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)2(0.2) = 1.352 2 4 6 8 0.1 0.3 0.4 0.2 Table 4.24 21. Identify the mistake in the probability distribution table. x P(x) x*P(x) 1 2 3 4 5 0.15 0.15 0.25 0.50 0.30 0.90 0.20 0.80 0.15 0.75 Table 4.25 22. Identify the mistake in the probability distribution table. x P(x) x*P(x) 1 2 3 4 5 0.15 0.15 0.25 0.40 0.25 0.65 0.20 0.85 0.15 1 Table 4.26 Use the following information to answer the next five exercises: A physics professor wants to know what percent of physics majors will spend the next several years doing post-graduate research. He has the following probability distribution. 264 CHAPTER 4 | DISCRETE RANDOM VARIABLES x P(x) x*P(x) 1 2 3 4 5 6 0.35 0.20 0.15 0.10 0.05 Table 4.27 23. Define the random variable X. 24. Define P(x), or the probability of x. 25. Find the probability that a physics major will do post-graduate research for four years. P(x = 4) = _______ 26. FInd the probability that a physics major will do post-graduate research for at most three years. P(x ≤ 3) = _______ 27. On average, how many years would you expect a physics major to spend doing post-graduate research? Use the following information to answer the next seven exercises: A ballet instructor is interested in knowing what percent of each year's class will continue on to the next, so that she can plan what classes to offer. Over the years, she has established the following probability distribution. • Let X = the number of years a student will study ballet with the teacher. • Let P(x) = the probability that a student will study ballet x years. 28. Complete Table 4.28 using the data provided. x P(x) x*P(x) 1 2 3 4 5 6 7 0.10 0.05 0.10 0.30 0.20 0.10 Table 4.28 29. In words, define the random variable X. 30. P(x = 4) = |
_______ 31. P(x < 4) = _______ 32. On average, how many years would you expect a child to study ballet with this teacher? 33. What does the column "P(x)" sum to and why? 34. What does the column "x*P(x)" sum to and why? 35. You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game? 36. You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. Should you play the game? 4.3 Binomial Distribution This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 Use the following information to answer the next eight exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly pick eight first-time, full-time freshmen from the survey. You are interested in the number that believes that same sex-couples should have the right to legal marital status. CHAPTER 4 | DISCRETE RANDOM VARIABLES 265 37. In words, define the random variable X. 38. X ~ _____(_____,_____) 39. What values does the random variable X take on? 40. Construct the probability distribution function (PDF). x P(x) Table 4.29 41. On average (μ), how many would you expect to answer yes? 42. What is the standard deviation (σ)? 43. What is the probability that at most five of the freshmen reply “yes”? 44. What is the probability that at least two of the freshmen reply “yes”? 4.4 Geometric Distribution Use the following information to answer the next six exercises |
: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies “yes.” You are interested in the number of freshmen you must ask. 45. In words, define the random variable X. 46. X ~ _____(_____,_____) 47. What values does the random variable X take on? 48. Construct the probability distribution function (PDF). Stop at x = 6. x P(x) 1 2 3 4 5 Table 4.30 266 CHAPTER 4 | DISCRETE RANDOM VARIABLES x P(x) 6 Table 4.30 49. On average (μ), how many freshmen would you expect to have to ask until you found one who replies "yes?" 50. What is the probability that you will need to ask fewer than three freshmen? 4.5 Hypergeometric Distribution Use the following information to answer the next five exercises: Suppose that a group of statistics students is divided into two groups: business majors and non-business majors. There are 16 business majors in the group and seven non-business majors in the group. A random sample of nine students is taken. We are interested in the number of business majors in the sample. 51. In words, define the random variable X. 52. X ~ _____(_____,_____) 53. What values does X take on? 54. Find the standard deviation. 55. On average (μ), how many would you expect to be business majors? 4.6 Poisson Distribution Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day. 56. Assume the event occurs independently in any given day. Define the random variable X. 57. What values does X take on? 58. What is the probability of getting 150 customers in one day? 59. What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day. 60. What is the probability that the store will have more than 12 customers in the first hour? 61. What is the probability that the store will have fewer than 12 customers in the first two hours? 62. Which type of distribution |
can the Poisson model be used to approximate? When would you do this? Use the following information to answer the next six exercises: On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age. 63. Assume the event occurs independently in any given day. In words, define the random variable X. 64. X ~ _____(_____,_____) 65. What values does X take on? 66. For the given values of the random variable X, fill in the corresponding probabilities. 67. Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S? Justify your answer numerically. 68. Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically. HOMEWORK 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable 69. Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given in Table 4.31. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 267 x P(x) 3 4 5 6 7 0.05 0.40 0.30 0.15 0.10 Table 4.31 In words, define the random variable X. a. b. What does it mean that the values zero, one, and two are not included for x in the PDF? 4.2 Mean or Expected Value and Standard Deviation 70. A theater group holds a fund-raiser. It sells 100 raffle tickets for $5 apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of $150. In words, define the random variable X. a. What are you interested in here? b. c. List the values that X may take on. d. Construct a PDF. e. If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle? 71. A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a fair |
coin and is equally likely to land on heads or tails. • • • If the card is a face card, and the coin lands on Heads, you win $6 If the card is a face card, and the coin lands on Tails, you win $2 If the card is not a face card, you lose $2, no matter what the coin shows. a. Find the expected value for this game (expected net gain or loss). b. Explain what your calculations indicate about your long-term average profits and losses on this game. c. Should you play this game to win money? 72. You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one $500 prize, two $100 prizes, and four $25 prizes. Find your expected gain or loss. 73. Complete the PDF and answer the questions. x P(x) xP(x) 0 1 2 3 0.3 0.2 0.4 Table 4.32 a. Find the probability that x = 2. b. Find the expected value. 74. Suppose that you are offered the following “deal.” You roll a die. If you roll a six, you win $10. If you roll a four or five, you win $5. If you roll a one, two, or three, you pay $6. In words, define the Random Variable X. a. What are you ultimately interested in here (the value of the roll or the money you win)? b. c. List the values that X may take on. d. Construct a PDF. e. Over the long run of playing this game, what are your expected average winnings per game? 268 CHAPTER 4 | DISCRETE RANDOM VARIABLES f. Based on numerical values, should you take the deal? Explain your decision in complete sentences. 75. A venture capitalist, willing to invest $1,000,000, has three investments to choose from. The first investment, a software company, has a 10% chance of returning $5,000,000 profit, a 30% chance of returning $1,000,000 profit, and a 60% chance of losing the million dollars. The second company, a hardware company, has a 20% chance of returning $3,000,000 profit, a 40% chance of returning $1,000,000 profit, and a 40% |
chance of losing the million dollars. The third company, a biotech firm, has a 10% chance of returning $6,000,000 profit, a 70% of no profit or loss, and a 20% chance of losing the million dollars. a. Construct a PDF for each investment. b. Find the expected value for each investment. c. Which is the safest investment? Why do you think so? d. Which is the riskiest investment? Why do you think so? e. Which investment has the highest expected return, on average? 76. Suppose that 20,000 married adults in the United States were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. Let X = the number of children married people have. x 0 1 2 3 4 5 P(x) xP(x) 0.10 0.20 0.30 0.10 0.05 6 (or more) 0.05 Table 4.33 In words, what does the expected value in this example represent? a. Find the probability that a married adult has three children. b. c. Find the expected value. d. Is it more likely that a married adult will have two to three children or four to six children? How do you know? 77. Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given as in Table 4.34. x P(x) 3 4 5 6 7 0.05 0.40 0.30 0.15 0.10 Table 4.34 On average, how many years do you expect it to take for an individual to earn a B.S.? 78. People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 269 x P(x) 0 1 2 3 4 5 0.03 0.50 0.24 0.70 0.04 Table 4.35 a. Describe the random variable X in words. b. Find the probability that a customer rents |
three DVDs. c. Find the probability that a customer rents at least four DVDs. d. Find the probability that a customer rents at most two DVDs. Another shop, Entertainment Headquarters, rents DVDs and video games. The probability distribution for DVD rentals per customer at this shop is given as follows. They also have a five-DVD limit per customer. x P(x) 0 1 2 3 4 5 0.35 0.25 0.20 0.10 0.05 0.05 Table 4.36 e. At which store is the expected number of DVDs rented per customer higher? f. If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form. If Video to Go expects 300 customers next week, and Entertainment HQ projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain. g. h. Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that? 79. A “friend” offers you the following “deal.” For a $10 fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift. • Ten of the coupons are for a free gift worth $6. • Eighty of the coupons are for a free gift worth $8. • Six of the coupons are for a free gift worth $12. • Four of the coupons are for a free gift worth $40. Based upon the financial gain or loss over the long run, should you play the game? a. Yes, I expect to come out ahead in money. b. No, I expect to come out behind in money. c. It doesn’t matter. I expect to break even. 80. Florida State University has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students. a. What is the average class size assuming each class is filled to capacity? b. Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable X equal the size of the student’s class. Define the PDF for X. 270 CHAPTER 4 | DISCRETE RANDOM |
VARIABLES c. Find the mean of X. d. Find the standard deviation of X. 81. In a lottery, there are 250 prizes of $5, 50 prizes of $25, and ten prizes of $100. Assuming that 10,000 tickets are to be issued and sold, what is a fair price to charge to break even? 4.3 Binomial Distribution 82. According to a recent article the average number of babies born with significant hearing loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf. Use the following information to answer the next four exercises. Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu. 83. Define the random variable and list its possible values. 84. State the distribution of X. 85. Find the probability that at least four of the 25 patients actually have the flu. 86. On average, for every 25 patients calling in, how many do you expect to have the flu? 87. People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given Table 4.37. There is five-video limit per customer at this store, so nobody ever rents more than five DVDs. x P(x) 0 1 2 3 4 5 0.03 0.50 0.24 0.07 0.04 Table 4.37 a. Describe the random variable X in words. b. Find the probability that a customer rents three DVDs. c. Find the probability that a customer rents at least four DVDs. d. Find the probability that a customer rents at most two DVDs. 88. A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students attend Tet festivities. We are interested in the number of students who will attend the festivities. In words, define the random variable X. a. b. List |
the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many of the 12 students do we expect to attend the festivities? e. Find the probability that at most four students will attend. f. Find the probability that more than two students will attend. Use the following information to answer the next two exercises: The probability that the San Jose Sharks will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games. 89. The expected number of wins for that upcoming month is: a. 1.67 b. 12 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 271 c. 382 1043 d. 4.43 Let X = the number of games won in that upcoming month. 90. What is the probability that the San Jose Sharks win six games in that upcoming month? a. 0.1476 b. 0.2336 c. 0.7664 d. 0.8903 91. What is the probability that the San Jose Sharks win at least five games in that upcoming month a. 0.3694 b. 0.5266 c. 0.4734 d. 0.2305 92. A student takes a ten-question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70% of the questions correct. 93. A student takes a 32-question multiple-choice exam, but did not study and randomly guesses each answer. Each question has three possible choices for the answer. Find the probability that the student guesses more than 75% of the questions correctly. 94. Six different colored dice are rolled. Of interest is the number of dice that show a one. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. On average, how many dice would you expect to show a one? e. Find the probability that all six dice show a one. f. Is it more likely that three or that |
four dice will show a one? Use numbers to justify your answer numerically. In words, define the random variable X. 95. More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. On average, how many schools would you expect to offer such courses? e. Find the probability that at most ten offer such courses. f. Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence. In words, define the random variable X. 96. Suppose that about 85% of graduating students attend their graduation. A group of 22 graduating students is randomly chosen. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many are expected to attend their graduation? e. Find the probability that 17 or 18 attend. f. Based on numerical values, would you be surprised if all 22 attended graduation? Justify your answer numerically. 97. At The Fencing Center, 60% of the fencers use the foil as their main weapon. We randomly survey 25 fencers at The Fencing Center. We are interested in the number of fencers who do not use the foil as their main weapon. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many are expected to not to use the foil as their main weapon? e. Find the probability that six do not use the foil as their main weapon. f. Based on numerical values, would you be surprised if all 25 did not use foil as their main weapon? Justify your answer numerically. 98. Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school. a. b. List the values that X may take on. In words, define the random variable X. 272 CHAPTER 4 | DISCRE |
TE RANDOM VARIABLES c. Give the distribution of X. X ~ _____(_____,_____) d. How many seniors are expected to have participated in after-school sports all four years of high school? e. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. f. Based upon numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. 99. The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many audits are expected in a 20-year period? e. Find the probability that a person is not audited at all. f. Find the probability that a person is audited more than twice. 100. It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. What is the probability that at least eight have adequate earthquake supplies? e. f. How many residents do you expect will have adequate earthquake supplies? Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why? 101. There are two similar games played for Chinese New Year and Vietnamese New Year. In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with a board with those numbers. In the Vietnamese version, fair dice with pictures of a gourd, fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also. We will play with bets being $1. The player places a bet on a number or object. The “house” rolls three dice |
. If none of the dice show the number or object that was bet, the house keeps the $1 bet. If one of the dice shows the number or object bet (and the other two do not show it), the player gets back his or her $1 bet, plus $1 profit. If two of the dice show the number or object bet (and the third die does not show it), the player gets back his or her $1 bet, plus $2 profit. If all three dice show the number or object bet, the player gets back his or her $1 bet, plus $3 profit. Let X = number of matches and Y = profit per game. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. List the values that Y may take on. Then, construct one PDF table that includes both X and Y and their probabilities. e. Calculate the average expected matches over the long run of playing this game for the player. f. Calculate the average expected earnings over the long run of playing this game for the player. g. Determine who has the advantage, the player or the house. 102. According to The World Bank, only 9% of the population of Uganda had access to electricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let X = the number of people who have access to electricity. a. What is the probability distribution for X? b. Using the formulas, calculate the mean and standard deviation of X. c. Use your calculator to find the probability that 15 people in the sample have access to electricity. d. Find the probability that at most ten people in the sample have access to electricity. e. Find the probability that more than 25 people in the sample have access to electricity. 103. The literacy rate for a nation measures the proportion of people age 15 and over that can read and write. The literacy rate in Afghanistan is 28.1%. Suppose you choose 15 people in Afghanistan at random. Let X = the number of people who are literate. a. Sketch a graph of the probability distribution of X. b. Using the formulas, calculate the (i) mean and (ii) standard deviation of X. c. Find the probability that more than five people in the sample are literate. Is it is more likely that three people or four people are literate. 4.4 Geometric Distribution |
104. A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call. In words, define the random variable X. a. b. List the values that X may take on. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 273 c. Give the distribution of X. X ~ _____(_____,_____) d. On average, how many dealerships would we expect her to have to call until she finds one that has the car? e. Find the probability that she must call at most four dealerships. f. Find the probability that she must call three or four dealerships. In words, define the random variable X. 105. Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many adults in America do you expect to survey until you find one who will watch the Super Bowl? e. Find the probability that you must ask seven people. f. Find the probability that you must ask three or four people. 106. It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. What is the probability that we must survey just one or two residents until we find a California resident who does not have adequate earthquake supplies? e. What is the probability that we must survey at least three California residents until we find a California resident who does not have adequate earthquake supplies? f. How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies |
? g. How many California residents do you expect to need to survey until you find a California resident who does have adequate earthquake supplies? 107. In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many pages do you expect to advertise footwear on them? e. f. What is the probability that fewer than ten will advertise footwear on them? g. Reminder: A page may be picked more than once. We are interested in the number of pages that we must randomly survey until we find one that has footwear advertised on it. Define the random variable X and give its distribution. Is it probable that all twenty will advertise footwear on them? Why or why not? h. What is the probability that you only need to survey at most three pages in order to find one that advertises footwear on it? i. How many pages do you expect to need to survey in order to find one that advertises footwear? 108. Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome. • p = probability of success (event F occurs) • q = probability of failure (event F does not occur) a. Write the description of the random variable X. b. What are the values that X can take on? c. Find the values of p and q. d. Find the probability that the first occurrence of event F (rolling a four or five) is on the second trial. 109. Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. What values does X take on? 110. The World Bank records the prevalence of HIV in countries around the world. According to their data, “Prevalence of HIV refers to the percentage of people ages 15 to 49 who |
are infected with HIV.”[1] In South Africa, the prevalence of HIV is 17.3%. Let X = the number of people you test until you find a person infected with HIV. 1. ”Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/ 274 CHAPTER 4 | DISCRETE RANDOM VARIABLES a. Sketch a graph of the distribution of the discrete random variable X. b. What is the probability that you must test 30 people to find one with HIV? c. What is the probability that you must ask ten people? d. Find the (i) mean and (ii) standard deviation of the distribution of X. 111. According to a recent Pew Research poll, 75% of millenials (people born between 1981 and 1995) have a profile on a social networking site. Let X = the number of millenials you ask until you find a person without a profile on a social networking site. a. Describe the distribution of X. b. Find the (i) mean and (ii) standard deviation of X. c. What is the probability that you must ask ten people to find one person without a social networking site? d. What is the probability that you must ask 20 people to find one person without a social networking site? e. What is the probability that you must ask at most five people? 4.5 Hypergeometric Distribution 112. A group of Martial Arts students is planning on participating in an upcoming demonstration. Six are students of Tae Kwon Do; seven are students of Shotokan Karate. Suppose that eight students are randomly picked to be in the first demonstration. We are interested in the number of Shotokan Karate students in that first demonstration. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many Shotokan Karate students do we expect to be in that first demonstration? 113. In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked at most once. In words, define the random variable X. a. b. List the |
values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many pages do you expect to advertise footwear on them? e. Calculate the standard deviation. 114. Suppose that a technology task force is being formed to study technology awareness among instructors. Assume that ten people will be randomly chosen to be on the committee from a group of 28 volunteers, 20 who are technically proficient and eight who are not. We are interested in the number on the committee who are not technically proficient. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many instructors do you expect on the committee who are not technically proficient? e. Find the probability that at least five on the committee are not technically proficient. f. Find the probability that at most three on the committee are not technically proficient. 115. Suppose that nine Massachusetts athletes are scheduled to appear at a charity benefit. The nine are randomly chosen from eight volunteers from the Boston Celtics and four volunteers from the New England Patriots. We are interested in the number of Patriots picked. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. Are you choosing the nine athletes with or without replacement? 116. A bridge hand is defined as 13 cards selected at random and without replacement from a deck of 52 cards. In a standard deck of cards, there are 13 cards from each suit: hearts, spades, clubs, and diamonds. What is the probability of being dealt a hand that does not contain a heart? a. What is the group of interest? b. How many are in the group of interest? c. How many are in the other group? d. Let X = _________. What values does X take on? e. The probability question is P(_______). f. Find the probability in question. g. Find the (i) mean and (ii) standard deviation of X. 4.6 Poisson Distribution SH.DYN.AIDS.ZS?order=wbapi_data_value_2011+wbapi_data_value+wbapi_data_value-last&sort=desc (accessed May 15, 2013). This content is available for free at http://textbooke |
quity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 275 117. The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let X = the number of calls received at noon. a. Find the mean and standard deviation of X. b. What is the probability that the office receives at most six calls at noon on Monday? c. Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon? d. What is the probability that the office receives more than eight calls at noon? 118. The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let X = the number of births in an hour. a. Find the mean and standard deviation of X. b. Sketch a graph of the probability distribution of X. c. What is the probability that the maternity ward will deliver three babies in one hour? d. What is the probability that the maternity ward will deliver at most three babies in one hour? e. What is the probability that the maternity ward will deliver more than five babies in one hour? 119. A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions. 120. The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. Find the probability that she has no children. e. Find the probability that she has fewer children than the Japanese average. f. Find the probability that she has more children than the Japanese average. 121. The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen. In words, define the Random Variable X. a. b. List the values that X may take on |
. c. Give the distribution of X. X ~ _____(_____,_____) d. Find the probability that she has no children. e. Find the probability that she has fewer children than the Spanish average. f. Find the probability that she has more children than the Spanish average. 122. Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. In one year, find the probability she produces: In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _______ d. Find the probability that she has no litters in one year. e. Find the probability that she has at least two litters in one year. f. Find the probability that she has exactly three litters in one year. 123. The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many cookies do we expect to have an extra fortune? e. Find the probability that none of the cookies have an extra fortune. f. Find the probability that more than three have an extra fortune. g. As n increases, what happens involving the probabilities using the two distributions? Explain in complete sentences. 124. According to the South Carolina Department of Mental Health web site, for every 200 U.S. women, the average number who suffer from anorexia is one. Out of a randomly chosen group of 600 U.S. women determine the following. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution ofX. X ~ _____(_____,_____) d. How many are expected to suffer from anorexia? 276 CHAPTER 4 | DISCRETE RANDOM VARIABLES e. Find the probability that no one suffers from anorexia. f. Find the probability that more than four suffer from anorexia. 125. The chance of an IRS audit for a tax return with over $25,000 in income is about |
2% per year. Suppose that 100 people with tax returns over $25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to anwer the following questions. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many are expected to be audited? e. Find the probability that no one was audited. f. Find the probability that at least three were audited. 126. Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. How many seniors are expected to have participated in after-school sports all four years of high school? e. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all In words, define the random variable X. four years of high school? Justify your answer numerically. f. Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically. 127. On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose that you buy one of his cakes. In words, define the random variable X. a. b. List the values that X may take on. c. Give the distribution of X. X ~ _____(_____,_____) d. On average, how many pieces of egg shell do you expect to be in the cake? e. What is the probability that there will not be any pieces of egg shell in the cake? f. Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not be any egg shell in any of the cakes? g. Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why? Use the following information to answer the next two exercises: The average number of times per week that |
Mrs. Plum’s cats wake her up at night because they want to play is ten. We are interested in the number of times her cats wake her up each week. 128. In words, the random variable X = _________________ a. b. c. d. the number of times Mrs. Plum’s cats wake her up each week. the number of times Mrs. Plum’s cats wake her up each hour. the number of times Mrs. Plum’s cats wake her up each night. the number of times Mrs. Plum’s cats wake her up. 129. Find the probability that her cats will wake her up no more than five times next week. a. 0.5000 b. 0.9329 c. 0.0378 d. 0.0671 REFERENCES 4.2 Mean or Expected Value and Standard Deviation Class Catalogue at SearchFormLegacy (accessed May 15, 2013). the Florida State University. Available online at https://apps.oti.fsu.edu/RegistrarCourseLookup/ “World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012. http://www.worldearthquakes.com/index.php?option=ethq_prediction (accessed May 15, 2013). 4.3 Binomial Distribution “Access to electricity (% of population),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/ EG.ELC.ACCS.ZS?order=wbapi_data_value_2009%20wbapi_data_value%20wbapi_data_value-first&sort=asc (accessed May 15, 2015). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 277 “Distance Education.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Distance_education (accessed May 15, 2013). “NBA Statistics – 2013,” ESPN NBA, 2013. Available online at http://espn.go.com/nba/statistics/_/seasontype/2 (accessed May 15, 2013). Newport, Frank. “Americans Still Enjoy Saving Rather than Spending |
: Few demographic differences seen in these views other than by income,” GALLUP® Economy, 2013. Available online at http://www.gallup.com/poll/162368/americansenjoy-saving-rather-spending.aspx (accessed May 15, 2013). Pryor, John H., Linda DeAngelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Fall 2011. Los Angeles: Cooperative Institutional Research Program at the Higher Education Research Institute at http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/ at Also TheAmericanFreshman2011.pdf (accessed May 15, 2013). available UCLA, online 2011. “The World FactBook,” Central Intelligence Agency. Available online at https://www.cia.gov/library/publications/theworld-factbook/geos/af.html (accessed May 15, 2013). “What are the key statistics about pancreatic cancer?” American Cancer Society, 2013. Available online at http://www.cancer.org/cancer/pancreaticcancer/detailedguide/pancreatic-cancer-key-statistics (accessed May 15, 2013). 4.4 Geometric Distribution “Millennials: A Portrait of Generation Next,” PewResearchCenter. Available online at http://www.pewsocialtrends.org/ files/2010/10/millennials-confident-connected-open-to-change.pdf (accessed May 15, 2013). “Millennials: Confident. Connected. Open to Change.” Executive Summary by PewResearch Social & Demographic Trends, 2013. Available online at http://www.pewsocialtrends.org/2010/02/24/millennials-confident-connected-open-tochange/ (accessed May 15, 2013). total “Prevalence of HIV, http://data.worldbank.org/indicator/ SH.DYN.AIDS.ZS?order=wbapi_data_value_2011+wbapi_data_value+wbapi_data_value-last&sort=desc (accessed May 15, 2013). (% of populations ages 15-49),” The World Bank, 2013. Available online at Pryor, John H., Linda De |
Angelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Fall 2011. Los Angeles: Cooperative Institutional Research Program at the Higher Education Research Institute at http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/ at Also TheAmericanFreshman2011.pdf (accessed May 15, 2013). available UCLA, online 2011. “Summary of the National Risk and Vulnerability Assessment 2007/8: A profile of Afghanistan,” The European Union and ICON-Institute. Available online at http://ec.europa.eu/europeaid/where/asia/documents/afgh_brochure_summary_en.pdf (accessed May 15, 2013). “The World FactBook,” Central Intelligence Agency. Available online at https://www.cia.gov/library/publications/theworld-factbook/geos/af.html (accessed May 15, 2013). “UNICEF reports on Female Literacy Centers in Afghanistan established to teach women and girls basic resading [sic] and writing skills,” UNICEF Television. Video available online at http://www.unicefusa.org/assets/video/afghan-femaleliteracy-centers.html (accessed May 15, 2013). 4.6 Poisson Distribution “ATL Fact Sheet,” Department of Aviation at the Hartsfield-Jackson Atlanta International Airport, 2013. Available online at http://www.atlanta-airport.com/Airport/ATL/ATL_FactSheet.aspx (accessed May 15, 2013). Center for Disease Control and Prevention. “Teen Drivers: Fact Sheet,” Injury Prevention & Control: Motor Vehicle Safety, October 2, 2012. Available online at http://www.cdc.gov/Motorvehiclesafety/Teen_Drivers/teendrivers_factsheet.html (accessed May 15, 2013). “Children and Childrearing,” Ministry of Health, Labour, and Welfare. Available online at http://www.mhlw.go.jp/english/ policy/children/children-childrearing/index.html (accessed May 15, 2013). “Eating Disorder Statistics,” South Carolina Department of Mental Health, 2006. Available online at http://www. |
state.sc.us/ dmh/anorexia/statistics.htm (accessed May 15, 2013). “Giving Birth in Manila: The maternity ward at the Dr Jose Fabella Memorial Hospital in Manila, the busiest in the Philippines, where at http://www.theguardian.com/world/gallery/2011/jun/08/philippines-health#/?picture=375471900&index=2 (accessed May 15, 2013). theguardian, 2013. Available online an average of 60 births a day,” there is “How Americans Use Text Messaging,” Pew Internet, 2013. Available online at http://pewinternet.org/Reports/2011/CellPhone-Texting-2011/Main-Report.aspx (accessed May 15, 2013). 278 CHAPTER 4 | DISCRETE RANDOM VARIABLES Lenhart, Amanda. “Teens, Smartphones & Testing: Texting volum is up while the frequency of voice calling is down. About one in four teens say they own smartphones,” Pew Internet, 2012. Available online at http://www.pewinternet.org/~/media/ Files/Reports/2012/PIP_Teens_Smartphones_and_Texting.pdf (accessed May 15, 2013). “One born every minute: the maternity unit where mothers are THREE to a bed,” MailOnline. Available online at http://www.dailymail.co.uk/news/article-2001422/Busiest-maternity-ward-planet-averages-60-babies-day-mothersbed.html (accessed May 15, 2013). Vanderkam, http://management.fortune.cnn.com/2012/10/08/stop-checking-your-email-now/ (accessed May 15, 2013). Email, Now.” CNNMoney, “Stop Checking Your 2013. Available Laura. online at “World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012. http://www.worldearthquakes.com/index.php?option=ethq_prediction (accessed May 15, 2013). SOLUTIONS 1 x P(x) 0 1 2 3 4 5 6 0.12 0.18 0.30 0.15 0.10 0.10 0.05 Table 4.38 3 0. |
10 + 0.05 = 0.15 5 1 7 0.35 + 0.40 + 0.10 = 0.85 9 1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45 11 x P(x) 0 1 2 3 0.03 0.04 0.08 0.85 Table 4.39 13 Let X = the number of events Javier volunteers for each month. 15 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 279 x P(x) 0 1 2 3 4 5 0.05 0.05 0.10 0.20 0.25 0.35 Table 4.40 17 1 – 0.05 = 0.95 19 0.2 + 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 post-graduate research. 25 1 – 0.35 – 0.20 – 0.15 – 0.10 – 0.05 = 0.15 27 1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years 29 X is the number of years a student studies ballet with the teacher. 31 0.10 + 0.05 + 0.10 = 0.25 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 43 0.4151 45 X = the number of freshmen selected from the study until one replied " |
yes" that same-sex couples should have the right to legal marital status. 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 0.0485 61 0.0214 280 CHAPTER 4 | DISCRETE RANDOM VARIABLES 63 X = the number of U.S. 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. 0.1 b |
. 1.6 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 281 75 a. Software Company x 5,000,000 1,000,000 –1,000,000 Table 4.42 P(x) 0.10 0.30 0.60 Hardware Company x 3,000,000 1,000,000 –1,000,00 Table 4.43 P(x) 0.20 0.40 0.40 Biotech Firm x 6,00,000 0 P(x) 0.10 0.70 –1,000,000 0.20 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 282 CHAPTER 4 | DISCRETE RANDOM VARIABLES 81 Let X = the amount of money to be won on a ticket. The following table shows the PDF for X. x 0 5 25 100 P(x) 0.969 250 10,000 50 10,000 10 10,000 = 0.025 = 0.005 = 0.001 Table 4.45 Calculate the expected value of X. 0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35 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 0.0165 87 a. X = the number of DVDs a Video to Go customer rents b. 0.12 c. 0.11 d. 0.77 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% of 32 questions correct. 75% 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% 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. 0.0135 f. P(x = 12) = 0.3186 P(x = 13) = 0.5882 More likely to get 13. 97 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 283 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(25,0.40) d. 10 e. 0.0442 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, 0.02) d. 0.4 e. 0.6676 f. 0.0071 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 −0.0787. You lose about eight cents, on average, per game. 7. The house has the advantage. 103 a. X ~ B(15, 0.281) Figure 4.10 b. i. Mean = μ = np = 15(0.281) = 4.215 ii. Standard Deviation = σ = npq = 15(0.281)(0.719) = 1.7409 284 CHAPTER 4 | DISCRE |
TE RANDOM VARIABLES c. P(x > 5) = 1 – P(x ≤ 5) = 1 – binomcdf(15, 0.281, 5) = 1 – 0.7754 = 0.2246 P(x = 3) = binompdf(15, 0.281, 3) = 0.1927 P(x = 4) = binompdf(15, 0.281, 4) = 0.2259 It is more likely that four people are literate that three people are. 105 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(0.40) c. 2.5 d. 0.0187 e. 0.2304 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. 0.9997 g. X = the number of pages we must survey until we find one that advertises footwear. X ~ G( 29 192 ) h. 0.3881 i. 6.6207 pages 109 0, 1, 2, and 3 111 a. X ~ G(0.25) b. i. Mean = μ = 1 p = 1 0.25 = 4 ii. Standard Deviation = σ = 1 − p p2 = 1 − 0.25 0.252 ≈ 3.4641 c. P(x = 10) = geometpdf(0.25, 10) = 0.0188 d. P(x = 20) = geometpdf(0.25, 20) = 0.0011 e. P(x ≤ 5) = geometcdf(0.25, 5) = 0.7627 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) This content is available for free at http://textbookequity.org/ |
introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 4 | DISCRETE RANDOM VARIABLES 285 d. Without replacement 117 a. X ~ P(5.5); μ = 5.5; σ = 5.5 ≈ 2.3452 b. P(x ≤ 6) = poissoncdf(5.5, 6) ≈ 0.6860 c. There is a 15.7% 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 – 0.8944 = 0.1056 119 Let X = the number of defective bulbs in a string. Using the Poisson distribution: • μ = np = 100(0.03) = 3 • X ~ P(3) • P(x ≤ 4) = poissoncdf(3, 4) ≈ 0.8153 Using the binomial distribution: • X ~ B(100, 0.03) • P(x ≤ 4) = binomcdf(100, 0.03, 4) ≈ 0.8179 The Poisson approximation is very good—the difference between the probabilities is only 0.0026. 121 a. X = the number of children for a Spanish woman b. 0, 1, 2, 3,... c. X ~ P(1.47) d. 0.2299 e. 0.5679 f. 0.4321 123 a. X = the number of fortune cookies that have an extra fortune b. 0, 1, 2, 3,... 144 c. X ~ B(144, 0.03) or P(4.32) d. 4.32 e. 0.0124 or 0.0133 f. 0.6300 or 0.6264 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. 0.1353 f. 0.3233 127 a. X = the number of shell pieces in one cake b. 0, 1, 2, 3,... c. X ~ P(1.5) d. 1. |
5 286 CHAPTER 4 | DISCRETE RANDOM VARIABLES e. 0.2231 f. 0.0001 g. Yes 129 d This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 287 5 | CONTINUOUS RANDOM VARIABLES Figure 5.1 The heights of these radish plants are continuous random variables. Introduction Chapter Objectives By the end of this chapter, the student should be able to: • 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 long distance 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. 288 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES NOTE 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 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, the formulas 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 using a continuous probability distribution to model probability, 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 content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 289 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. Figure 5.5 290 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 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 1 20 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 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. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS |
RANDOM VARIABLES 291 Figure 5.7 ⎛ AREA = (15 – 4) ⎝ ⎛ AREA = (15 – 4) ⎝ ⎞ ⎠ = 0.55 ⎞ ⎠ = 0.55 1 20 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) (can 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 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). 292 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 is inclusive or exclusive. Example 5.2 The data in Table 5.1 are 55 smiling times, in seconds, of an eight-week-old baby. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 293 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 µ = a + b 2 and σ = (b − a)2 12 For this problem, the theoretical mean and standard deviation are 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. 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 5.2 294 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 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 a. Find P(2 < x < 18). 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 P(x < k) = 0.90 ⎛ ⎝base⎞ ⎛ ⎠ ⎝height⎞ ⎠ = 0.90 ⎛ (k − 0) ⎝ 1 23 ⎞ ⎠ = 0.90 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 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 295 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. 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 X ~ U (0, 23): P(A|B) = P(A AND B) P(B) For this problem, A is (x > 12) and B is (x > 8). 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 5.14 296 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 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 less than 12.5 minutes is 0.8333. 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 time a person must wait falls below what value? NOTE This asks for the 90th percentile. Solution 5.4 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 297 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. 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 donut. Then X ~ U (0.5, 4). a. The probability that a randomly selected nine-year old child eats a donut 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 donut in more than two minutes given that the child has already been eating the don |
ut for more than 1.5 minutes. 298 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES The second question has a conditional probability. You are asked to find 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 donut for more than 1.5 minutes. Solve the problem two different ways (see Example 5.2). You must reduce the sample space. First way: Since you know the child has already been eating the donut 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): f(x) = 1 4 − 1.5 = 2 5 for 1.5 ≤ x ≤ 4. Find P(x > 2|x > 1.5). Draw a graph. Figure 5.17 P(x > 2|x > 1.5) = (base)(new height) = (4 − 2) ⎛ ⎝ =? ⎞ ⎠ 2 5 Solution 5.5 b. 4 5 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 donut 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). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 299 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% of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25% of repair times.) What percentile does this represent? e. Find the mean and standard deviation 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; since X ~ U (1.5, 4), x can not 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. 300 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES Figure 5.20 Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% 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% 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% 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% of furnace repairs take at least 3.375 hours (3.375 hours or longer). Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times. Solution 5.6 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 301 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 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. The exponential distribution is widely used in the field of reliability. Reliability deals with the amount 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 postal clerk spends five minutes with the customers. The graph is as follows: 302 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 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 Exercise 5.0, 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 Figure 5.23 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 303 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: 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) = 2.8 minutes. The calculator simplifies the calculation for percentile k. See the Solve for k: k = ln(0. |
50) -0.25 following two notes. NOTE A formula for the percentile k is k = ln(1 − AreaToTheLe f t) −m where ln is the natural log. 304 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 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 ten 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 $.40 in his or her pocket or purse. (Shade P(x < 0.40)). Example 5.9 On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed. a. What is the probability that a computer part lasts more than 7 years? Solution 5.9 a. Let x = the amount of time (in years) a computer part lasts. μ = 10 so m = 1 = 0.1 µ = 1 10 Find P(x > 7). Draw the graph. 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. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 305 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 ten 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. Figure 5.26 Solve for k: k = ln(1 – 0.80) – 0.1 = 16.1 years Eighty percent of the computer parts last at most 16.1 years. 306 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 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. 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 =. If another person arrives at a public telephone just before you, find the probability that you will have to wait 1 12 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 _______. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 307 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. 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(–0.5x)e. Therefore P(X < 1) = 1 – e(–0.5)(1) ≈ 0.3935. 1 - e^(–0.5) ≈ 0.3935 308 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES Figure 5.28 d. P(X > 5) = 1 – P(X < 5) = 1 – (1 – e(–5)(0.5)) = e–2.5 ≈ 0.0821. Figure 5.29 1 – (1 – e^( – 5*0.5)) or e^( – 5*0.5) 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, seventy 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 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 309 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 busier 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 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 that P (X > r + t | X > r) = P (X > t) for all r ≥ 0 and t ≥ 0 For example, if five minutes has 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 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. 310 CHAPTER 5 |
| CONTINUOUS RANDOM VARIABLES 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*2)) = e^(–0.25*2). 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 over 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!. Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. (k! = k*(k-1*)(k–2)*(k-3)…3*2*1) This content is available for free at http://textbookequity. |
org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 311 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 ten seconds. c. Find the probability that exactly five calls occur within a minute. d. Find the probability that less 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 there are four calls occur per minute, so 15 seconds, or 15 60 successive calls on average. = 0.25 minutes occur between b. Let T = time elapsed between calls. From part a, μ = 0.25, so m = 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 ten seconds (ten seconds = 1/6 minute) is P⎛ ⎝ – 41 6 ≈ 0.4866. 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! ≈ 0.1563. (5! = (5)(4)(3)(2)(1)) 312 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES poissonpdf(4, 5) = 0.1563. 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 = 0.0707. 1 – poissoncdf(32, 40). = 0.0707 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 there are at most 2 accidents occur in any given week. b. What is the probability that there is at least two weeks between any 2 accidents? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 313 5.1 Continuous Distribution Class Time: Names: 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. 314 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? Plot the Data 1. Construct a box plot of the data. Be sure to use a ruler to scale accurately and draw straight edges. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 315 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? 316 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 m and the standard deviation is σ = 1 notation is X ~ Exp(m). The mean is μ = 1 m. The probability density function is f(x) = me−mx, x ≥ 0 and the cumulative distribution function is P(X ≤ x) = 1 − e−mx. memoryless property For an exponential random variable X, 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). the memoryless property is the statement Poisson distribution 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; it is often referred as the rectangular distribution because the graph of the pdf has the form of a rectangle. 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). 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 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 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 =. The standard deviation of X is a + b 2 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 317 (b − a)2 σ = 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 If X has an exponential distribution with mean μ, then the decay parameter is m = 1 x ≥ 0 and m > 0. The probability density function of X is f(x) = me-mx (or equivalently f (x) = 1 distribution function of X is P(X ≤ x) = 1 – e–mx. µ, and we write X ∼ Exp(m) where µe − 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 PX is (X = k) = λk e−k k!. This may be computed 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 The standard deviation is σ = (b – a)2 12 Probability density function (pdf) f(x): • f(x) ≥ 0 Probability density function: f (x) = 1 b − a for • The total area under the curve f(x) is one. a ≤ X ≤ b Cumulative distribution function (cdf): P( |
X ≤ x) 5.2 The Uniform Distribution X = a real number between a and b (in some instances, X can take on the values a and b). a = smallest X; b = largest X X ~ U (a, b) The mean is µ = a + b 2 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 318 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES • pdf: f (x) = 1 b − a for a ≤ x ≤ b • mean µ = 1 m • 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) ( 1 b – a) standard deviation σ = µ • • 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) 5.3 The Exponential Distribution • Poisson probability: P(X = k) = Exponential: X ~ Exp(m) where m = the decay parameter λ • pdf: f(x) = me(–mx) where x ≥ 0 and m > 0 • k! = k*(k-1)*(k-2)*(k-3)…3*2*1 λk e−k k! with mean • cdf: P(X ≤ x) = 1 – e(–mx) PRACTICE 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? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 319 Figure 5.39 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)? 12. Find the probability that x falls in the shaded area., and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < 320 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 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 and the function is restricted to 1 ≤ x ≤ 4. Describe P⎛ ⎝x > 3 2 ⎞ ⎠. 5.2 The Uniform Distribution Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes. 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 Table 5.4 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 321 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 eleven 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. 40. Graph the probability distribution. a. Sketch the graph of the probability distribution. 322 CHAPTER 5 | CONTINUOUS RANDOM |
VARIABLES 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) = _______ 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. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 323 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%, 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 Use the following information to answer the next ten 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? 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. 324 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES Use the following information to answer the next seven 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 16 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. This means, 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) = __________ 70. Find the percentage of carbon-14 lasting longer than 10,000 years. a. Sketch the graph, and shade the area of interest. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 Figure 5.50 b. Find the probability. P(x > 10,000) = ________ 71. Thirty percent (30%) of carbon-14 will decay within how many years? a. Sketch the graph, and shade the area of interest. CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 325 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 the |
percent 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. 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) = _________ 326 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES i. Find the 90th percentile. 76. According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena Hospital, 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 ten pounds in a month. h. Suppose it is known that the individual lost more than ten 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 in a probability question, similarly to parts g and h, draw the picture, and find the probability. 77. A subway train on the Red Line 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 in a probability question, similarly to parts g and h, draw the picture, and find the probability. 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 b. two c. three d |
. four 80. Find the 30th percentile for the waiting times (in minutes). a. two b. 2.4 c. 2.75 d. three 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 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 327 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 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% 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 is between $19 and $22. c. Find the upper quartile - 25% 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. 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 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 drivers goes between 400 and 650 miles in a day. c. At least how many miles does the truck driver travel on the furthest 10% of days? 5.3 The Exponential Distribution 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. a. Define the random variable. X = _________________________________. b. c. X ~ ________ Is X continuous or |
discrete? 328 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES d. μ = ________ e. σ = ________ f. Draw a graph of the probability distribution. Label the axes. g. Find the probability that the percent is less than 12. h. Find the probability that the percent 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%? ii. What would make this number higher than 9.848%? 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. The decay rate is: 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. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 329 b. Find the probability that a light bulb lasts between six and ten 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. 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 less 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 3 no-hitters in a single season? 98. During the years 1998–2012, a total of 29 earthquakes of magnitude greater than 6.5 have occurred in Papua New Guinea. Assume that the time spent waiting between earthquakes is exponential. 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 U.S. 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. If 100 people arrive, how many on average would be expected to have Type B blood? a. b. What is the probability that over 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 web site 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 web site is more than ten minutes. b. The top 25% of durations between visits are at least how long? c. Suppose that 20 minutes have passed since the last visit to the web site. What is the probability that the next visit will occur within the next 5 minutes? d. Find the probability that less than 7 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 2 minutes. b. Find the probability that the time between two successive visits to the urgent care facility is more than 15 minutes. c. If 10 minutes have passed since the last arrival, what is the probability that the next person will arrive within the 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, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995. 5.3 The Exponential Distribution Data from the United States Census Bureau. Data from World Earthquakes, 2013. Available online at http://www.world-earthquakes.com/ (accessed June 11, 2013). 330 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES “No-hitter.” Baseball-Reference.com, 2013. Available online at http://www.baseball-reference.com/bullpen/No-hitter (accessed June 11, 2013). Zhou, Rick. “Exponential Distribution lecture slides.” Available online at www.public.iastate.edu/~riczw/stat330s11/ lecture/lec13.pdf (accessed June 11, 2013). 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 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 331 where x is between 0.5 and 9.5, inclusive. 39 f(x) = 1 9 41 μ = |
5 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 332 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 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) 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 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 333 Figure 5.54 b. P(19 < X < 22) = (22 – 19 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 = 500 400 = 1 8 = 0.125. b. P(400 < X < 650) = 700 − 650 700 − 300 = 250 400 = 0.625 c. 0.10 = width 700 − 300, so width = 400(0.10) = 40. Since 700 – 40 = 660, the drivers travel at least 660 miles on the furthest 10% of days. 87 a. X = the useful life of a particular car battery, measured in months. b. X is continuous. c. X ~ Exp(0.025) 334 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 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. e. five 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 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES 335 c. We want to find 0.70 = P(T > t) = 1 – ⎛ ⎜. ⎞ ⎟ = e ⎠ Solving for t, e – t 8 = 0.70, so – Or use t = ln(area_to_the_right) ( – m) t 8 = = ln(0.70), and t = –8ln(0.70) ≈ 2.85 years. ln(0.70) – 1 8 ≈ 2.85 years. Figure 5.57 d. We want to find 0.02 = P(T < t.98, so – Solving for t, e – t 8 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 3 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. 336 CHAPTER 5 | CONTINUOUS RANDOM VARIABLES c. Let X = the number of no-hitters is a season. Assume that X is Poisson 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. 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 ⎛ ⎝ 20 ⎞ ⎠ 8 9 ≈ 0.0948. (The geometric distribution is more appropriate than the exponential because the number of people between Type |
B people is discrete instead of continuous.) 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 content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 337 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: • 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. Some of these include 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 them. 338 CHAPTER 6 | THE NORMAL DISTRIBUTION The normal distribution has two parameters (two numerical descriptive measures), 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 probability density function is a rather complicated function. Do not memorize it. It is not necessary. f(x The cumulative distribution function is P(X < x). It is calculated either by a calculator or a computer, or it is looked up in a table. Technology has made the tables virtually obsolete. For that reason, as well as the fact that there are various table formats, we are not including table instructions. The curve is symmetrical 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. As the notation indicates, the normal distribution depends only on the mean and the standard deviation. 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. This means there are an infinite number of normal probability distributions. One 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 one 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 standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard deviation. For example, |
if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows: x = μ + (z)(σ) = 5 + (3)(2) = 11 The z-score is three. The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation z = x − µ σ produces the distribution Z ~ N(0, 1). The value x comes from a normal distribution with mean μ and standard deviation σ. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 339 Z-Scores 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. Example 6.1 Suppose X ~ N(5, 6). This 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. The standard deviation is σ = 6. 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. Notice that: 5 + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1) 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. 6.1 What is the z-score of x, when x = 1 and X ~ N(12,3)? Example 6.2 Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. 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 ten 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. 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? 340 CHAPTER 6 | THE NORMAL DISTRIBUTION 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 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 ten 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 says the following: • About 68% of the x values lie between –1σ and +1σ of the mean µ (within one standard deviation of the mean). • About 95% of the x values lie between –2σ and +2σ of the mean µ (within two standard deviations of the mean). • About 99.7% 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. • The z-scores for +2σ and –2σ are +2 and –2, respectively. • The z-scores for +3σ and –3σ are +3 and –3 respectively. 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 to 2010. Then X ~ N(170, 6.28). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 341 a. Suppose a 15 to 18-year |
-old male from Chile was 168 cm tall from 2009 to 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 from 2009 to 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, 1.27, right 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 to 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 from 2009 to 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 to 1985. Then Y ~ N(172.36, 6.34). 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 to 2010. 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? Solution 6.4 The z-score for x = 160.58 is z = –1.5. The z-score for y = 162.85 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? 342 CHAPTER 6 | THE NORMAL DISTRIBUTION Example 6.5 Suppose x has a normal distribution with mean 50 and standard deviation 6. • About 68% 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 of the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively. • About 95% 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 of the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively. • About 99.7% 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 of 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% of the values lie? Example 6.6 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 in 1984 to 1985. Then Y ~ N(172.36, 6.34). a. About 68% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. b. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________ respectively. c. About 99.7% of the y values lie between what two values? These values are ________________. The z- scores are ________________, respectively. Solution 6.6 a. About 68% of the values lie between 166.02 and 178.7. The z-scores are –1 and 1. b. About 95% of the values lie between 159.68 and 185.04. The z-scores are –2 and 2. c. About 99.7% of the values lie between 153.34 and 191.38. 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% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. b. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. c. About 99.7% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 6.2 | Using the Normal Distribution The shaded area in the following graph indicates the area to the left of x. This area is represented by the probability P(X < x). Normal tables, computers, and calculators provide or calculate the probability P(X < x). CHAPTER 6 | THE NORMAL DISTRIBUTION 343 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. 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. 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. 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. 344 CHAPTER 6 | THE NORMAL DISTRIBUTION 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 are 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. 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 (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the zscore was used. You calculate the z-score and look up the area to the left. The probability is the area to the right. z = 65 – 63 5 = 0.4 Area to the left is 0.6554. P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 345 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%). c. Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% 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. 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 ten percent are the same or higher. The variable k is often called a critical value. k = 69.4 Figure 6.6 346 CHAPTER 6 | THE NORMAL DISTRIBUTION The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this 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% of scores are below k and 30% 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% of the test scores fall at or below 65.5 and 30% 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 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). The probability for which you are looking is the area between x = 1.8 and x = 2.75. P(1.8 < x < 2.75) = 0.5886 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content |
/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 347 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. Figure 6.8 invNorm(0.25,2,0.5) = 1.66 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. 348 CHAPTER 6 | THE NORMAL DISTRIBUTION Example 6.10 There are approximately one billion smartphone users in the world today. 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. 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 b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old. 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% 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 There are approximately one billion smartphone users in the world today. 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. invNorm(0.75,36.9,13.9) = Q3 = 46.2754 invNorm(0.25,36.9,13.9) = Q1 = 27.5246 IQR = Q3 – Q1 = 18.7508 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 349 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. 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 = 40.42. Forty percent of the ages that range from 13 to 55+ are at least 40.42 years. 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% 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% of mandarin oranges from this farm have diameters between ______ and ______. Solution 6.12 b. 1 – 0.20 = 0.80 The tails of the graph of the normal distribution each have an area of 0.40. 350 CHAPTER 6 | THE NORMAL DISTRIBUTION Find k1, the 40th percentile, and k2, the 60th percentile (0.40 + 0.20 = 0.60). k1 = invNorm(0.40,5.85,0.24) = 5.79 cm k2 = invNorm(0.60,5.85,0.24) = 5.91 cm c. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence. Solution 6.12 c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm. 6.12 Using the information from Example 6.12, answer the following: a. The middle 45% of mandarin oranges from this farm are between ______ and ______. b. Find the 16th percentile and interpret it in a complete sentence. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 351 6.1 Normal Distribution (Lap Times) Class Time: Names: 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. Collect the Data 1. Use the data from Terri Vogel’s Log Book. 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 2. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 6.10 3. Calculate the following: a. b. x¯ = _______ s = _______ 352 CHAPTER 6 | THE NORMAL DISTRIBUTION 4. 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 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. This content is available for free at http://text |
bookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 353 6.2 Normal Distribution (Pinkie Length) Class Time: Names: 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 pinky finger (in centimeters). 1. Randomly survey 30 adults for their pinky 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 354 CHAPTER 6 | THE NORMAL DISTRIBUTION Using your sample mean, sample standard deviation, and histogram, what was the approximate theoretical 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 pinky length is more than 6.5 cm? • Explain the meaning of the 85th percentile of this 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 pinky length is more than 6.5 cm? • Explain the meaning of the 85th percentile of this 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 Describe the Data and Theoretical Distribution, explain why or why not. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 355 KEY TERMS Normal Distribution a continuous random variable (RV) with pdf f(x) = 2 – (x – m) 2σ 2 1 σ 2π e, 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) Z = the random variable for z-scores Z ~ N(0, 1) μ = the mean; σ = the standard deviation 6.2 Using the Normal Distribution Normal Distribution: X ~ N(µ, σ) where µ is the mean and σ is the standard deviation. 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) 6.1 The Standard Normal Distribution 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-scores 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? 356 CHAPTER 6 | THE NORMAL DISTRIBUTION 3. X ~ N(1, 2) σ = _______ 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)? 34. Suppose X ~ N(15, 3). Between what x values does 68.27% 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% of the data lie? The range of x values is centered at the mean of the distribution(i.e., –3). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 357 36. Suppose X ~ N(–3, 1). Between what x values does 34.14% 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 betwween 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? 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? 358 CHAPTER 6 | THE NORMAL DISTRIBUTION 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. 57. Use the following information to answer the next three exercise: 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 six years. a. Sketch the situation. Label and scale the axes. Shade the region corresponding to the probability. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 359 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%. Figure 6.18 b. P(x < k) = __________ Therefore, k = _________ 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 ten days to recover? a. 1.5 b. 0.2 c. 2.2 d. 7.3 62. The length of time to find 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 360 CHAPTER 6 | THE NORMAL DISTRIBUTION 63. The heights of the 430 National Basketball Association 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 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 an NBA 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, but 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. 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 the 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. 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 alternate to the SAT 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 they 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 upon 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 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 361 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 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 over 72 inches? Explain why or why not, and justify your answer numerically. d. The middle 40% 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% 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% 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 America 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 = percent of fat calories. a. X ~ _____(_____,_____) b. Find the probability that the percent 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 fewer 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 the child spends alone per day. a. b. X ~ _____(_____,_____) c. Find the probability that the child spends less than one hour per day unsupervised. Sketch the graph, and write the In words, define the random variable |
X. probability statement. d. What percent of the children spend over ten 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 President Clinton. The standard deviation was 572.3. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton 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 President Clinton. Sketch the Is 1,956.8 a population mean or a sample mean? How do you know? graph and write the probability statement. 362 CHAPTER 6 | THE NORMAL DISTRIBUTION d. Find the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton. e. Find the third quartile for votes for President Clinton. 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% of her laps are under _____. e. The middle 80% 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 ten customers has been 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. 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. This content is available for free at http://textbookequ |
ity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 6 | THE NORMAL DISTRIBUTION 363 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 (the data). 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. An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. |
The birth was uncomplicated, and the child needed no medical intervention. 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% 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 are 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% (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% 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 $1 scratch off lotto ticket will be a winner one out of five times. Out of a shipment of n = 190 lotto tickets, find the probability for the lotto tickets that there are somewhere between 34 and 54 prizes. somewhere between 54 and 64 prizes. a. b. c. more than 64 prizes. 88. Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site. On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent. 364 CHAPTER 6 | THE NORMAL DISTRIBUTION a. Find the probability that the percent of 18 to 34-year-olds who check Facebook 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 |
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