Chapter 07Continuous Probability Distributions True / False Questions 1. A continuous uniform distribution is always symmetric. True False 2. The height and width of a continuous uniform distribution's PDF are the same. True False 3. A continuous uniform distribution U(0,800) will have μ = 400 and σ = 230.94. True False 4. A continuous uniform distribution U(100,200) will have the same standard deviation as a continuous uniform distribution U(200,300). True False 5. For a continuous uniform distribution U(200,400), the parameters are μ = 300 and σ = 100. True False 6. The exponential distribution describes the number of arrivals per unit of time. True False 7. The exponential distribution is always skewed right. True False 8. If arrivals follow a Poisson distribution, waiting times follow the exponential distribution. True False 9. The triangular distribution is used in "what-if" analysis for business planning. True False 10 The triangular distribution is symmetric. . True False 11 The triangular distribution T(0,10,20) is skewed left. . True False 12 A triangular distribution can be skewed either left or right. . True False 13 For a continuous random variable, the total area beneath the PDF will . be greater than zero but less than one. True False 14 The exponential distribution is continuous and the Poisson distribution . is discrete, yet the two distributions are closely related. True False 15 The mean, median, and mode of a normal distribution will always be . the same. True False 16 There is a simple formula for normal areas, but we prefer a table for . greater accuracy. True False 17 Normal distributions differ only in their means and variances. . True False 18 Any normal distribution has a mean of 0 and a standard deviation of 1. . True False 19 We would use a normal distribution to model the waiting time until the . next Florida hurricane strike. True False 20 Experience suggests that 4 percent of all college students have had a . tonsillectomy. In a sample of 300 college students, we need to find the probability that at least 10 have had a tonsillectomy. It is acceptable to use the normal distribution to estimate this probability. True False 21 The normal is a good approximation to the binomial when n is greater . than or equal to 10. True False 22 The true proportion of accounts receivable with some kind of error is 4 . percent for Venal Enterprises. If an auditor randomly samples 50 accounts receivable, it is acceptable to use the normal approximation to estimate the probability that fewer than two will contain errors. True False 23 The normal distribution is a good approximation to the binomial if both . π ≥ 10 and n ≥ 10. True False 24 The normal distribution is a good approximation to the binomial if n = . 200 and π = .03. True False 25 The normal distribution is a good approximation to the binomial if n = . 25 and π = .50. True False 26 The exponential distribution can be either right-skewed or left-skewed, . depending on λ. True False 27 The number of lightning strikes in a day in Miami is a continuous . random variable. True False 28 The area under a normal curve is 1 only if the distribution is . standardized N(0,1). True False 29 The area under an exponential curve can exceed 1 because the . distribution is right-skewed. True False Multiple Choice Questions 30 A machine dispenses water into a glass. Assuming that the amount of . water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces, the average amount of water dispensed by the machine is: A. B. C. D. 12 ounces. 13 ounces. 14 ounces. 16 ounces. 51 ounces. Assuming that the amount of .31 A machine dispenses water into a glass.5.00 ounces. D. the standard deviation of the amount of water dispensed is about: A.5000 .57 ounce. B. . 34 A random variable X is best described by a continuous uniform . 32. 0. B. distribution from 20 to 45 inclusive. C.85.1. water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces. 52. 6.5. 30. B. 3.6666 33 A random variable X is best described by a continuous uniform . 3. 32.3333 .22. D. what is the probability that 13 or more ounces will be dispensed in a given glass? A. Assuming that the amount of . water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces. 1. . B.5. 33.5. 7.1666 . D. The mean of this distribution is: A. 31. The standard deviation of this distribution is approximately: A. D. 32 A machine dispenses water into a glass. C. C. C.5.73 ounces. distribution from 20 to 45 inclusive. C.1653. standard deviation approximately equal to A. . 13. B. 20. distribution from 20 to 45 inclusive. What is P(30 ≤ X ≤ 40)? A.1222. 11.5000. D.2407. 188. B. 37 The Excel function =40*RAND() would generate random numbers with . . D.55.5 hour for the next arrival is: A. probability of waiting more than 0. B. . 38 If arrivals occur at a mean rate of 3. .20 .33. standard deviation approximately equal to: A. C. . C. 200. 231.6 events per hour.35 A random variable X is best described by a continuous uniform .80 36 The Excel function =800*RAND() would generate random numbers with . D. D.60 . C. 400.27. . the exponential . 19. B.00.40 . .6 events per minute. the exponential . is Bob right? A.8347.5 minutes for the next arrival is: A. . . B. "Oh. No.0535.7809. . . .7104. Yes. . probability of waiting less than 0.6 events per minute. good. D." Bob said. Must have n to answer.0202.2019.8812. . B.7981. B. Axolotl's class BIO 417. 41 If arrivals occur at a mean rate of 1.8105. C. .39 If arrivals occur at a mean rate of 3.2564. B. .7122. C." Assuming a normal distribution of scores.0122. the . D.5 hour for the next arrival is: A. exponential probability of waiting more than 1. .6 events per hour. the . my score is in the top 10 percent.52 in Prof. "Life Cycle of the Ornithorhynchus. . C. exponential probability of waiting less than 1 minute for the next arrival is: A. . 42 Bob's z-score for the last exam was 1. 40 If arrivals occur at a mean rate of 2. D. C. . The first quartile for the lengths of brook trout would be: A. C.43 The lengths of brook trout caught in a certain Colorado stream are .00 inches 9. D. B. C. 12.65 inches. What lower limit should the State Game Commission set on length if it is desired that 80 percent of the catch may be kept by fishers? A. D.6826 .98 inches. B. D. normally distributed with a mean of 14 inches and a standard deviation of 3 inches.2486 . B. 45 The lengths of brook trout caught in a certain Colorado stream are .4082 44 The lengths of brook trout caught in a certain Colorado stream are . 11. .01 inches. 11. What proportion of brook trout caught will be between 12 and 18 inches in length? A.22 inches . normally distributed with a mean of 14 inches and a standard deviation of 3 inches. normally distributed with a mean of 14 inches and a standard deviation of 3 inches.80 inches 11.6563 . 10. 16.00 inches. C.48 inches 12. For a simple haircut. C. normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. What percent of customers require less than 32 minutes for a simple haircut? A.72 percent 45.8 minutes. the time to complete a simple haircut is . between 17.99 percent 99. B. A. normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. the middle 90 percent of the customers will require: A. 3(n + 1)/4 minutes 26 minutes 25. 95. between 19. the time to complete a simple haircut is .6 minutes. C. C.46 In Melanie's Styling Salon. ______________ the area under the normal curve between z = 1 and z = 2. 49 The area under the normal curve between z = 0 and z = 1 is .9 and 30.45 percent 97. The slowest quartile of customers will require longer than how many minutes for a simple haircut? A. B. D.99 percent 47 In Melanie's Styling Salon. B.0 and 30.4 and 31.1 minutes. the time to complete a simple haircut is . normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes. B. D. C. between 18.2 and 32.7 minutes 27. D. between 20. less than greater than equal to .0 minutes.7 minutes 48 In Melanie's Styling Salon. 8.67. B.28. What is the probability that the MPG for a randomly selected compact car would be less than 32? A.45. 40 percent. The value of z is approximately: A. 1. 0. C. D.8944 0. C. 50 percent. D. D. B.80 that a standard normal random variable is . 16 percent 34 percent 68 percent 75 percent 51 A student's grade on an examination was transformed to a z value of . a firm whose price-earnings ratio has a standardized value of z = 1. 52 The MPG (miles per gallon) for a certain compact car is normally . 1. A.5596 53 The probability is . we know that she scored approximately in the top: A. 1.35. 15 percent. 0. D. between -z and +z. distributed with a mean of 31 and a standard deviation of 0.96. Assuming a normal distribution.1056 0. . distribution. 1. B. B.00 is approximately in the highest ______ percent of firms in the industry.3944 0. 25 percent.50 The price-earnings ratio for firms in a given industry follows the normal . C. C. In this industry. C. 44.2 35.S.75 46.75 47. Census "long" . B.6 56 The time required for a citizen to complete the 2010 U. 68th 75th 78th 84th . B.50 52. D. 27.S. D. C.4772 0. 0. The slowest 10 percent of the citizens would need at least how many minutes to complete the form? A.9974 0. What percentile is he in? A. Census "long" . Jason's exam score was one . form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. standard deviation above the mean. C.8 59. D. What is the third quartile (in minutes) for the time required to complete the form? A.9772 0.50 57 Exam scores were normal in BIO 200. C. What proportion of the citizens will require less than one hour? A. form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. Census "long" .S.54 The time required for a citizen to complete the 2010 U.8 52. B. B. D. form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes.9997 55 The time required for a citizen to complete the 2010 U. 25 in the standard normal distribution will be: A. What fraction of the applicants would you expect to have a score of 400 or above? A. between z = 2. D. business are given an aptitude test. the area . smaller. 0.00 and z = 1.7734 0. Scores are normally distributed with a mean of 460 and standard deviation of 80. B.0401 0. C. impossible to compare without knowing μ and σ. C. business are given an aptitude test.5401 0.7500 . 59 A large number of applicants for admission to graduate study in .25. What fraction of applicants would you expect to have scores of 600 or above? A. 0.4599 0. D. D. the same. B.00 and z = 2. C. Scores are normally distributed with a mean of 460 and standard deviation of 80.0852 60 A large number of applicants for admission to graduate study in . larger.7266 0.2734 0.58 Compared to the area between z = 1. B. $5.4700. Given that . B.17.0906. business are given an aptitude test.25 . D.76. 0. 64 Assume that X is normally distributed with a mean μ = $64. 0.0853. ≤ Z ≤ 2. C.5016. then P(1.37) is: A.4147. 646.83. $13. then P(Z ≤ . 0. 0. 62 If the random variable Z has a standard normal distribution.05. $20. B. B. D. The top 2. 600. 0.17) is: A. D.61 A large number of applicants for admission to graduate study in .5 percent of the applicants would have a score of at least (choose the nearest integer): A. 0. 617.9147. C.2981. -1. 606.9200. we can calculate that the standard deviation of X is approximately: A. D. 63 If the random variable Z has a standard normal distribution. $7. 0. C.3944. Scores are normally distributed with a mean of 460 and standard deviation of 80. . 0. B. P(X ≥ $75) = 0. C. 68 The variable in a normal distribution can assume any value between . The 67th percentile of the distribution is: A.00. $28.5000. C. B. B. 67 The area under the normal curve between the 20th and 70th percentiles . B.00. $20. Given . 0. -3 and +3 -4 and +4 -1 and +1 -∞ and +∞ . D. From this we can determine that the mean of the distribution is equal to: A. A. 95.9193. 0. C.64. 90. variance of 36.7000. $26. 0. 72. is equal to: A. 66 The random variable X is normally distributed with mean of 80 and .84. 82. B. D. that P(X ≤ $10) = 0. C.1841.65 The standard deviation of a normal random variable X is $20. C. D. $13. D. D. distribution? A. B. μ = 0. standard deviations. . subtract the mean from the standard deviation and divide by the variance.72 percent. D. 71 Light bulbs are normally distributed with an average lifetime of 1000 . C. 75. The probability that a light bulb picked at random will last less than 1500 hours is about: A. standard deviations and means. standard deviations. B. D. 68. hours and a standard deviation of 250 hours. subtract the mean from the original observation and divide the result by the variance. B. would: A. B. σ = 0 μ = 1. then divide by the variance. 97. σ = 1 70 Any two normal curves are the same except for their: . means. A. subtract the mean from the original observation and divide the result by the standard deviation. C. σ = 1 μ = 1. C. add the mean and the original observation. and kurtosis. σ = 0 μ = 0.00 percent. C. 95.00 percent.44 percent. 72 To convert a normally distributed variable X into a standard Z score we .69 What are the mean and standard deviation for the standard normal . means. skewness. Triangular Uniform Normal Exponential 75 On average.0952 . . a major earthquake (Richter scale 6. C. 23. C. B.724 hours. C. The normal distribution is sometimes skewed. D. 28. B. the 90th percentile of waiting times to the next shutdown will be: A.180 hours. nonworking web URL ("This page cannot be displayed") as you click on various websites for Florida condo rentals? A. times a decade in a certain California county. B.8607 .782 hours. The exponential distribution is always skewed right. B. D. . 18. The uniform distribution is never skewed.500 . 20. operating hours. The triangular distribution may be skewed left or right.1393 . D. which statement is . incorrect? A. C.0 or above) occurs 3 . 74 Which model best describes your waiting time until you get the next .9048 76 If the mean time between in-flight aircraft engine shutdowns is 12. What is the probability that less than six months will pass before the next earthquake? A.733 hours.73 Regarding continuous probability distributions. D. C. The median time between breakdowns is: A. What is the probability that less than 6 minutes will elapse before the next breakdown? A. minutes between breakdowns. a hospital's CT scan facility is 4. D.1353 .77 On average. 35.000 operating hours. B. .1813 .000 hours? A. 15 minutes elapse between discoveries of fraudulent . What is the probability that less than 30 minutes will elapse before the next fraudulent corporate tax return is discovered? A. . D.8000 . . what is the probability of unscheduled maintenance in the next 5.2865 .8187 .5000 79 A certain assembly line at Vexing Manufacturing Company averages 30 . minutes between breakdowns. 30.7 minutes. C.6044 . .8647 78 If the mean time between unscheduled maintenance of LCD displays in .2224 80 A certain assembly line at Vexing Manufacturing Company averages 30 . C.7389 . D. B.7135 . D.0 minutes.0488 . corporate tax returns in a certain IRS office. B. 20. 25.4 minutes. C.8 minutes. B. What percentile is he in? 68th 75th 84th 92nd A. C. "Yipe! My score is within the bottom quartile. Yes No Must know the class size to answer 83 Exam scores were normal in MIS 200. time (working days) until an office photocopier breaks down (i.50 and z = 0.15 in FIN 417. D." Bob said. B.50 and z = 1. 84 Compared to the area between z = 0. C. smaller larger the same . requires unscheduled maintenance)? A. "Capital Budgeting .81 Which probability model is most appropriate to describe the waiting .75. B. between z = 1. standard deviations above the mean.e.. C. C.41 . is Bob right? A. Jason's exam score was 1. B. Normal Uniform Exponential Poisson 82 Bob's z-score for the last exam was -1." Assuming a normal distribution. Strategies. the area . B. D.75 in the standard normal distribution will be: A. 87. 0. 0. 0.2118. 0. that are uniformly distributed between 0 and 100. D. 608. B.00.1203. -1. D. B. then P(Z ≤ . D. 33.0446. 0. C.9573. 50. then P(1. The standard deviation of this distribution is approximately: A. 87 If the random variable Z has a standard normal distribution. 88 Excel's =100*RAND() function produces continuous random numbers . 25. C. D.00.72) is: A. C. 0. 575.85 If GMAT scores for applicants at Oxnard Graduate School of Business . are N(500. 0.1091.33. 0. B. 86 If the random variable Z has a standard normal distribution.17 . 582. ≤ Z ≤ 2.0427. C.3944.26) is: A. B. 601. . 28. 50) then the top 5 percent of the applicants would have a score of at least (choose the nearest integer): A.5016. C. C. distribution that is N(475.155 minutes (69. B. The mean of this distribution is approximately A. distributed between 0 and 1.240 minute (14.S. distributed from 0 to 1. What is the probability that the random number exceeds . B.DIST(80.75? A.3333 . 75 percent below). minute.3 seconds) 0. 75 percent 50 percent 25 percent 91 Which is the correct Excel formula for the 80th percentile of a .919 minutes (115.2500 . 33. D. A.2 arrivals per .INV(0. B. .89 Excel's =RAND() function produces random numbers that are uniformly .e. find the 75th percentile of waiting times until the next arrival (i.4 seconds) 1. 33)? A.475)/33) 92 If arrivals follow a Poisson distribution with mean 1. C. 475.2887 90 Excel's =RAND() function produces random numbers that are uniformly . C. 1. =NORM.INV((80 .1 seconds) .80. 475.5000 .1) =NORM.. 33) =NORM. B. Based on past .1797 . no answer. experience. B.022.20.2305 0. the probability of an undeliverable letter is 0.06.0368 0. B.93 A software developer makes 175 phone calls to its current customers. B. . or answering machine). C.063. D. with some kind of error is .9632 0.1587 .2097 . .0544 95 A letter is mailed to a sample of 500 homeowners. If an auditor randomly samples 225 accounts receivable. 94 For Gardyloo Manufacturing. . There is an 8 percent chance of reaching a given customer (instead of a busy signal. The normal approximation of the probability of reaching at least 20 customers is: A. The normal approximation to the binomial probability of 40 or more undeliverable letters is: A.7695 . 0. D. what is the approximate normal probability that 39 or fewer will contain errors? A. . the true proportion of accounts receivable . C. C. D. .937. .007. percent. π = 0. C. B. Each question has five .0336 .0088 0. π = 0. What is the approximate normal probability that a "guesser" will score at least 60 points? A.2000 0. what is the approximate probability that at least 26 will result in default? A.0668 0. B. 0. C.4913 98 For which binomial distribution would a normal approximation be most . . If a bank makes 100 of these loans. B. D. passing requires a score of at least . π = 0.4713 . C. 60. B. D. C. π = 0.96 In a T-F exam with 100 questions. acceptable? A.04 n = 40.25 n = 400. the default rate on a certain type of commercial loan is 20 . D.05 n = 100.02 99 Historically.0062 0.0251 . What would be the approximate probability that a "guesser" could achieve a score of 30 or more? A.0846 0.0015 0.0287 . n = 50.0377 97 A multiple choice exam has 100 questions. 0. D. choices. 100 A company employs 300 employees. are rented from Hertz. A. the mean is ______ and the standard deviation is _____. 7. B. .9 percent of adult males will not have to stoop as they enter? A. 63 90. D.3 cm 201. If 25.2577 . 30 90.1335 . C. B. turnover rate for employees.1128 . D. and a standard deviation of 7 cm. what is the normal approximation to the probability that fewer than 20 will be stolen? A. 15 101 The probability that a rental car will be stolen is 0. D. Each year. 195. 90. C. C.001. We want to do a normal approximation to the binomial distribution of the number of employees who leave each year. there is a 30 percent . For this normal approximation. how high should an aircraft lavatory door be to ensure that 99. B.937 90.7 cm 201.000 cars .8335 102 If adult male heights are normally distributed with a mean of 180 cm .6 cm 207.4 cm . 33. 105 The triangular distribution T(0. B. 23. with a safety margin of 15 cm on each end of the bassinet? A. A. 3. .103 TotCo is developing a new deluxe baby bassinet. 18. newborn baby is normally distributed with a mean of 50 cm and a standard deviation of 5 cm. 14. 95. D. 13. what should be the interior length of the bassinet to ensure that 99 percent of newborn babies will fit. 3.775. C.082. 23. 35. 10. 12. D. 106 The triangular distribution T(5. B. D. C. 2.22 cm 91.63 cm 98. 12. 62) has a mean of: . If the length of a .024. C. A. 20) has a standard deviation of: . B. C. B. D. 4. A. 26) has a mean of: .45 cm 85.92 cm 104 The triangular distribution T(4.994. 30. The midrange of the distribution is 40. B. D. D. A. The triangular distribution always has a single mode. daily commute time (minutes). commute time (minutes). The triangular distribution is right-skewed.710. 8. C. P(X < 5) is more than P(X ≤ 5). The mean of the triangular distribution is (a + b + c)/3. P(X < 5) is the same as P(X ≤ 5). B.5.107 The triangular distribution T(10. . C.498. The mode of the distribution is at the mean. 61) to represent his daily . 15. 108 Which statement is incorrect? . The mean of the distribution is 37.498. Which statement is incorrect? A. 110 Phyllis used a triangular distribution of T(10. B. 30. A. C. B. A. 9. Which statement is incorrect? A. The midrange of the distribution is 15. C. D. The distribution is right-skewed. 20. 109 Bob used a triangular distribution of T(20. The distribution is right-skewed. 50) has a standard deviation of: . 7. 9. 111 In a continuous distribution: . 20) to represent her .225. C. The mean of the distribution is 15. B. The mode of the distribution exceeds the mean. P(X < 5) is less than P(X ≤ 5). C. what is the . CDF is usually the same as the PDF. B. B. 113 If the mean waiting time for the next arrival is 12 minutes. D. B. 7.112 In a continuous distribution the .3 minutes 9. D.1 minutes 12 minutes 114 If the mean waiting time for the next arrival is 18 minutes.2 minutes 3. D. first quartile (25th percentile) for waiting times? A. PDF shows the area under the curve. A. median waiting time? A. what is the . C. 13 minutes 7. PDF is usually higher than the CDF. CDF is used to find left-tail probabilities. C.2 minutes 8.9 minutes 5.1 minutes . B. Yes. C. . No. 116 Could this function be a PDF? . C.115 Could this function be a PDF? . Yes. It depends on x. A. It depends on x. No. A. B. 117 The ages of job applicants for a security guard position are uniformly . distributed between 25 and 65. Could a 25-year-old job applicant be two standard deviations below the mean (or more than two standard deviations)? A. B. C. Yes. No. Impossible to determine from given information. Chapter 07 Continuous Probability Distributions Answer Key True / False Questions 1. A continuous uniform distribution is always symmetric. TRUE The PDF is the same height for all X values. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-02 Calculate uniform probabilities. Topic: Uniform Continuous Distribution 2. The height and width of a continuous uniform distribution's PDF are the same. FALSE The PDF height must be 1/(b - a) so that the total area is unity. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-02 Calculate uniform probabilities. Topic: Uniform Continuous Distribution 3. A continuous uniform distribution U(0,800) will have μ = 400 and σ = 230.94. TRUE Apply the formulas for the mean and standard deviation. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities. Topic: Uniform Continuous Distribution 4. A continuous uniform distribution U(100,200) will have the same standard deviation as a continuous uniform distribution U(200,300). TRUE In the standard deviation formula, (b - a)2 is the same for both these examples. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities. Topic: Uniform Continuous Distribution 5. For a continuous uniform distribution U(200,400), the parameters are μ = 300 and σ = 100. FALSE The standard deviation is [(400 - 200)2/12]1/2 = 57.7. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities. Topic: Uniform Continuous Distribution 6. The exponential distribution describes the number of arrivals per unit of time. FALSE Arrivals per unit of time would be Poisson (but waiting time is exponential). AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-07 Find the exponential probability for a given x. Topic: Exponential Distribution 7. The exponential distribution is always skewed right. TRUE The PDF clearly shows extreme right-skewness. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-07 Find the exponential probability for a given x. Topic: Exponential Distribution 8. If arrivals follow a Poisson distribution, waiting times follow the exponential distribution. TRUE Review the definition of an exponential distribution. AACSB: Analytic Blooms: Understand Difficulty: 1 Easy Learning Objective: 07-07 Find the exponential probability for a given x. Topic: Exponential Distribution 9. The triangular distribution is used in "what-if" analysis for business planning. TRUE Simplicity in visualizing planning scenarios is an attraction of the triangular distribution. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Topic: Triangular Distribution (Optional) FALSE Only left-skewed if the mode is right of the axis midpoint. The triangular distribution T(0. TRUE Left-skewed if the mode is right of the axis midpoint. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). FALSE Triangular distribution is symmetric only if the mode is at the axis midpoint. Topic: Triangular Distribution (Optional) . A triangular distribution can be skewed either left or right.10. and vice versa.10. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). Topic: Triangular Distribution (Optional) 12. Topic: Triangular Distribution (Optional) 11.20) is skewed left. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). The triangular distribution is symmetric. TRUE Poisson arrivals (discrete) imply exponential waiting times (continuous). FALSE The total area must be 1 if it is a PDF. The mean. AACSB: Analytic Blooms: Remember Difficulty: 2 Medium Learning Objective: 07-03 Know the form and parameters of the normal distribution. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-01 Define a continuous random variable. AACSB: Analytic Blooms: Understand Difficulty: 2 Medium Learning Objective: 07-07 Find the exponential probability for a given x. For a continuous random variable. The exponential distribution is continuous and the Poisson distribution is discrete. yet the two distributions are closely related. Topic: Describing a Continuous Distribution 14. TRUE A normal distribution is perfectly symmetric. Topic: Normal Distribution .13. the total area beneath the PDF will be greater than zero but less than one. median. and mode of a normal distribution will always be the same. Topic: Exponential Distribution 15. Topic: Normal Distribution . AACSB: Analytic Blooms: Understand Difficulty: 1 Easy Learning Objective: 07-03 Know the form and parameters of the normal distribution.16. AACSB: Analytic Blooms: Understand Difficulty: 2 Medium Learning Objective: 07-03 Know the form and parameters of the normal distribution. FALSE Only the standardized normal is N(0. but we prefer a table for greater accuracy. but there is no exact formula for areas under the curve. Normal distributions differ only in their means and variances. FALSE We have a formula for the PDF. Topic: Normal Distribution 18. Topic: Normal Distribution 17. Any normal distribution has a mean of 0 and a standard deviation of 1. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-03 Know the form and parameters of the normal distribution. There is a simple formula for normal areas.1). TRUE All normal distributions look the same except for scaling. π) ≥ 10. The normal is a good approximation to the binomial when n is greater than or equal to 10. In a sample of 300 college students.π) ≥ 10. so waiting times are exponential. We would use a normal distribution to model the waiting time until the next Florida hurricane strike. FALSE The quick rule is nπ ≥ 10 and n(1 . Topic: Normal Approximations . Experience suggests that 4 percent of all college students have had a tonsillectomy. FALSE Hurricane arrivals might be regarded as Poisson events. TRUE The quick rule is nπ ≥ 10 and n(1 . Topic: Normal Approximations 21. we need to find the probability that at least 10 have had a tonsillectomy. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. which is the case in this example.19. It is acceptable to use the normal distribution to estimate this probability. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-07 Find the exponential probability for a given x. Topic: Exponential Distribution 20. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. it is acceptable to use the normal approximation to estimate the probability that fewer than two will contain errors.22. The normal distribution is a good approximation to the binomial if both π ≥ 10 and n ≥ 10. FALSE The quick rule is nπ ≥ 10 and n(1 . If an auditor randomly samples 50 accounts receivable.03. Topic: Normal Approximations 24. which is not fulfilled in this case. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. The normal distribution is a good approximation to the binomial if n = 200 and π = . FALSE The quick rule is nπ ≥ 10 and n(1 . AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Topic: Normal Approximations 23.π) ≥ 10.π) ≥ 10. which is not fulfilled in this case. Topic: Normal Approximations .π) ≥ 10. The true proportion of accounts receivable with some kind of error is 4 percent for Venal Enterprises. FALSE The quick rule is nπ ≥ 10 and n(1 . Topic: Exponential Distribution 27. TRUE The quick rule is nπ ≥ 10 and n(1 . Topic: Normal Approximations 26. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.25. depending on λ. Topic: Describing a Continuous Distribution .π) ≥ 10. The exponential distribution can be either right-skewed or leftskewed. FALSE The PDF of the exponential shows that it is always right-skewed. The number of lightning strikes in a day in Miami is a continuous random variable. which is fulfilled in this case. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-07 Find the exponential probability for a given x. The normal distribution is a good approximation to the binomial if n = 25 and π = . .50. . FALSE The "number of ." anything is discrete. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-01 Define a continuous random variable. Topic: Normal Distribution 29. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-07 Find the exponential probability for a given x. The area under a normal curve is 1 only if the distribution is standardized N(0. Topic: Exponential Distribution Multiple Choice Questions .1).28. the total area under the PDF is one. FALSE If it's a PDF. The area under an exponential curve can exceed 1 because the distribution is right-skewed. FALSE Any normal distribution has a total area of one under the PDF. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-03 Know the form and parameters of the normal distribution. 3.10)2/12]1/2 = 1. The standard deviation is [(16 . 1. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces. ounces. The mean is halfway between the end points of the distribution.51 ounces.73 ounces.57 ounce. the average amount of water dispensed by the machine is: A. 12 13 14 16 ounces. 3. Topic: Uniform Continuous Distribution . AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-02 Calculate uniform probabilities. ounces.30. the standard deviation of the amount of water dispensed is about: A.73. D. B. C.00 ounces. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces. Topic: Uniform Continuous Distribution 31. B. ounces. D. 0. A machine dispenses water into a glass. A machine dispenses water into a glass. C. . B. 32. 30.5000 .5. The mean is halfway between the end points of the distribution. B. Topic: Uniform Continuous Distribution . C.32. D. 31. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities. Topic: Uniform Continuous Distribution 33.5.5. A machine dispenses water into a glass. A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-02 Calculate uniform probabilities.3333 .5. what is the probability that 13 or more ounces will be dispensed in a given glass? A. D. The mean of this distribution is: A. Assuming that the amount of water dispensed follows a continuous uniform distribution from 10 ounces to 16 ounces.1666 .6666 Half the area is above 13. 33. C. 22.60 . A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive.1.20)2/12]1/2 = 7. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities. The standard deviation is [(45 .40. 32. C. B. 52. Topic: Uniform Continuous Distribution . C. The standard deviation of this distribution is approximately: A.22.40 . Topic: Uniform Continuous Distribution 35.20 .34. B. D. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities. . A random variable X is best described by a continuous uniform distribution from 20 to 45 inclusive.80 The desired area is 10/25 = . What is P(30 ≤ X ≤ 40)? A. 6. 7.85.5. D. Topic: Uniform Continuous Distribution . 188.33. 200.00. C. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities. 20. 231. B. 19. 11.27. The standard deviation is [(800 . The standard deviation is [(40 .0)2/12]1/2 = 11. The Excel function =800*RAND() would generate random numbers with standard deviation approximately equal to: A.94.55. B. C.0)2/12]1/2 = 230. D.36.55. The Excel function =40*RAND() would generate random numbers with standard deviation approximately equal to A. 13. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities. Topic: Uniform Continuous Distribution 37. 400. D. the exponential probability of waiting less than 0.5 hour for the next arrival is: A. D.8347.5000. If arrivals occur at a mean rate of 3.1653.7122. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-07 Find the exponential probability for a given x.. .6 × 0. . Topic: Exponential Distribution .7809.1653 = .50) = exp(-3. P(X < . . C.2407. C. . .8347.38.6 × 0.1653.6 events per hour. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-07 Find the exponential probability for a given x. If arrivals occur at a mean rate of 3. .8105. B. Topic: Exponential Distribution 39.6 events per hour.50) = 1 .exp(-3. D.50) = . the exponential probability of waiting more than 0. P(X > .1222. .5 hour for the next arrival is: A.50) = 1 . B. . 0535. B.7981. the exponential probability of waiting more than 1. ..6 events per minute.2564.6 × 1. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-07 Find the exponential probability for a given x. C. . .40. If arrivals occur at a mean rate of 2. Topic: Exponential Distribution 41. . .exp(-1. P(X > 1.5 minutes for the next arrival is: A.7981.2019. If arrivals occur at a mean rate of 1. . AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-07 Find the exponential probability for a given x. C.5) = exp(-2. Topic: Exponential Distribution .0202.6 × 1) = 1 .7104. . the exponential probability of waiting less than 1 minute for the next arrival is: A.0202. (X < 1) = 1 .2019 = .0122. D.50) = . B.8812.6 events per minute. D. . No. 6563 using Excel. C." Assuming a normal distribution of scores.6568 (from Appendix C) or .6826 . "Oh. Topic: Standard Normal Distribution . The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches.6563 . D.67 < Z < 1. Bob's z-score for the last exam was 1. C. is Bob right? A.33) = . AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. good. . Topic: Standard Normal Distribution 43.9357.2486 . Axolotl's class BIO 417. What proportion of brook trout caught will be between 12 and 18 inches in length? A.52) = .52 in Prof. P(Z < 1.42. Must have n to answer.4082 P(12 < X < 18) = P(-. B. Yes. B. "Life Cycle of the Ornithorhynchus. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. my score is in the top 10 percent." Bob said. 01 11.98.65 inches.INV(.3) = 11. or Q1 = 14 . D. B. or X = 14 .22 inches Using Excel =NORM.675(3) = 11. D. inches.3) = 11. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. What lower limit should the State Game Commission set on length if it is desired that 80 percent of the catch may be kept by fishers? A.80 inches 11.48 inches 12. 12. The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches.20.14.475.0.84(3) = 11.00 inches 9. Using Excel =NORM.00 11. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.44. inches.975 using Appendix C.INV(. The first quartile for the lengths of brook trout would be: A.14. inches. Topic: Standard Normal Distribution 45. C. Topic: Standard Normal Distribution . The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. B.48 using Appendix C. 16.98 10. C.25.0. B.4) = 27.25.75.4. Topic: Standard Normal Distribution 47. D. In Melanie's Styling Salon.25)/4 = 1. What percent of customers require less than 32 minutes for a simple haircut? A.72 45.46. In Melanie's Styling Salon. C. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. B. the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes.675(4) = 27. or Q3 = 25 + 0.698. 95.45 97.7 using Appendix C. Topic: Standard Normal Distribution .1) = 0.7 minutes 27. C. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.99 percent percent percent percent Using Excel =NORMDIST(32.75 with Appendix C. 3(n + 1)/4 minutes 26 minutes 25.99 99. the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes.9599. The slowest quartile of customers will require longer than how many minutes for a simple haircut? A.7 minutes Using Excel =NORM.INV(.25. or use z = (32 . D. between between between between 18. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.0 32. Topic: Standard Normal Distribution . A. B. The area under the normal curve between z = 0 and z = 1 is ______________ the area under the normal curve between z = 1 and z = 2. D. C. Topic: Standard Normal Distribution 49. The 90 percent range is μ ± 1. In Melanie's Styling Salon.0 17.2 and and and and 31. minutes.8 minutes. the time to complete a simple haircut is normally distributed with a mean of 25 minutes and a standard deviation of 4 minutes.6 30. B. minutes. For a simple haircut.9 20.48. less than greater than equal to The standard normal PDF grows closer to the axis as z increases to the right of zero. the middle 90 percent of the customers will require: A. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.1 30. minutes.4 19. C.645σ. In this industry. 15 50 40 25 percent.86 percent of the area is above one standard deviation.7486. P(Z > 0. C. we know that she scored approximately in the top: A. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. percent. a firm whose price-earnings ratio has a standardized value of z = 1.P(Z < 0. D. 16 34 68 75 percent percent percent percent About 15.67) = 1 . A student's grade on an examination was transformed to a z value of 0. D. A.2514 = . B. Topic: Standard Normal Distribution . Assuming a normal distribution. B.00 is approximately in the highest ______ percent of firms in the industry. The price-earnings ratio for firms in a given industry follows the normal distribution. percent. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.. Topic: Standard Normal Distribution 51. C. percent.67.67) = 1 .50. D.52. C. D. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.8944 0.25) = .3944 0. What is the probability that the MPG for a randomly selected compact car would be less than 32? A. 0. B. 1.5596 P(X < 32) = P(Z < 1. 1.8. B.1056 0. C.28.45.80 that a standard normal random variable is between -z and +z. The value of z is approximately: A. 1. The probability is .282.96. For tail areas of .8944.1000 we would use z = 1. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. 1. Topic: Standard Normal Distribution 53.35. The MPG (miles per gallon) for a certain compact car is normally distributed with a mean of 31 and a standard deviation of 0. Topic: Standard Normal Distribution . C.9772. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.9974 0.S. Topic: Standard Normal Distribution .6 Using Excel =NORM.282(10) = 52.INV(. D.4772 0.82 using Appendix C. or 40 + 1.2 35.8 52. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes.9997 P(X < 60) = P(Z < 2. B. What proportion of the citizens will require less than one hour? A. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. 0. C. The slowest 10 percent of the citizens would need at least how many minutes to complete the form? A. 27.40.82. D.90.9772 0. The time required for a citizen to complete the 2010 U.8 59.S. The time required for a citizen to complete the 2010 U.00) = .54. B. Topic: Standard Normal Distribution 55.10) = 52. 75 46. D. Jason's exam score was one standard deviation above the mean.75 using Appendix C AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. B. 68th 75th 78th 84th About 15. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. What percentile is he in? A. Census "long" form is normally distributed with a mean of 40 minutes and a standard deviation of 10 minutes. C.50 52. The time required for a citizen to complete the 2010 U.10) = 46.75.75 47.675(10) = 46. C. Topic: Standard Normal Distribution 57. D. What is the third quartile (in minutes) for the time required to complete the form? A.INV(. Topic: Standard Normal Distribution . 44.50 Using Excel =NORM.75. Exam scores were normal in BIO 200.40. or Q3 = 40 + 0.56.87 percent of the area lies above one standard deviation.S. B. The normal PDF approaches the axis as z increases beyond zero. impossible to compare without knowing μ and σ.4599 0. smaller. Topic: Standard Normal Distribution 59. C.00 and z = 2. D. B. D. B. A large number of applicants for admission to graduate study in business are given an aptitude test.0401. Compared to the area between z = 1. Scores are normally distributed with a mean of 460 and standard deviation of 80. Topic: Standard Normal Distribution . the same. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.58.75) = . What fraction of applicants would you expect to have scores of 600 or above? A. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. the area between z = 2.5401 0.0852 P(X > 600) = P(Z > 1.0401 0. C. 0. larger.25.00 and z = 1.25 in the standard normal distribution will be: A. 8.7734 0. A large number of applicants for admission to graduate study in business are given an aptitude test. A large number of applicants for admission to graduate study in business are given an aptitude test.7734. D. 600.96. The top 2. Topic: Standard Normal Distribution .60. B. 617. Scores are normally distributed with a mean of 460 and standard deviation of 80. B.2734 0.75) = . 606.7266 0. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.5 percent of the applicants would have a score of at least (choose the nearest integer): A. C. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. D. 646. we get X = 460 + 1. What fraction of the applicants would you expect to have a score of 400 or above? A. Using z = 1. 0. Topic: Standard Normal Distribution 61.7500 P(X > 400) = P(Z > -0. Scores are normally distributed with a mean of 460 and standard deviation of 80. C.96 × 80 = 616. 25) = .3944. If the random variable Z has a standard normal distribution. then P(1.0853. D.0906.62. then P(Z ≤ -1. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.17) . From Appendix C we get the left tail area of . Topic: Standard Normal Distribution 63.9147. If the random variable Z has a standard normal distribution.0853.4147. 0.37) is: A. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. P(Z ≤ 2. 0.0906. B. B.4700. C. 0.25 ≤ Z ≤ 2. C. 0.P(Z ≤ 1. 0.17) is: A. D.9200. 0. 0. Topic: Standard Normal Distribution . 0.5016. so with x = 75 we set z = (x . For a right-tail area of . $5. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. From this we can determine that the mean of the distribution is equal to: A. The standard deviation of a normal random variable X is $20. B.76. $20. C.μ)/σ and solve for μ.53. D.83.05. C. Given that P(X ≥ $75) = 0. Topic: Standard Normal Distribution 65. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. $13.μ)/σ and solve for σ. Topic: Standard Normal Distribution .17.2981. Given that P(X ≤ $10) = 0. For a left-tail area of .1841 we need z = -.64.90. D. $7. $20.2981 we need z = -. $26. $28. B.1841. so with x = 10 we set z = (x . $13. we can calculate that the standard deviation of X is approximately: A. Assume that X is normally distributed with a mean μ = $64. 9193. this must be . C. B.44(6) = 82. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.66. Logically. Topic: Standard Normal Distribution 67.84. Since P(Z < 0. B. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.5000.64. 90.64. The area under the normal curve between the 20th and 70th percentiles is equal to: A.00.50. 95. The 67th percentile of the distribution is: A. (from Appendix C) we get 80 + 0.7000. as you can verify from Appendix C. 82.6700.00..44) = . 0. 72. The random variable X is normally distributed with mean of 80 and variance of 36. D. 0. 0.70 .20 = . C. Topic: Normal Distribution . μ μ μ μ = = = = 0. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-03 Know the form and parameters of the normal distribution. The variable in a normal distribution can assume any value between A. σ σ σ σ = = = = 0 1 0 1 We have standardized so the mean must be zero. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-03 Know the form and parameters of the normal distribution. B. but the curve never quite touches the z-axis. 0. D. What are the mean and standard deviation for the standard normal distribution? A.68. C. Topic: Standard Normal Distribution . 1. C. D. -3 and +3 -4 and +4 -1 and +1 -∞ and +∞ Almost all the area is within -3 and +3. B. 1. Topic: Normal Distribution 69. D.00 percent. B. B. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel.00) = . Topic: Standard Normal Distribution . means. Topic: Normal Distribution 71. percent. AACSB: Analytic Blooms: Remember Difficulty: 1 Easy Learning Objective: 07-03 Know the form and parameters of the normal distribution. percent. C.9772 from Appendix C (or from Excel). standard deviations and means. and kurtosis. The probability that a light bulb picked at random will last less than 1500 hours is about: A.σ) so show the similarity of all normals. Light bulbs are normally distributed with an average lifetime of 1000 hours and a standard deviation of 250 hours.00 68.44 75. We write N(μ. C. percent. skewness. Any two normal curves are the same except for their: A. standard deviations. P(Z < 1500) = P(Z < 2. 97. standard deviations.72 95.70. means. D. subtract the mean from the original observation and divide the result by the variance. To convert a normally distributed variable X into a standard Z score we would: A. Review the z-score transformation. AACSB: Analytic Blooms: Understand Difficulty: 2 Medium Learning Objective: 07-03 Know the form and parameters of the normal distribution. The triangular distribution may be skewed left or right. which statement is incorrect? A. B. C. add the mean and the original observation. B. The normal distribution is sometimes skewed. D. AACSB: Analytic Blooms: Understand Difficulty: 1 Easy Learning Objective: 07-03 Know the form and parameters of the normal distribution.72. D. subtract the mean from the original observation and divide the result by the standard deviation. The uniform distribution is never skewed. C. subtract the mean from the standard deviation and divide by the variance. The exponential distribution is always skewed right. Topic: Standard Normal Distribution 73. Regarding continuous probability distributions. Review the characteristics of these four distributions. Topic: Normal Distribution . then divide by the variance. 0952 .0 or above) occurs 3 times a decade in a certain California county.025 × 6) = 1 . .. C. Which model best describes your waiting time until you get the next nonworking web URL ("This page cannot be displayed") as you click on various websites for Florida condo rentals? A. On average.025 earthquake/month so P(X < 6) = 1 . AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-07 Find the exponential probability for a given x.exp(0. D. Topic: Exponential Distribution .8607 . Topic: Exponential Distribution 75.8607 = . Triangular Uniform Normal Exponential Waiting time until the next event resembles an exponential distribution.9048 Set λ = 3/120 = 0. D.74. C. What is the probability that less than six months will pass before the next earthquake? A. a major earthquake (Richter scale 6.1393 . B. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-07 Find the exponential probability for a given x. B.1393. 782 23. Set λ = 1/12500 and solve for x in left-tail area of 1 . Topic: Exponential Distribution .. On average. 15 minutes elapse between discoveries of fraudulent corporate tax returns in a certain IRS office. the 90th percentile of waiting times to the next shutdown will be: A. .180 28. D.1353 = .8647.exp(-λx) = .8647 P(X < 30) = 1 . If the mean time between in-flight aircraft engine shutdowns is 12.6044 . B.1353 .90 by taking logs of both sides.724 hours. What is the probability that less than 30 minutes will elapse before the next fraudulent corporate tax return is discovered? A. C. D. hours. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-07 Find the exponential probability for a given x.7389 . 20.500 operating hours. Topic: Exponential Distribution 77. C. hours.exp(-(1/15) × 30) = 1 .exp(-λx) = 1 .733 18.76. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-08 Solve for x for a given exponential probability. B. hours. D.2865 .8187 . B. .2224 (X < 6) = 1 . What is the probability that less than 6 minutes will elapse before the next breakdown? A.exp(-λx) = 1 . B.5000 P(X < 5000) = 1 . If the mean time between unscheduled maintenance of LCD displays in a hospital's CT scan facility is 4. .exp(-(1/4000) × 5000) = 1 .. Topic: Exponential Distribution 79.7135.000 hours? A.0488 .exp(-λx) = 1 . AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-07 Find the exponential probability for a given x.8187 = . A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns.. D. C.7135 . AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-07 Find the exponential probability for a given x.8000 .1813 .000 operating hours. C.2865 = .1813.78. Topic: Exponential Distribution .exp(-(1/30) × 6) = 1 . what is the probability of unscheduled maintenance in the next 5. exp(-λx) = .0 35. A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. D.80. Set λ = 1/30 and solve for x in right-tail area of 1 . Topic: Exponential Distribution 81. Normal Uniform Exponential Poisson Poisson breakdowns suggest exponential waiting time. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-07 Find the exponential probability for a given x. B. C. The median time between breakdowns is: A.50 by taking logs of both sides. requires unscheduled maintenance)? A. D.4 20. C. minutes. 30. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-08 Solve for x for a given exponential probability.. Topic: Exponential Distribution . minutes.e. minutes.8 minutes.7 25. B. Which probability model is most appropriate to describe the waiting time (working days) until an office photocopier breaks down (i. Jason's exam score was 1.41) = . B.9207. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. Topic: Standard Normal Distribution 83. Exam scores were normal in MIS 200.82." Assuming a normal distribution. Topic: Standard Normal Distribution . "Capital Budgeting Strategies. D.41 standard deviations above the mean. B.675 so Bob is indeed below that point. P(Z < 1. Bob's z-score for the last exam was -1. "Yipe! My score is within the bottom quartile. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. C. is Bob right? A." Bob said. C.15 in FIN 417. What percentile is he in? 68th 75th 84th 92nd A. Yes No Must know the class size to answer The bottom quartile would be below z = -. 608. If GMAT scores for applicants at Oxnard Graduate School of Business are N(500. B.84. smaller larger the same The normal PDF approaches the axis as z increases beyond zero so areas get smaller. C. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. B.75 in the standard normal distribution will be: A. 575. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. The top 5 percent would require z = 1.645 so x = 500 + 1.25. C.50 and z = 0. Topic: Standard Normal Distribution 85. 50) then the top 5 percent of the applicants would have a score of at least (choose the nearest integer): A. 582. Topic: Standard Normal Distribution . the area between z = 1. D.645(50) = 582. Compared to the area between z = 0.75.50 and z = 1. 601. 0. 0. Topic: Standard Normal Distribution 87.72) is: A. If the random variable Z has a standard normal distribution.26) is: A. then P(1.3944.2118.0427.1091. 0.17 ≤ Z ≤ 2.P(Z ≤ 1. If the random variable Z has a standard normal distribution.86. B. 0.17). 0. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. D.0446. 0. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-04 Find the normal probability for given z or x using tables or Excel. C. 0.9573.26) . C. B.1203. 0.S.DIST(-1. Subtract P(Z ≤ 2. Topic: Standard Normal Distribution .5016. then P(Z ≤ -1. D.1). Use Appendix C or Excel =NORM.72. C. The standard deviation is [(b . The standard deviation of this distribution is approximately: A. B. 28. 50.2500 . 25. Excel's =RAND() function produces random numbers that are uniformly distributed between 0 and 1.87. . AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-02 Calculate uniform probabilities.2887 The mean is halfway between the end points. C. B. D.0)2/12]1/2.a)2/12]1/2 = [(100 . AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-02 Calculate uniform probabilities.00. Topic: Normal Distribution 89. D. Topic: Uniform Continuous Distribution . The mean of this distribution is approximately A.3333 .88.5000 . Excel's =100*RAND() function produces continuous random numbers that are uniformly distributed between 0 and 100. 33.00.33. B. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-02 Calculate uniform probabilities.75? A. 75 percent 50 percent 25 percent This is the upper 25 percent. C. Topic: Uniform Continuous Distribution 91. 33) =NORM. What is the probability that the random number exceeds . B. 475. C.S. 33)? A.1) =NORM. 475. 33. Which is the correct Excel formula for the 80th percentile of a distribution that is N(475. AACSB: Technology Blooms: Remember Difficulty: 2 Medium Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel. Excel's =RAND() function produces random numbers that are uniformly distributed from 0 to 1.DIST(80.INV((80 .90. =NORM.INV(0. Topic: Standard Normal Distribution .80.475)/33) Review Excel functions in Appendix J. 08) = 14 and σ = [nπ(1 .063. A software developer makes 175 phone calls to its current customers.2 arrivals per minute. find the 75th percentile of waiting times until the next arrival (i..08)]1/2 = 3.007.25 by taking logs of both sides.1 seconds) Set λ = 1. no answer. If arrivals follow a Poisson distribution with mean 1. Topic: Normal Approximations .532515) using z = (x . C.2 and solve for x in right-tail area of exp(-λx) = .5 (with the continuity correction) and calculate the binomial P(X ≥ 20) ≈ P(z ≥ 1. The normal approximation of the probability of reaching at least 20 customers is: A. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Set n = 175 and π = . 75 percent below).e.532515. Use x = 19.92. There is an 8 percent chance of reaching a given customer (instead of a busy signal.μ)/σ = 1.08)(1 . . C.. . A.π)]1/2 = [175(. Topic: Exponential Distribution 93. or answering machine). B.155 minutes (69. D.4 seconds) 1.919 minutes (115.588872. B. 1.3 seconds) 0. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-08 Solve for x for a given exponential probability.937. .08. Calculate μ = nπ = (175)(.022.240 minute (14. . 5 (with the continuity correction) and calculate the binomial P(X ≤ 39) ≈ P(z ≤ -. B. what is the approximate normal probability that 39 or fewer will contain errors? A.20.06)]1/2 = 5..78895.31037.916667) using z = (x . D. The normal approximation to the binomial probability of 40 or more undeliverable letters is: A.0544 Set μ = nπ and σ = [nπ(1 . the true proportion of accounts receivable with some kind of error is .π)]1/2 = [225(.π)]1/2 = [500(.5 (with the continuity correction) and calculate the binomial P(X ≥ 40) ≈ P(z ≥ 1.μ)/σ = -.π)]1/2 and convert x = 39. C. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.916667.μ)/σ = 1.06)(1 .06) = 30 and σ = [nπ(1 .78895) using z = (x .06. the probability of an undeliverable letter is 0. B. Use x = 39.20. A letter is mailed to a sample of 500 homeowners.7695 Set n = 500 and π = .20)]1/2 = 6.06.2097 .μ)/σ.1797 . Set n = 225 and π = ..20)(1 .94.5 (using the continuity correction) to a z score with z = (x . For Gardyloo Manufacturing. C. Use x = 39. 0. If an auditor randomly samples 225 accounts receivable.000. AACSB: Analytic Blooms: Apply .9632 0.0368 0. Topic: Normal Approximations 95. Calculate μ = nπ = (225)(. D. Based on past experience. Calculate μ = nπ = (500)(. .20) = 45 and σ = [nπ(1 .1587 .2305 0. C.4713 . B.0088 0. so we calculate μ = nπ = (100)(.0251 . A multiple choice exam has 100 questions.μ)/σ = 2.π)]1/2 = [100(. so we calculate μ = nπ = (100)(. There are n = 100 questions.90.4913 A guesser would have a 20 percent chance of a correct answer (1 out of 5) so we set π = . 0. Topic: Normal Approximations 97. In a T-F exam with 100 questions.50)(1 . passing requires a score of at least 60.0287 . What would be the approximate probability that a "guesser" could achieve a score of 30 or more? A.0377 A guesser would have a 50 percent chance of a correct answer.50. D.. Use x = 29.5 (with the continuity correction) and calculate the binomial P(X ≥ 30) ≈ P(z ≥ 2. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. D.π)]1/2 = [100(.375) using z = (x ..20)]1/2 = 4.0062 0.90) using z = (x . What is the approximate normal probability that a "guesser" will score at least 60 points? A. so we set π = .20.50) = 50 and σ = [nπ(1 . AACSB: Analytic . C.20) = 20 and σ = [nπ(1 .μ)/σ = 1. Topic: Normal Approximations 96.5 (with the continuity correction) and calculate the binomial P(X ≥ 60) ≈ P(z ≥ 1. .50)]1/2 = 5. There are n = 100 questions.0015 0.375.20)(1 .Difficulty: 3 Hard Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Use x = 59. Each question has five choices. B. the default rate on a certain type of commercial loan is 20 percent.20)(1 .05 n = 100. Use x = 25. Calculate μ = nπ = (100)(. B. Historically. If a bank makes 100 of these loans. 0.20)]1/2 = 4. π = 0.0336 Set n = 100 and π = . π = 0.04 n = 40.0846 0. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. π = 0.20.π) ≥ 10. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. Topic: Normal Approximations 98.Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.25 n = 400. Topic: Normal Approximations . what is the approximate probability that at least 26 will result in default? A. For which binomial distribution would a normal approximation be most acceptable? A. n = 50.π)]1/2 = [100(.375) using z = (x . C. Topic: Normal Approximations 99.5 (with the continuity correction) and calculate the binomial P(X ≥ 26) ≈ P(z ≥ 1. B.μ)/σ = 1.375. π = 0. C. D.02 We want nπ ≥ 10 and n(1 .2000 0. D.0668 0.20) = 20 and σ = [nπ(1 .. 9975 Use x = 19.000 cars are rented from Hertz.10055) using z = (x .π)]1/2 = [25000(.000 and π = .001)(1 . B. Calculate μ = nπ = (25000)(. 30 90. D.001) = 25 and σ = [nπ(1 . what is the normal approximation to the probability that fewer than 20 will be stolen? A. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson.. B. Topic: Normal Approximations 101. the mean is ______ and the standard deviation is _____. 63 90.μ)/σ = -1.30. there is a 30 percent turnover rate for employees.1128 .937 90.001. 15 Use n = 300 and π = . Topic: Normal Approximations . C.001)]1/2 = 4.100. 90.5 (with the continuity correction) and calculate the binomial P(X < 20) ≈ P(z <-1.8335 Set n = 25. Each year.1335 . A company employs 300 employees.2577 . For this normal approximation. 7.001. A. We want to do a normal approximation to the binomial distribution of the number of employees who leave each year. C.10055. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-06 Use the normal approximation to a binomial or a Poisson. If 25. The probability that a rental car will be stolen is 0. D. . and then calculate μ = nπ and σ = [nπ(1 π)]1/2. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.92 cm cm cm cm With Excel we get =NORM.50. 195. Topic: Standard Normal Distribution 103.5) + 30 = 91.7 201.6 207. TotCo is developing a new deluxe baby bassinet.INV(.22 91. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-05 Solve for z or x for a given normal probability using tables or Excel.3 201. If the length of a newborn baby is normally distributed with a mean of 50 cm and a standard deviation of 5 cm.INV(. D.63.4 cm cm cm cm With Excel we get =NORM.9 percent of adult males will not have to stoop as they enter? A. or Appendix C with z = 3.09.63 98. C. C.99.63. D.180.45 85. with a safety margin of 15 cm on each end of the bassinet? A. Topic: Standard Normal Distribution .102. B. B.999. or Appendix C with z = 2. what should be the interior length of the bassinet to ensure that 99 percent of newborn babies will fit. 95. If adult male heights are normally distributed with a mean of 180 cm and a standard deviation of 7 cm.7) = 201. how high should an aircraft lavatory door be to ensure that 99.33. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). 2. 13. 12.082. Mean is (4 + 12 + 26)/3. C. 14. 26) has a mean of: A. Set a = 0.775. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional).bc)/18]1/2 = 4. 3. 3. The triangular distribution T(4. 4.024. Topic: Triangular Distribution (Optional) 105. 18. The triangular distribution T(0. Topic: Triangular Distribution (Optional) . c = 20 and use the triangular standard deviation formula σ = [(a2 + b2 + c2 . 10. b = 10. 12. C. 20) has a standard deviation of: A. D.082. B.994. D.ac . B.104.ab . 23.ac . 9.bc)/18]1/2 = 8. 7. 9. b = 20. Topic: Triangular Distribution (Optional) . The triangular distribution T(10. 62) has a mean of: A. c = 50 and use the triangular standard deviation formula σ = [(a2 + b2 + c2 .498. 50) has a standard deviation of: A. The triangular distribution T(5.498. B. AACSB: Analytic Blooms: Apply Difficulty: 1 Easy Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). 33. C.710. 20. Set a = 10. C. D. 35. 8.ab . B.498. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). 30. Mean is (5 + 23 + 62)/3.225. D. Topic: Triangular Distribution (Optional) 107. 23.106. and is symmetric only if the mode is halfway between a and c. The mean is (20 + 30 + 61)/3 = 37. The triangular distribution may be skewed right or left. The mode of the distribution exceeds the mean. Which statement is incorrect? A. which exceeds the mode b = 30. This triangular distribution is right-skewed. Which statement is incorrect? A. 30. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). C. Topic: Triangular Distribution (Optional) 109. D. Topic: Triangular Distribution (Optional) .5. The triangular distribution always has a single mode. The mean of the triangular distribution is (a + b + c)/3. AACSB: Analytic Blooms: Remember Difficulty: 2 Medium Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). The mean of the distribution is 37. The triangular distribution is right-skewed. It would be helpful to sketch a graph of the PDF. C. B. The midrange of the distribution is 40. The distribution is right-skewed. 61) to represent his daily commute time (minutes). B.108. Review properties of the triangular distribution. Bob used a triangular distribution of T(20. The distribution is right-skewed. B. The mode of the distribution is at the mean. 15. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-09 Use the triangular distribution for "what-if" analysis (optional). (a + c)/2 = (10 + 20)/2 = 15 so the mode b = 15 is halfway between the minimum and maximum. Topic: Triangular Distribution (Optional) 111. P(X < 5) is more than P(X ≤ 5). Topic: Describing a Continuous Distribution . A point has no area in a continuous CDF. The midrange of the distribution is 15. C. Which statement is incorrect? A. B. The distribution is symmetric if the mode b lies halfway between the end points. The mean of the distribution is 15. In a continuous distribution: A.110. P(X < 5) is the same as P(X ≤ 5). AACSB: Analytic Blooms: Understand Difficulty: 2 Medium Learning Objective: 07-01 Define a continuous random variable. C. D. In this example. Phyllis used a triangular distribution of T(10. 20) to represent her daily commute time (minutes). P(X < 5) is less than P(X ≤ 5). Topic: Describing a Continuous Distribution 113. what is the median waiting time? A. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-08 Solve for x for a given exponential probability. AACSB: Analytic Blooms: Understand Difficulty: 2 Medium Learning Objective: 07-01 Define a continuous random variable. The CDF shows P(X ≤ x). In a continuous distribution the A. C.3 9. 7. D. PDF is usually higher than the CDF. PDF shows the area under the curve. B.2 8.112. If the mean waiting time for the next arrival is 12 minutes. Topic: Exponential Distribution .50 to solve for x. B. C. Review definitions of PDF and CDF. CDF is used to find left-tail probabilities. D.1 12 minutes minutes minutes minutes Set λ = 1/12 minute per arrival and take logs of both sides of exp(-λx) = . CDF is usually the same as the PDF. D. which is not 1 so it cannot be a PDF. It depends on x. If the mean waiting time for the next arrival is 18 minutes. No. 13 7. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-08 Solve for x for a given exponential probability.9 5. AACSB: Analytic Blooms: Apply . B. Topic: Exponential Distribution 115. Yes.1 minutes minutes minutes minutes Set λ = 1/18 minute per arrival and take logs of both sides of exp(-λx) = . Could this function be a PDF? A.2 3. what is the first quartile (25th percentile) for waiting times? A. C. B.75 to solve for x. Area = 1/2 × base × height = .114.500. C. Difficulty: 2 Medium Learning Objective: 07-01 Define a continuous random variable. AACSB: Analytic Blooms: Apply Difficulty: 2 Medium Learning Objective: 07-01 Define a continuous random variable. which is not 1 so it cannot be a PDF. Topic: Describing a Continuous Distribution 116. No. C. Yes. Topic: Describing a Continuous Distribution . It depends on x. Could this function be a PDF? A. Area = base × height = 2. B. we can see that 25 is not 2σ below the mean of 45.54. Since σ = [(65 . B. Could a 25-year-old job applicant be two standard deviations below the mean (or more than two standard deviations)? A.117. Yes. Topic: Uniform Continuous Distribution . No. AACSB: Analytic Blooms: Apply Difficulty: 3 Hard Learning Objective: 07-02 Calculate uniform probabilities. Impossible to determine from given information. C.25)2/12]1/2 = 11. The ages of job applicants for a security guard position are uniformly distributed between 25 and 65.