 6.9 A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other |
things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment 390 Chapter 6 | The Normal Distribution are normally distributed and the standard deviation for the times is half an hour. a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day. Solution 6.9 a. Let X = the amount of time, in hours, a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5. Find P(1.8 < x < 2.75). First, calculate the z-scores for each x-value. z = 1.8 − 2 0.5 z = 2.75 − 2 0.5 = −0.2 0.5 = 0.75 0.5 = − 0.40 = 1.5 Now, use the Z-table to locate the area under the normal curve to the left of each of these z-scores. The area to the left of the z-score of −0.40 is 0.3446. The area to the left of the z-score of 1.5 is 0.9332. The area between these scores will be the difference in the two areas, or 0.9332 − 0.3446, which equals 0.5886. Figure 6.7 normalcdf(1.8,2.75,2,0.5) = 0.5886 The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886. b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment. Solution 6.9 b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25th percentile, k, where P(x < k) = 0.25. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 391 Figure 6.8 invNorm(0.25,2,0.5) = 1.66 We use invNorm because we are looking for the k-value. The maximum number of hours per day that the bottom quartile of households uses a personal computer |
for entertainment is 1.66 hours. 6.9 The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70. Example 6.10 In the United States smartphone users between the ages of 13 and 55+ between the ages of 13 and 55+ approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively. a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old. Solution 6.10 a. normalcdf(23,64.7,36.9,13.9) = 0.8186 The z-scores are calculated as z = 23 − 36.9 13.9 z = 64.7 − 36.9 13.9 = −13.9 13.9 = 27.8 13.9 = − 1 = 2 The Z-table shows the area to the left of a z-score with an absolute value of 1 to be 0.1587. It shows the area to the left of a z-score of 2 to be 0.9772. The difference in the two areas is 0.8185. This is slightly different than the area given by the calculator, due to rounding. b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old. 392 Chapter 6 | The Normal Distribution Solution 6.10 b. normalcdf(–1099,50.8,36.9,13.9) = 0.8413 c. Find the 80th percentile of this distribution, and interpret it in a complete sentence. Solution 6.10 c. invNorm(0.80,36.9,13.9) = 48.6 The 80th percentile is 48.6 years. 80 percent of the smartphone users in the age range 13–55+ are 48.6 years old or less. 6.10 Use the information in Example 6.10 to answer the following questions: a. Find the 30th percentile, and interpret it in a complete sentence. b. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old? Example 6.11 In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate |
mean and standard deviation of 36.9 years and 13.9 years, respectively. Using this information, answer the following questions. —Round answers to one decimal place. a. Calculate the interquartile range (IQR). Solution 6.11 a. IQR = Q3 – Q1 Calculate Q3 = 75th percentile and Q1 = 25th percentile. Recall that we can use invNorm to find the k-value. We can use this to find the quartile values. invNorm(0.75,36.9,13.9) = Q3 = 46.2754 invNorm(0.25,36.9,13.9) = Q1 = 27.5246 IQR = Q3 – Q1 = 18.8 b. Forty percent of the ages that range from 13 to 55+ are at least what age? Solution 6.11 b. Find k where P(x ≥ k) = 0.40. At least translates to greater than or equal to. 0.40 = the area to the right The area to the left = 1 – 0.40 = 0.60. The area to the left of k = 0.60 invNorm(0.60,36.9,13.9) = 40.4215 k = 40.4. Forty percent of the ages that range from 13 to 55+ are at least 40.4 years. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 393 6.11 Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean μ = 81 points and standard deviation σ = 15 points. a. Calculate the first- and third-quartile scores for this exam. b. The middle 50 percent of the exam scores are between what two values? Example 6.12 A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm. a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph. Solution 6.12 a. normalcdf(6,10^99,5.85,0.24) = 0.2660 Figure 6.9 b. The middle 20 percent |
of mandarin oranges from this farm have diameters between ______ and ______. Solution 6.12 b. 1 – 0.20 = 0.80. Outside of the middle 20 percent will be 80 percent of the values. The tails of the graph of the normal distribution each have an area of 0.40. Find k1, the 40th percentile, and k2, the 60th percentile (0.40 + 0.20 = 0.60). This leaves the middle 20 percent, in the middle of the distribution. k1 = invNorm(0.40,5.85,0.24) = 5.79 cm k2 = invNorm(0.60,5.85,0.24) = 5.91 cm So, the middle 20 percent of mandarin oranges have diameters between 5.79 cm and 5.91 cm. c. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence. 394 Chapter 6 | The Normal Distribution Solution 6.12 c. 6.16, Ninety percent of the diameter of the mandarin oranges is at most 6.16 cm. 6.12 Using the information from Example 6.12, answer the following: a. The middle 45 percent of mandarin oranges from this farm are between ______ and ______. b. Find the 16th percentile, and interpret it in a complete sentence. 6.3 | Normal Distribution—Lap Times This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 395 6.1 Normal Distribution (Lap Times) Student Learning Outcome • The student will compare and contrast empirical data and a theoretical distribution to determine if Terry Vogel's lap times fit a continuous distribution. Directions Round the relative frequencies and probabilities to four decimal places. Carry all other decimal answers to two places. Use the data from Appendix C. Use a stratified sampling method by lap— races 1 to 20— and a random number generator to pick six lap times from each stratum. Record the lap times below for laps two to seven. _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ |
_______ Table 6.1 Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 6.10 a. Calculate the following: x¯ = _______ s = _______ b. Draw a smooth curve through the tops of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. (Keep it simple. Does the graph go straight across, does it have a V-shape, does it have a hump in the middle or at either end, and so on?) Analyze the Distribution 396 Chapter 6 | The Normal Distribution Using your sample mean, sample standard deviation, and histogram to help, what is the approximate theoretical distribution of the data? • X ~ _____(_____,_____) • How does the histogram help you arrive at the approximate distribution? Describe the Data Use the data you collected to complete the following statements. • The IQR goes from __________ to __________. IQR = __________. (IQR = Q3 – Q1) • • The 15th percentile is _______. • The 85th percentile is _______. • The median is _______. • The empirical probability that a randomly chosen lap time is more than 130 seconds is _______. • Explain the meaning of the 85th percentile of this data. Theoretical Distribution Using the theoretical distribution, complete the following statements. You should use a normal approximation based on your sample data. • The IQR goes from __________ to __________. IQR = _______. • • The 15th percentile is _______. • The 85th percentile is _______. • The median is _______. • The probability that a randomly chosen lap time is more than 130 seconds is _______. • Explain the meaning of the 85th percentile of this distribution. Discussion Questions Do the data from the section titled Collect the Data give a close approximation to the theoretical distribution in the section titled Analyze the Distribution? In complete sentences and comparing the result in the sections titled Describe the Data and Theoretical Distribution, explain why or why not. 6.4 | Normal Distribution—Pinkie Length This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 397 6.2 Normal Distribution (Pinkie Length) Student Learning Outcomes • The student will compare empirical data and a theoretical distribution |
to determine if data from the experiment follow a continuous distribution. Collect the Data Measure the length of your pinkie finger, in centimeters. 1. Randomly survey 30 adults for their pinkie finger lengths. Round the lengths to the nearest 0.5 cm. _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ _______ Table 6.2 2. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 6.11 3. Calculate the following: a. b. x¯ = _______ s = _______ 4. Draw a smooth curve through the top of the bars of the histogram. Write one to two complete sentences to describe the general shape of the curve. Keep it simple. Does the graph go straight across, does it have a V-shape, does it have a hump in the middle or at either end, and so on? Analyze the Distribution Using your sample mean, sample standard deviation, and histogram, what was the approximate theoretical distribution 398 Chapter 6 | The Normal Distribution of the data you collected? • X ~ _____ (_____, _____) • How does the histogram help you arrive at the approximate distribution? Describe the Data Using the data you collected complete the following statements. Hint—Order the data. REMEMBER (IQR = Q3 – Q1) IQR = _______ • • The 15th percentile is _______. • The 85th percentile is _______. • Median is _______. • What is the theoretical probability that a randomly chosen pinkie length is more than 6.5 cm? • Explain the meaning of the 85th percentile of these data. Theoretical Distribution Using the theoretical distribution, complete the following statements. Use a normal approximation based on the sample mean and standard deviation. IQR = _______ • • The 15th percentile is _______. • The 85th percentile is _______. • Median is _______. • What is the theoretical probability that a randomly chosen pinkie length is more than 6.5 cm? • Explain the meaning of the 85th percentile of these data. Discussion Questions Do the data you collected give a close approximation to the theoretical distribution? In complete sentences and comparing the results in the sections titled Desc |
ribe the Data and Theoretical Distribution, explain why or why not. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 399 KEY TERMS normal distribution a continuous random variable (RV) where μ is the mean of the distribution and σ is the standard deviation; notation: X ~ N(μ, σ). If μ = 0 and σ = 1, the RV is called the standard normal distribution. standard normal distribution a continuous random variable (RV) X ~ N(0, 1); when X follows the standard normal distribution, it is often noted as Z ~ N(0, 1). z-score the linear transformation of the form z = x – μ σ ; if this transformation is applied to any normal distribution X ~ N(μ, σ), the result is the standard normal distribution Z ~ N(0, 1); If this transformation is applied to any specific value x of the RV with mean μ and standard deviation σ, the result is called the z-score of x. The z-score allows us to compare data that are normally distributed but scaled differently. CHAPTER REVIEW 6.1 The Standard Normal Distribution A z-score is a standardized value. Its distribution is the standard normal, Z ~ N(0, 1). The mean of the z-scores is zero and the standard deviation is one. If z is the z-score for a value x from the normal distribution N(µ, σ), then z tells you how many standard deviations x is above—greater than—or below—less than—µ. 6.2 Using the Normal Distribution The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bellshaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean µ and the standard deviation σ. A special normal distribution, called the standard normal distribution, is the distribution of z-scores. Its mean is zero, and its standard deviation is one. FORMULA REVIEW 6.0 Introduction X ∼ N(μ, σ) μ = the mean, σ = the standard deviation Z = the random variable for z-scores 6.2 Using the Normal Distribution Normal Distribution: X ~ N(µ, � |
�), where µ is the mean and σ is the standard deviation 6.1 The Standard Normal Distribution Standard Normal Distribution: Z ~ N(0, 1). Calculator function for probability: normalcdf (lower x value of the area, upper x value of the area, mean, standard deviation) Calculator function for the kth percentile: k = invNorm (area to the left of k, mean, standard deviation) Z ~ N(0, 1) z = a standardized value (z-score) mean = 0, standard deviation = 1 To find the kth percentile of X when the z-score is known, k = μ + (z)σ z-score: z = x – μ σ PRACTICE 6.1 The Standard Normal Distribution 1. A bottle of water contains 12.05 fluid ounces with a standard deviation of 0.01 ounces. Define the random variable X in words. X = ____________. 2. A normal distribution has a mean of 61 and a standard deviation of 15. What is the median? 400 3. X ~ N(1, 2) σ = _______ Chapter 6 | The Normal Distribution 4. A company manufactures rubber balls. The mean diameter of a ball is 12 cm with a standard deviation of 0.2 cm. Define the random variable X in words. X = ______________. 5. X ~ N(–4, 1) What is the median? 6. X ~ N(3, 5) σ = _______ 7. X ~ N(–2, 1) μ = _______ 8. What does a z-score measure? 9. What does standardizing a normal distribution do to the mean? 10. Is X ~ N(0, 1) a standardized normal distribution? Why or why not? 11. What is the z-score of x = 12, if it is two standard deviations to the right of the mean? 12. What is the z-score of x = 9, if it is 1.5 standard deviations to the left of the mean? 13. What is the z-score of x = –2, if it is 2.78 standard deviations to the right of the mean? 14. What is the z-score of x = 7, if it is 0.133 standard deviations to the left of the mean? 15. Suppose X ~ N(2, 6). What value of x has a z-score of three? 16. Suppose X ~ N |
(8, 1). What value of x has a z-score of –2.25? 17. Suppose X ~ N(9, 5). What value of x has a z-score of –0.5? 18. Suppose X ~ N(2, 3). What value of x has a z-score of –0.67? 19. Suppose X ~ N(4, 2). What value of x is 1.5 standard deviations to the left of the mean? 20. Suppose X ~ N(4, 2). What value of x is two standard deviations to the right of the mean? 21. Suppose X ~ N(8, 9). What value of x is 0.67 standard deviations to the left of the mean? 22. Suppose X ~ N(–1, 2). What is the z-score of x = 2? 23. Suppose X ~ N(12, 6). What is the z-score of x = 2? 24. Suppose X ~ N(9, 3). What is the z-score of x = 9? 25. Suppose a normal distribution has a mean of six and a standard deviation of 1.5. What is the z-score of x = 5.5? 26. In a normal distribution, x = 5 and z = –1.25. This tells you that x = 5 is ____ standard deviations to the ____ (right or left) of the mean. 27. In a normal distribution, x = 3 and z = 0.67. This tells you that x = 3 is ____ standard deviations to the ____ (right or left) of the mean. 28. In a normal distribution, x = –2 and z = 6. This tells you that x = –2 is ____ standard deviations to the ____ (right or left) of the mean. 29. In a normal distribution, x = –5 and z = –3.14. This tells you that x = –5 is ____ standard deviations to the ____ (right or left) of the mean. 30. In a normal distribution, x = 6 and z = –1.7. This tells you that x = 6 is ____ standard deviations to the ____ (right or left) of the mean. 31. About what percent of x values from a normal distribution lie within one standard deviation, left and right, of the mean of that distribution? 32. About what percent of the x values from a normal distribution |
lie within two standard deviations, left and right, of the mean of that distribution? 33. About what percent of x values lie between the second and third standard deviations, both sides? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 401 34. Suppose X ~ N(15, 3). Between what x values does 68.27 percent of the data lie? The range of x values is centered at the mean of the distribution (i.e., 15). 35. Suppose X ~ N(–3, 1). Between what x values does 95.45 percent of the data lie? The range of x values is centered at the mean of the distribution (i.e., –3). 36. Suppose X ~ N(–3, 1). Between what x values does 34.14 percent of the data lie? 37. About what percent of x values lie between the mean and three standard deviations? 38. About what percent of x values lie between the mean and one standard deviation? 39. About what percent of x values lie between the first and second standard deviations from the mean, both sides? 40. About what percent of x values lie between the first and third standard deviations, both sides? Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. 41. Define the random variable X in words. X = _______________. 42. X ~ _____(_____, _____) 6.2 Using the Normal Distribution 43. How would you represent the area to the left of one in a probability statement? Figure 6.12 44. What is the area to the right of one? Figure 6.13 45. Is P(x < 1) equal to P(x ≤ 1)? Why or why not? 402 Chapter 6 | The Normal Distribution 46. How would you represent the area to the left of three in a probability statement? Figure 6.14 47. What is the area to the right of three? Figure 6.15 48. If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x? 49. If the area to the right of x in a normal distribution is 0 |
.543, what is the area to the left of x? Use the following information to answer the next four exercises: X ~ N(54, 8) 50. Find the probability that x > 56. 51. Find the probability that x < 30. 52. Find the 80th percentile. 53. Find the 60th percentile. 54. X ~ N(6, 2) Find the probability that x is between three and nine. 55. X ~ N(–3, 4) Find the probability that x is between one and four. 56. X ~ N(4, 5) Find the maximum of x in the bottom quartile. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 403 57. Use the following information to answer the next three exercises: The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Find the probability that a CD player will break down during the guarantee period. a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. Figure 6.16 b. P(0 < x < ____________) = ___________. Use zero for the minimum value of x. 58. Find the probability that a CD player will last between 2.8 and 6 years. a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. Figure 6.17 b. P(__________ < x < __________) = __________ 59. Find the 70th percentile of the distribution for the time a CD player lasts. a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the lower 70 percent. Figure 6.18 b. P(x < k) = __________. Therefore, k = _________. 404 Chapter 6 | The Normal Distribution HOMEWORK 6.1 The Standard Normal Distribution Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. 60. What is the median recovery time? a. 2.7 b. 5.3 c. 7.4 d. 2.1 61. What is the z |
-score for a patient who takes 10 days to recover? a. 1.5 b. 0.2 c. 2.2 d. 7.3 62. The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true? I. The data cannot follow the uniform distribution. II. The data cannot follow the exponential distribution. III. The data cannot follow the normal distribution. a. b. c. d. I only II only III only I, II, and III 63. The heights of the 430 basketball players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with a mean, µ = 79 inches, and a standard deviation, σ = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences: a. 77 inches b. 85 inches c. If a player reported his height had a z-score of 3.5, would you believe him? Explain your answer. 64. The systolic blood pressure, given in millimeters, of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. Systolic blood pressure for males follows a normal distribution. a. Calculate the z-scores for the male systolic blood pressures 100 and 150 millimeters. b. If a male friend of yours said he thought his systolic blood pressure was 2.5 standard deviations below the mean, and that he believed his blood pressure was between 100 and 150 millimeters, what would you say to him? 65. Kyle’s doctor told him that the z-score for his systolic blood pressure is 1.75. Which of the following is the best interpretation of this standardized score? The systolic blood pressure, given in millimeters, of males has an approximately normal distribution with mean µ = 125 and standard deviation σ = 14. If X = a systolic blood pressure score, then X ~ N (125, 14). a. Which answer(s) is/are correct? i. Kyle’s systolic blood pressure is 175. ii. Kyle’s systolic blood pressure is 1.75 times the average blood pressure of men his age. iii. Kyle |
’s systolic blood pressure is 1.75 above the average systolic blood pressure of men his age. iv. Kyles’s systolic blood pressure is 1.75 standard deviations above the average systolic blood pressure for men. b. Calculate Kyle’s blood pressure. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 405 66. Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and same gender. In 2009, weights for all 80 cm girls in the reference population had a mean µ = 10.2 kg and standard deviation σ = 0.8 kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that correspond to the following weights and interpret them: a. 11 kg b. 7.9 kg c. 12.2 kg 67. In 2005, 1,475,623 students heading to college took the SAT exam. The distribution of scores in the math section of the SAT follows a normal distribution with mean µ = 520 and standard deviation σ = 115. a. Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence. b. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score? c. For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternative to the SAT math test, and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test that each person took? 6.2 Using the Normal Distribution Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. 68. What is the probability of spending more than two days in recovery? a. 0.0580 b. 0.8447 c. 0.0553 d. 0.9420 69. The 90th percentile for recovery times is – a. 8.89 b. 7.07 c. |
7.99 d. 4.32 Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. 70. Based on the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space? a. Yes b. No c. Unable to determine 71. Find the probability that it takes at least eight minutes to find a parking space. a. 0.0001 b. 0.9270 c. 0.1862 d. 0.0668 72. Seventy percent of the time, it takes more than how many minutes to find a parking space? a. 1.24 b. 2.41 c. 3.95 d. 6.05 406 Chapter 6 | The Normal Distribution 73. According to a study done by De Anza students, the height for Asian adult males is normally distributed with an average of 66 inches and a standard deviation of 2.5 inches. Suppose one Asian adult male is randomly chosen. Let X = height of the individual. a. X ~ _____ (_____,_____) b. Find the probability that the person is between 65 and 69 inches. Include a sketch of the graph, and write a probability statement. c. Would you expect to meet many Asian adult males taller than 72 inches? Explain why or why not, and numerically justify your answer. d. The middle 40 percent of heights fall between what two values? Sketch the graph, and write the probability statement. 74. IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. a. X ~ _____ (_____, _____) b. Find the probability that the person has an IQ greater than 120. Include a sketch of the graph, and write a probability statement. c. MENSA is an organization whose members have the top 2 percent of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. Sketch the graph, and write the probability statement. d. The middle 50 percent of IQs fall between what two values? Sketch the graph, and write the probability statement. 75. The percent of fat calories that a person in the United States consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10. Suppose that one individual is randomly chosen |
. Let X = percentage of fat calories. a. X ~ _____ (_____, _____) b. Find the probability that the percentage of fat calories a person consumes is more than 40. Graph the situation. Shade in the area to be determined. c. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement. 76. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. a. b. If X = distance in feet for a fly ball, then X ~ _____ (_____, _____) If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled less than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade the region corresponding to the probability. Find the probability. c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the probability statement. 77. In China, four-year-olds average three hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.5 hours and the amount of time spent alone is normally distributed. We randomly select one Chinese four-year-old living in a rural area. We are interested in the amount of time that child spends alone per day. In words, define the random variable X. a. b. X ~ _____ (_____, _____) c. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the probability statement. d. What percentage of the children spend more than 10 hours per day unsupervised? e. Seventy percent of the children spend at least how long per day unsupervised? 78. In the 1992 presidential election, Alaska’s 40 election districts averaged 1,956.8 votes per district for a candidate. The standard deviation was 572.3. There are only 40 election districts in Alaska. The distribution of the votes per district for the candidate was bell-shaped. Let X = number of votes for the candidate for an election district. a. State the approximate distribution of X. b. c. Find the probability that a randomly selected district had fewer than 1,600 votes for the candidate. Sketch the Is 1,956.8 a population mean or a sample mean? How do you know? graph |
, and write the probability statement. d. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for the candidate. e. Find the third quartile for votes for the candidate. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 407 79. Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 21 days and a standard deviation of seven days. In words, define the random variable X. a. b. X ~ _____ (_____, _____) c. If one of the trials is randomly chosen, find the probability that it lasted at least 24 days. Sketch the graph and write the probability statement. d. Sixty percent of all trials of this type are completed within how many days? 80. Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5-mile lap, in a seven-lap race, with a standard deviation of 2.28 seconds. The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps. In words, define the random variable X. a. b. X ~ _____ (_____, _____) c. Find the percent of her laps that are completed in less than 130 seconds. d. The fastest 3 percent of her laps are under _____. e. The middle 80 percent of her laps are from _______ seconds to _______ seconds. 81. Thuy Dau, Ngoc Bui, Sam Su, and Lan Voung conducted a survey as to how long customers at Lucky claimed to wait in the checkout line until their turn. Let X = time in line. Table 6.3 displays the ordered real data, in minutes. 0.50 4.25 5 6 1.75 4.25 5.25 6 7.25 7.25 2 4.25 5.25 6.25 7.25 2.25 4.25 5.5 6.25 7.75 2.25 4.5 5.5 2.5 4.75 5.5 6.5 6.5 8 8.25 2.75 4.75 5.75 6.5 9.5 3.25 4.75 5.75 6.75 9.5 3.75 5 3.75 5 6 6 6.75 9.75 6.75 10.75 |
Table 6.3 a. Calculate the sample mean and the sample standard deviation. b. Construct a histogram. c. Draw a smooth curve through the midpoints of the tops of the bars. d. In words, describe the shape of your histogram and smooth curve. e. Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by X ~ _____ (_____, _____) f. Use the distribution in part e to calculate the probability that a person will wait fewer than 6.1 minutes. g. Determine the cumulative relative frequency for waiting less than 6.1 minutes. h. Why aren’t the answers to part f and part g exactly the same? i. Why are the answers to part f and part g as close as they are? j. If only 10 customers were surveyed rather than 50, do you think the answers to part f and part g would have been closer together or farther apart? Explain your conclusion. 82. Suppose that Ricardo and Anita attend different colleges. Ricardo’s GPA is the same as the average GPA at his school. Anita’s GPA is 0.70 standard deviations above her school average. In complete sentences, explain why each of the following statements may be false: a. Ricardo’s actual GPA is lower than Anita’s actual GPA. b. Ricardo is not passing because his z-score is zero. c. Anita is in the 70th percentile of students at her college. 408 Chapter 6 | The Normal Distribution 83. Table 6.4 shows a sample of the maximum capacity—maximum number of spectators—of sports stadiums. The table does not include horse-racing or motor-racing stadiums. 40,000 40,000 45,050 45,500 46,249 48,134 49,133 50,071 50,096 50,466 50,832 51,100 51,500 51,900 52,000 52,132 52,200 52,530 52,692 53,864 54,000 55,000 55,000 55,000 55,000 55,000 55,000 55,082 57,000 58,008 59,680 60,000 60,000 60,492 60,580 62,380 62,872 64,035 65,000 65,050 65,647 66,000 66,161 67,428 68,349 68,976 69,372 70,107 70, |
585 71,594 72,000 72,922 73,379 74,500 75,025 76,212 78,000 80,000 80,000 82,300 Table 6.4 a. Calculate the sample mean and the sample standard deviation for the maximum capacity of sports stadiums. b. Construct a histogram. c. Draw a smooth curve through the midpoints of the tops of the bars of the histogram. d. e. Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can In words, describe the shape of your histogram and smooth curve. then be approximated by X ~ _____ (_____, _____). f. Use the distribution in part e to calculate the probability that the maximum capacity of sports stadiums is less than 67,000 spectators. g. Determine the cumulative relative frequency that the maximum capacity of sports stadiums is less than 67,000 spectators. Hint—Order the data and count the sports stadiums that have a maximum capacity less than 67,000. Divide by the total number of sports stadiums in the sample. h. Why aren’t the answers to part f and part g exactly the same? 84. The length of a pregnancy of a certain female animal is normally distributed with a mean of 280 days and a standard deviation of 13 days. The father was not present from 240 to 306 days before the birth of the offspring, so the pregnancy would have been less than 240 days or more than 306 days long, if he was the father. What is the probability that he was NOT the father? What is the probability that he could be the father? Calculate the z-scores first, and then use those to calculate the probability. 85. A NUMMI assembly line, which has been operating since 1984, has built an average of 6,000 cars and trucks a week. Generally, 10 percent of the cars were defective coming off the assembly line. Suppose we draw a random sample of n = 100 cars. Let X represent the number of defective cars in the sample. What can we say about X in regard to the 68–95–99.7 empirical rule—one standard deviation, two standard deviations, and three standard deviations from the mean being referred to? Assume a normal distribution for the defective cars in the sample. 86. We flip a coin 100 times (n = 100) and note that it only comes up heads 20 percent (p = 0.20) of the time. The mean and standard deviation |
for the number of times the coin lands on heads is µ = 20 and σ = 4—verify the mean and standard deviation. Solve the following: a. There is about a 68 percent chance that the number of heads will be somewhere between ___ and ___. b. There is about a ____chance that the number of heads will be somewhere between 12 and 28. c. There is about a ____ chance that the number of heads will be somewhere between eight and 32. 87. A child playing a carnival game will be a winner one out of five times. If 190 games are played, what is the probability that there are a. b. c. more than 64 wins somewhere between 34 and 54 wins somewhere between 54 and 64 wins This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 409 88. A social media site provides a variety of statistics on its website that detail the growth and popularity of the site. On average, 28 percent of 18- to 34-year-olds check their social media profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent. a. Find the probability that the percentage of 18- to 34-year-olds who check the social media website before getting out of bed in the morning is at least 30. b. Find the 95th percentile, and express it in a sentence. REFERENCES 6.1 The Standard Normal Distribution CollegeBoard. group http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf College-bound seniors: (2012). Total 2012 profile report. Retrieved from Joyce, C. A., Janssen, S., & Liu, M. L. (2010). The world almanac and book of facts, 2010. New York, NY: World Almanac Books. London School http://conflict.lshtm.ac.uk/page_125.htm of Hygiene and Tropical Medicine. (2009). Calculation of z-scores. Retrieved from National Center for Education Statistics. (2009). ACT score averages and standard deviations, by sex and race/ethnicity, and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009. Retrieved from http://nces.ed.gov/programs/digest/d09/tables/ |
dt09_147.asp NBA.com. (2013). NBA Media Ventures. Retrieved from http://www.nba.com StatCrunch. viewreport.php?reportid=11960 (2010). Blood pressure of males and females. Retrieved from http://www.statcrunch.com/5.0/ The Mercury News. (n.d.). Retrieved from http://www.mercurynews.com/ Wikipedia. (2013). List of List_of_stadiums_by_capacity stadiums by capacity - Wikipedia. Retrieved from https://en.wikipedia.org/wiki/ 6.2 Using the Normal Distribution Chicago Public Media & Ira Glass. (2013). 403: NUMMI. Retrieved from http://www.thisamericanlife.org/radioarchives/ episode/403/nummi lauramitchell347. (2012, Dec. 28). Smart phone users, by the numbers. Visually. Retrieved from http://visual.ly/smartphone-users-numbers Statistics Brain Research Institute. http://www.statisticbrain.com/facebook-statistics/ (2013). Facebook company statistics – statistic brain. Retrieved from Wikipedia (2013). Naegele's rule. Retrieved from http://en.wikipedia.org/wiki/Naegele's_rule Win at the Lottery. (2013). Scratch-off lottery ticket playing tips. Retrieved from www.winatthelottery.com/public/ department40.cfm SOLUTIONS 1 ounces of water in a bottle 3 2 5 –4 7 –2 9 The mean becomes zero. 11 z = 2 Chapter 6 | The Normal Distribution 410 13 z = 2.78 15 x = 20 17 x = 6.5 19 x = 1 21 x = 1.97 23 z = –1.67 25 z ≈ –0.33 27 0.67, right 29 3.14, left 31 about 68 percent 33 about 4 percent 35 between –5 and –1 37 about 50 percent 39 about 27 percent 41 The lifetime of a Sunshine CD player measured in years 43 P(x < 1) 45 Yes, because they are the same in a continuous distribution: P(x = 1) = 0 47 1 – P(x < 3) or P(x > 3) 49 1 – 0.543 = 0.457 51 0.0013 53 56.03 55 0.1186 57 a. Check student’s solution b |
. 3, 0.1979 59 a. Check student’s solution b. 0.70, 4.78 years 61 c 63 a. Use the z-score formula. z = –0.5141. The height of 77 inches is 0.5141 standard deviations below the mean. An NBA player whose height is 77 inches is shorter than average. b. Use the z-score formula. z = 1.5424. The height 85 inches is 1.5424 standard deviations above the mean. An NBA player whose height is 85 inches is taller than average. c. Height = 79 + 3.5(3.89) = 90.67 inches, which is over 7.7 feet tall. There are very few NBA players this tall; so, the answer is no, not likely. 65 a. iv b. Kyle’s blood pressure is equal to 125 + (1.75)(14) = 149.5. 67 Let X = an SAT math score and Y = an ACT math score. a. X = 720 720 – 520 = 1.74 The exam score of 720 is 1.74 standard deviations above the mean of 520. 15 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 6 | The Normal Distribution 411 b. z = 1.5 The math SAT score is 520 + 1.5(115) ≈ 692.5. The exam score of 692.5 is 1.5 standard deviations above the mean of 520. c. X – μ σ = 700 – 514 117 ≈ 1.59, the z-score for the SAT. Y – μ σ = 30 – 21 5.3 ≈ 1.70, the z-scores for the ACT. With respect to the test they took, the person who took the ACT did better—has the higher z-score). 69 c 71 d 73 a. X ~ N(66, 2.5) b. 0.5404 c. No, the probability that an Asian male is over 72 inches tall is 0.0082. 75 a. X ~ N(36, 10) b. The probability that a person consumes more than 40 percent of their calories as fat is 0.3446. c. Approximately 25 percent of people consume less than 29.26 percent of their calories as fat. 77 a. X = number of hours that a Chinese four-year |
-old in a rural area is unsupervised during the day. b. X ~ N(3, 1.5) c. The probability that the child spends less than one hour a day unsupervised is 0.0918. d. The probability that a child spends over 10 hours a day unsupervised is less than 0.0001. e. 2.21 hours 79 a. X = the distribution of the number of days a particular type of criminal trial will take b. X ~ N(21, 7) c. The probability that a randomly selected trial will last more than 24 days is 0.3336. d. 22.77 81 a. mean = 5.51, s = 2.15 b. Check student's solution. c. Check student's solution. d. Check student's solution. e. X ~ N(5.51, 2.15) f. 0.6029 g. The cumulative frequency for less than 6.1 minutes is 0.64. h. The answers to part f and part g are not exactly the same, because the normal distribution is only an approximation to the real one. i. The answers to part f and part g are close, because a normal distribution is an excellent approximation when the sample size is greater than 30. j. The approximation would have been less accurate, because the smaller sample size means that the data does not fit a normal curve as well. Chapter 6 | The Normal Distribution 412 83 1. mean = 60,136 s = 10,468 2. Answers will vary 3. Answers will vary 4. Answers will vary 5. X ~ N(60136, 10468) 6. 0.7440 7. The cumulative relative frequency is 43/60 = 0.717. 8. The answers for part f and part g are not the same because the normal distribution is only an approximation. 85 n = 100; p = 0.1; q = 0.9 μ = np = (100)(0.10) = 10 σ = npq = (100)(0.1)(0.9) = 3 i. z = ±1: x1 = µ + zσ = 10 + 1(3) = 13 and x2 = µ – zσ = 10 – 1(3) = 7. 68 percent of the defective cars will fall between seven and 13 ii. iii. z = ±2: x1 = µ + zσ = 10 + 2(3) = 16 |
and x2 = µ – zσ = 10 – 2(3) = 4. 95 percent of the defective cars will fall between four and 16 z = ±3: x1 = µ + zσ = 10 + 3(3) = 19 and x2 = µ – zσ = 10 – 3(3) = 1. 99.7 percent of the defective cars will fall between one and 19 87 n = 190; p = 1 5 = 0.2; q = 0.8 μ = np = (190)(0.2) = 38 σ = npq = (190)(0.2)(0.8) = 5.5136 a. For this problem: P(34 < x < 54) = normalcdf(34,54,48,5.5136) = 0.7641 b. For this problem: P(54 < x < 64) = normalcdf(54,64,48,5.5136) = 0.0018 c. For this problem: P(x > 64) = normalcdf(64,1099,48,5.5136) = 0.0000012 (approximately 0) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 413 7 | THE CENTRAL LIMIT THEOREM Figure 7.1 If you want to figure out the distribution of the change people carry in their pockets, using the central limit theorem and assuming your sample is large enough, you will find that the distribution is normal and bell-shaped. (credit: John Lodder) Introduction Chapter Objectives By the end of this chapter, the student should be able to do the following: • Recognize central limit theorem problems • Classify continuous word problems by their distributions • Apply and interpret the central limit theorem for means • Apply and interpret the central limit theorem for sums Why are we so concerned with means? Two reasons are they give us a middle ground for comparison, and they are easy to calculate. In this chapter, you will study means and the central limit theorem. The central limit theorem (clt) is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem, and both alternatives are concerned with drawing a finite samples size n from a population with a 414 Chapter 7 | The Central Limit Theorem known mean, μ, and a known standard deviation, � |
�. The first alternative says that if we collect samples of size n with a large enough n, calculate each sample's mean, and create a histogram of those means, then the resulting histogram will tend to have an approximate normal bell shape. The second alternative says that if we again collect samples of size n that are large enough, calculate the sum of each sample and create a histogram, then the resulting histogram will again tend to have a normal bell shape. The central limit theorem for sample means is more discussed in the world of statistics, but it is important to note that taking each sample's sum and graphing the sums will also result in a normal histogram. There are instances where one wishes to calculate the sum of a sample, as opposed to its mean. In either case, it does not matter what the distribution of the original population is, or whether you even need to know it. The important fact is that the distributions of sample means and the sums tend to follow the normal distribution. The size of the sample, n, that is required in order to be large enough depends on the original population from which the samples are drawn (the sample size should be at least 30 or the data should come from a normal distribution). If the original population is far from normal, then more observations are needed for the sample means or sums to be normal. Sampling is done with replacement. Suppose eight of you roll one fair die ten times, seven of you roll two fair dice ten times, nine of you roll five fair dice ten times, and 11 of you roll ten fair dice ten times. Each time a person rolls more than one die, he or she calculates the sample mean of the faces showing. For example, one person might roll five fair dice and get 2, 2, 3, 4, and 6 on one roll. The mean is.4. The 3.4 is one mean when five fair dice are rolled. This same person would roll the five dice nine more times and calculate nine more means for a total of ten means. Your instructor will pass out the dice to several people. Roll your dice ten times. For each roll, record the faces, and find the mean. Round to the nearest 0.5. Your instructor (and possibly you) will produce one graph (it might be a histogram) for one die, one graph for two dice, one graph for five dice, and one graph for ten dice. Because the mean when you roll one die is just the face on the die, what |
distribution do these means appear to be representing? Draw the graph for the means using two dice. Do the sample means show any kind of pattern? Draw the graph for the means using five dice. Do you see any pattern emerging? Finally, draw the graph for the means using ten dice. Do you see any pattern to the graph? What can you conclude as you increase the number of dice? As the number of dice rolled increases from one to two to five to ten, the following is happening: 1. The mean of the sample means remains approximately the same. 2. The spread of the sample means (the standard deviation of the sample means) gets smaller. 3. The graph appears steeper and thinner. You have just demonstrated the central limit theorem (clt). The central limit theorem tells you that as you increase the number of dice, the sample means tend toward a normal distribution (the sampling distribution). 7.1 | The Central Limit Theorem for Sample Means (Averages) Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose a. μ x = the mean of X This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 415 b. σ x = the standard deviation of X If you draw random samples of size n, then as n increases, the random variable X to be normally distributed and ¯, which consists of sample means, tends ¯ X ∼ N ⎛ ⎝μ x, σ x n ⎞ ⎠ The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculating their means, the sample means form their own normal distribution (the sampling distribution). The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. The variable n is the number of values that are averaged together, not the number of times the experiment is done. ¯, which consists To put it more formally, if you draw random samples of size n, the distribution of the random variable X of sample means, is called the sampling distribution of the mean. The sampling distribution of the mean approaches a normal distribution as n, the sample size, increases. The random variable X ¯ has a different z-score |
associated with it from that of the random variable X. The mean x¯ is the ¯ value of X in one sample. z = x¯ − μ x σ x n ⎞ ⎠ ⎛ ⎝, ¯. μX is the average of both X and X σ x¯ = σx n = standard deviation of X ¯ and is called the standard error of the mean. To find probabilities for means on the calculator, follow these steps. 2nd DISTR 2:normalcdf normalcd f ⎛ ⎝lower value o f the area, upper value o f the area, mean, standard deviation sample size ⎞ ⎠ where • mean is the mean of the original distribution • • standard deviation is the standard deviation of the original distribution sample size = n Example 7.1 A distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25 are drawn randomly from the population. a. Find the probability that the sample mean is between 85 and 92. Solution 7.1 a. Let X = one value from the original unknown population. The probability question asks you to find a probability for the sample mean. 416 Chapter 7 | The Central Limit Theorem Let X ¯ = the mean of a sample of size 25. Because μx = 90, σx = 15, and n = 25, ¯ X ∼ N ⎛ ⎝μ x, σ x n ⎞ ⎠ Find P(85 < x¯ < 92). Draw a graph. P(85 < x¯ < 92) = 0.6997 The probability that the sample mean is between 85 and 92 is 0.6997. Figure 7.2 Find P(85 < x¯ < 92). Draw a graph. P(85 < x¯ < 92) = 0.6997 normalcdf(lower value, upper value, mean, standard error of the mean) The parameter list is abbreviated (lower value, upper value, μ, σ n ). normalcdf(85,92,90, 15 25 ) = 0.6997 b. Find the value that is two standard deviations above the expected value, 90, of the sample mean. Solution 7.1 b. To find the value that is two standard deviations above the expected value 90, use the following formula ⎛ value = µ x + ( # ofSTDEVs) ⎝ σ x n � |
� ⎠ value = 90 + 2 ⎛ ⎝ 15 25 ⎞ ⎠ = 96. The value that is two standard deviations above the expected value is 96. The standard error of the mean is σx n = 15 25 = 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size n. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 417 7.1 An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size n = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50. Example 7.2 The length of time, in hours, it takes a group of people, 40 years old and older, to play one soccer match is normally distributed with a mean of 2 hours and a standard deviation of 0.5 hours. A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours. Solution 7.2 Let X = the time, in hours, it takes to play one soccer match. The probability question asks you to find a probability for the sample mean time, in hours, it takes to play one soccer match. ¯ = the mean time, in hours, it takes to play one soccer match. Let X If μX = _________, σX = __________, and n = ___________, then X ~ N(______, ______) by the central limit theorem for means. μX = 2, σX = 0.5, n = 50, and X ~ N ⎛ ⎝2, 0.5 50 ⎞ ⎠ Find P(1.8 < x¯ < 2.3). Draw a graph. normalcdf P(1.8 < x¯ < 2.3) = 0.9977 ⎛ ⎝1.8,2.3,2,.5 50 ⎞ ⎠= 0.9977 The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977. 7.2 The length of time taken on the SAT exam for a group of students is |
normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of n = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours. To find percentiles for means on the calculator, follow these steps. 2nd DIStR 3:invNorm k = invNorm ⎛ ⎝area to the left of k, mean, standard deviation sample size ⎞ ⎠ 418 Chapter 7 | The Central Limit Theorem where • k = the kth percentile • mean is the mean of the original distribution • • standard deviation is the standard deviation of the original distribution sample size = n Example 7.3 In a recent study reported Oct. 29, 2012, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n = 100. a. What are the mean and standard deviation for the sample mean ages of tablet users? b. What does the distribution look like? c. Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study). d. Find the 95th percentile for the sample mean age (to one decimal place). Solution 7.3 a. Because the sample mean tends to target the population mean, we have μχ = μ = 34. The sample standard = 1.5. deviation is given by σχ = σ n = 15 100 = 15 10 b. The central limit theorem states that for large sample sizes (n), the sampling distribution will be approximately normal. c. The probability that the sample mean age is more than 30 is given by P(Χ > 30) = normalcdf(30,E99,34,1.5) = 0.9962. d. Let k = the 95th percentile. ⎝0.95,34, 15 100 k = invNorm ⎛ ⎞ ⎠ = 36.5 7.3 A gaming marketing gap for men between the ages of 30 to 40 has been identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article’s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take |
a sample of 100 randomly selected gamers. If your target market is 29- to 35-year-olds, should you continue with your development strategy? Example 7.4 The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60. a. What are the mean and standard deviation for the sample mean number of app engagement minutes by a tablet user? b. What is the standard error of the mean? c. Find the 90th percentile for the sample mean time for app engagement for a tablet user. Interpret this value This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 419 in a complete sentence. d. Find the probability that the sample mean is between eight minutes and 8.5 minutes. Solution 7.4 a. μ x¯ = μ = 8.2 σ x¯ = σ n = 1 60 = 0.13 b. This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60. k = invNorm ⎛ c. Let k = the 90th percentile. ⎝0.90,8.2, 1 60 time for table users is less than 8.37 minutes. ⎞ ⎠ = 8.37. This values indicates that 90 percent of the average app engagement d. P(8 < x¯ < 8.5) = normalcdf ⎛ ⎝8,8.5,8.2, 1 60 ⎞ ⎠ = 0.9293 7.4 Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, x¯ = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces? 7.2 | The Central Limit Theorem for Sums (Optional) Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose: a. μX = the mean of Χ b. σΧ = the standard deviation |
of X If you draw random samples of size n, then as n increases, the random variable ΣX consisting of sums tends to be normally distributed and ΣΧ ~ N[(n)(μΧ), ( n )(σΧ)]. The central limit theorem for sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size. The random variable ΣX has the following z-score associated with it: a. Σx is one sum. b. z = Σx – (n)(μ X) ( n)(σ X) i. ii. (n)(μX) = mean of ΣX ( n)(σ X) = standard deviation of ΣX To find probabilities for sums on the calculator, follow these steps: 2nd DISTR 420 Chapter 7 | The Central Limit Theorem 2:normalcdf normalcdf(lower value of the area, upper value of the area, (n)(mean), ( n )(standard deviation)) where, • mean is the mean of the original distribution, • • standard deviation is the standard deviation of the original distribution, and sample size = n. Example 7.5 An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population. a. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500. b. Find the sum that is 1.5 standard deviations above the mean of the sums. Solution 7.5 Let X = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values. ΣX = the sum or total of 80 values. Because μX = 90, σX = 15, and n = 80, ΣX ~ N[(80)(90), ( 80 )(15)] • mean of the sums = (n)(μX) = (80)(90) = 7200 • • standard deviation of the sums = ( n)(σ X ) = ( 80) (15) sum of 80 values = Σx = 7500 a. |
Find P(Σx > 7500) P(Σx > 7500) = 0.0127 Figure 7.3 normalcdf(lower value, upper value, mean of sums, stdev of sums) The parameter list is abbreviated(lower, upper, (n)(μX, ( n) (σX)) normalcdf (7500,1E99,(80)(90), ⎛ ⎝ 80⎞ ⎠ (15)) = 0.0127 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 421 REMINDER 1E99 = 1099. Press the EE key for E. b. Find Σx where z = 1.5. Σx = (n)(μX) + (z) ( n) (σΧ) = (80)(90) + (1.5)( 80 )(15) = 7401.2 7.5 An unknown distribution has a mean of 45 and a standard deviation of 8. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400. To find percentiles for sums on the calculator, follow these steps: 2nd DIStR 3:invNorm k = invNorm (area to the left of k, (n)(mean), ( n) (standard deviation)) where, • k is the kth percentile, • mean is the mean of the original distribution, • • standard deviation is the standard deviation of the original distribution, and sample size = n. Example 7.6 In a recent study reported Oct. 29, 2012, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample size is 50. a. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution? b. Find the probability that the sum of the ages is between 1,500 and 1,800 years. c. Find the 80th percentile for the sum of the 50 ages. Solution 7.6 a. μΣx = nμx = 50(34) = 1,700 and σΣx = n σx = ( 50 ) (15) = 106.01 The distribution is normal for sums by the central limit theorem. b. P( |
1500 < Σx < 1800) = normalcdf (1500, 1800, (50)(34), ( 50 ) (15)) = 0.7974 c. Let k = the 80th percentile. k = invNorm(0.80,(50)(34), ( 50 ) (15)) = 1789.3 422 Chapter 7 | The Central Limit Theorem 7.6 In a recent study reported Oct.29, 2012, the mean age of tablet users is 35 years. Suppose the standard deviation is 10 years. The sample size is 39. a. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution? b. Find the probability that the sum of the ages is between 1,400 and 1,500 years. c. Find the 90th percentile for the sum of the 39 ages. Example 7.7 The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70. a. What are the mean and standard deviation for the sums? b. Find the 95th percentile for the sum of the sample. Interpret this value in a complete sentence. c. Find the probability that the sum of the sample is at least 10 hours. Solution 7.7 a. μΣx = nμx = 70(8.2) = 574 minutes and σΣx = ( n)(σ x) = ( 70 ) (1) = 8.37 minutes b. Let k = the 95th percentile. k = invNorm (0.95,(70)(8.2), ( 70) (1)) = 587.76 minutes Ninety-five percent of the app engagement times are at most 587.76 minutes. c. 10 hours = 600 minutes P(Σx ≥ 600) = normalcdf(600,E99,(70)(8.2), ( 70) (1)) = 0.0009 7.7 The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70. a. What is the probability that the sum of the sample is between seven hours and 10 hours? What does this mean in context of the problem? b. Find the 84th and 16th percentiles for the sum of the sample. Interpret these values in context. 7.3 | Using the |
Central Limit Theorem It is important for you to understand when to use the central limit theorem. If you are being asked to find the probability of the mean, use the clt for the means. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums. NOTE If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 423 Examples of the Central Limit Theorem Law of Large Numbers The law of large numbers says that if you take samples of larger and larger sizes from any population, then the mean x¯ of the samples tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that.) This means that the sample mean x¯ must be close to the population mean μ. We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers. ¯ the standard deviation for X is σ n Central Limit Theorem for the Mean and Sum Examples Example 7.8 A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find: a. b. c. d. the probability that the mean stress score for the 75 students is less than 2 the 90th percentile for the mean stress score for the 75 students the probability that the total of the 75 stress scores is less than 200 the 90th percentile for the total stress score for the 75 students Let X = one stress score. Problems (a) and (b) ask you to find a probability or a percentile for a mean. Problems (c) and (d) ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75. Because the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (see Continuous Random Variables for an explanation of a uniform distribution), µ X =a + |
b 2 (b – a)2 12 σ X = = 1 + 5 2 (5 – 1)2 12 = = 3 = 1.15 In the formula above, the denominator is understood to be 12, regardless of the endpoints of the uniform distribution. For problems (a) and (b), let X ¯ = the mean stress score for the 75 students. Then, ¯ X ~ N ⎛ ⎝3, 1.15 75 ⎞ ⎠ where n = 75. a. Find P( x¯ < 2). Draw the graph. Solution 7.8 a. P( x¯ < 2) = 0 The probability that the mean stress score is less than 2 is about zero. 424 Chapter 7 | The Central Limit Theorem Figure 7.4 normalcdf ⎛ ⎝1,2,3,1.15 75 ⎞ ⎠ = 0 REMINDER The smallest stress score is one. b. Find the 90th percentile for the mean of 75 stress scores. Draw a graph. Solution 7.8 b. Let k = the 90th precentile. Find k, where P( x¯ < k) = 0.90. k = 3.2 Figure 7.5 The 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90 percent of all the means of 75 stress scores are at most 3.2, and that 10 percent are at least 3.2. invNorm ⎛ ⎝0.90,3,1.15 75 ⎞ ⎠ = 3.2 For problems (c) and (d), let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N[(75)(3), ( 75) (1.15)]. c. Find P(Σx < 200). Draw the graph. Solution 7.8 c. The mean of the sum of 75 stress scores is (75)(3) = 225. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 425 The standard deviation of the sum of 75 stress scores is ( 75) (1.15) = 9.96. P(Σx < 200) = 0 Figure 7.6 The probability that the total of 75 scores is less than 200 is about zero. normalcdf |
(75,200,(75)(3), ( 75) (1.15)). REMINDER The smallest total of 75 stress scores is 75, because the smallest single score is one. d. Find the 90th percentile for the total of 75 stress scores. Draw a graph. Solution 7.8 d. Let k = the 90th percentile. Find k where P(Σx < k) = 0.90. k = 237.8 Figure 7.7 The 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90 percent of all the sums of 75 scores are no more than 237.8 and 10 percent are no less than 237.8. invNorm(0.90,(75)(3), ( 75) (1.15)) = 237.8 426 Chapter 7 | The Central Limit Theorem 7.8 Use the information in Example 7.8, but use a sample size of 55 to answer the following questions. a. Find P( x¯ < 7). b. Find P(Σx > 170). c. Find the 80th percentile for the mean of 55 scores. d. Find the 85th percentile for the sum of 55 scores. Example 7.9 Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract. The analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes. Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract. Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance. X ∼ Exp ⎛ ⎝ ⎞ ⎠ 1 22. From previous chapters, we know that μ = 22 and σ = 22. ¯ = the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance. Let X ¯ ~ N ⎛ X ⎝22, 22 80 ⎞ ⎠ by the central limit theorem for sample means. Using the clt to find probability a. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find P( x¯ > 20). Draw the graph. b. Suppose that one customer who exceeds the time limit for his cell phone contract |
is randomly selected. Find the probability that this individual customer's excess time is longer than 20 minutes. This is asking us to find P(x > 20). c. Explain why the probabilities in parts (a) and (b) are different. Solution 7.9 a. Find: P( x¯ > 20) P( x¯ > 20) = 0.79199 using normalcdf ⎛ ⎝20,1E99,22, 22 80 ⎞ ⎠ The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 427 Figure 7.8 REMINDER 1E99 = 1099 and –1E99 = –1099. Press the EE key for E. Or just use 1099 instead of 1E99. ⎛ b. Find P(x > 20). Remember to use the exponential distribution for an individual. X ~ Exp ⎝ ⎞ ⎠. 1 22 P(x > 20) = e ⎛ ⎝− ⎛ ⎝ 1 22 ⎞ ⎞ ⎠(20) ⎠ ( – 0.04545(20)) or e = 0.4029 c. 1. P(x > 20) = 0.4029, but P( x¯ > 20) = 0.7919 2. The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means. 3. When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the clt. Use the clt with the normal distribution when you are being asked to find the probability for a mean. Using the clt to find percentiles Find the 95th percentile for the sample mean excess time for a sample of 80 customers who exceed their basic contract time allowances. Draw a graph. Solution 7.9 Let k = the 95th percentile. Find k where P( x¯ < k) = 0.95. k = 26.0 using invNorm ⎛ = 26.0 ⎝0.95,22, 22 80 ⎞ ⎠ 428 Chapter 7 | The Central Limit Theorem Figure 7. |
9 The 95th percentile for the sample mean excess time used is about 26.0 minutes for a random sample of 80 customers who exceed their contractual allowed time. 95 percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes. 7.9 Use the information in Example 7.9, but change the sample size to 144. a. Find P(20 < x¯ < 30). b. Find P(Σx is at least 3000). c. Find the 75th percentile for the sample mean excess time of 144 customers. d. Find the 85th percentile for the sum of 144 excess times used by customers. Example 7.10 U.S. scientists studying a certain medical condition discovered that a new person is diagnosed every two minutes, on average. Suppose the standard deviation is 0.5 minutes and the sample size is 100. a. Find the median, the first quartile, and the third quartile for the sample mean time of diagnosis in the United States. b. Find the median, the first quartile, and the third quartile for the sum of sample times of diagnosis in the United States. c. Find the probability that a diagnosis occurs on average between 1.75 and 1.85 minutes. d. Find the value that is two standard deviations above the sample mean. e. Find the IQR for the sum of the sample times. Solution 7.10 a. We have μx = μ = 2 and σx = σ n 1. 50th percentile = μx = μ = 2, 2. 25th percentile = invNorm(0.25,2,0.05) = 1.97, and = 0.5 10 = 0.05. Therefore, This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 429 3. 75th percentile = invNorm(0.75,2,0.05) = 2.03. b. We have μΣx = n(μx) = 100(2) = 200 and σμx = n (σx) = 10(0.5) = 5. Therefore, 1. 50th percentile = μΣx = n(μx) = 100(2) = 200, 2. 25th percentile = invNorm(0.25,200,5) = 196.63, and 3 |
. 75th percentile = invNorm(0.75,200,5) = 203.37. c. P(1.75 < x¯ < 1.85) = normalcdf(1.75,1.85,2,0.05) = 0.0013 d. Using the z-score equation, z = x¯ – μ x¯ σ x¯, and solving for x, we get x = 2(0.05) + 2 = 2.1. e. The IQR is 75th percentile – 25th percentile = 203.37 – 196.63 = 6.74. 7.10 Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follows a normal distribution. a. b. c. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120. If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120. If the sample was four women between the ages of 18–24 and we did not know the original distribution, could the central limit theorem be used? Example 7.11 A study was done about a medical condition that affects a certain group of people. The age range of the people was 14–61. The mean age was 30.9 years with a standard deviation of nine years. a. b. c. In a sample of 25 people, what is the probability that the mean age of the people is less than 35? Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results. In a sample of 49 people, what is the probability that the sum of the ages is no less than 1,600? Is it likely that the sum of the ages of the 49 people are at most 1,595? Interpret the results. d. e. Find the 95th percentile for the sample mean age of 65 people. Interpret the results. f. Find the 90th percentile for the sum of the ages of 65 people. Interpret the results. Solution 7.11 a. P( x¯ < 35) = normalcdf(-E99,35,30.9,1.8) = 0.9886 b. |
P( x¯ > 50) = normalcdf(50, E99,30.9,1.8) ≈ 0. For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50. c. P(Σx ≥ 1,600) = normalcdf(1600,E99,1514.10,63) = 0.0864 d. P(Σx ≤ 1,595) = normalcdf(-E99,1595,1514.10,63) = 0.9005. This means that there is a 90 percent chance that the sum of the ages for the sample group n = 49 is at most 1,595. 430 Chapter 7 | The Central Limit Theorem e. The 95th percentile = invNorm(0.95,30.9,1.1) = 32.7. This indicates that 95 percent of the people in the sample of 65 are younger than 32.7 years, on average. f. The 90th percentile = invNorm(0.90,2008.5,72.56) = 2101.5. This indicates that 90 percent of the people in the sample of 65 have a sum of ages less than 2,101.5 years. 7.11 According to data from an aerospace company, the 757 airliner carries 200 passengers and has doors with a mean height of 72 inches. Assume for a certain population of men we have a mean of 69 inches inches and a standard deviation of 2.8 inches. a. What mean doorway height would allow 95 percent of men to enter the aircraft without bending? b. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men? c. For engineers designing the 757, which result is more relevant: the height from part (a) or part (b)? Why? HISTORICAL NOTE Normal Approximation to the Binomial Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n (say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n, you had to use the binomial formula, which could be very |
complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the following conditions for a binomial distribution: • There are a certain number, n, of independent trials. • The outcomes of any trial are success or failure. • Each trial has the same probability of a success, p. Recall that if X is the binomial random variable, then X ~ B(n, p). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5; the approximation is better if they are both greater than or equal to 10. The product >5 is more or less accepted as the norm here.). This is another accepted rule. So, for whatever value of x we are looking at (the number of successes). We add 0.5 if we are looking for the probability that is less than or equal to that number. We subtract 0.5 if we are looking for the probability that is greater than or equal to that number. Then the binomial can be approximated by the normal distribution with mean μ = np and standard deviation σ = npq. Remember that q = 1 – p. In order to get the best approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x – 0.5). This is another accepted rule. So, for whatever value of x we are looking at (the number of successes). We add 0.5 if we are looking for the probability that is less than or equal to that number. We subtract 0.5 if we are looking for the probability that is greater than or equal to that number. The number 0.5 is called the continuity correction factor and is used in the following example. Example 7.12 Suppose in a local kindergarten through 12th grade (K–12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed. a. Find the probability that at least 150 favor a charter school. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 7 | The Central Limit Theorem 431 b. Find the probability that at most 160 favor a charter school. c. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.