Chapter 04

Chapter 4 Discrete Random Variables True/False 1. The binomial experiment consists of n independent, identical trials, each of which results in either success or failure and is such that the probability of success on any trial is the same. Answer: True Difficulty: Medium 2. A Poisson random variable is a continuous variable that can be used to describe the number of occurrences of an event over a specified interval of time or space. Answer: False Difficulty: Medium 3. The hypergeometric distribution is appropriate when sampling with replacement from small populations Answer: False Difficulty: Medium 4. A discrete random variable may assume a countable number of outcome values. Answer: True Difficulty: Easy 5. The variable “home ownership” can take on one of two values, one if the person living in a home owns the home and zero if the person living in a home does not own the home is an example of a discrete random variable. Answer: True Difficulty: Easy 6. If the number of surface nonconformities on a specific size of a metal piece is the discrete random variable in question, then the appropriate probability distribution that can describe the probability of a specific size metal sheet containing 3 nonconformities is most likely given by the binomial distribution. Answer: False Difficulty: Medium 7. The mean of the binomial distribution is np(1-p) Answer: False Difficulty: Easy 8. In a binomial experiment, the results of one trial are dependent on the results of other trials. Answer: False Difficulty: Easy

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9. In a binomial distribution the random variable X is continuous. Answer: False Difficulty: Easy Multiple Choice 10. When using hypergeometric distribution, it is assumed that the trials are __________ and the probability of success ________ from trial to trial. A) Independent, does not change B) Dependent, does not change C) Independent, changes D) Dependent, changes Answer: D Difficulty: Medium 11. If p = .1 and n = 5, then the corresponding binomial distribution is A) Right skewed B) Left skewed C) Symmetric D) Bimodal Answer: A Difficulty: Medium 12. If p = .5 and n = 4, then the corresponding binomial distribution is A) Right skewed B) Left skewed C) Symmetric D) Bimodal Answer: C Difficulty: Medium 13. The requirement that the probability of success remains constant from trial to trial is a property of the _________________ distribution. A) Binomial B) Uniform C) Normal D) Poisson E) Hypergeometric Answer: A Difficulty: Medium (REF)

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14. If the number of surface nonconformities on a specific size of metal piece is the discrete random variable in question, then the appropriate probability distribution that can describe the probability of a specific size metal sheet containing 3 defectives is given most likely by ___________________ distribution A) Binomial B) Poisson C) Hypergeometric D) Both A and B Answer: B Difficulty: Medium (REF) 15. Which of the following distributions can be used to solve the following problem? The average number of cars arriving at a drive-thru fast food restaurant is 3 in ten minutes, what is the probability that exactly 4 cars will arrive in a five minute interval? A) Binomial B) Poisson C) Both of the above D) None of the above Answer: B Difficulty: Medium (REF) 16. The mean of the binomial distribution is equal to: A) p B) np C) px(1-p)n-x D) (n)(p)(1-p) E) n. p (1 − p ) Answer: B Difficulty: Medium 17. A company's annual profit, y, is described by y = -1,500,000 + 60x where x is the number of units sold. If the expected number of units sold is 100,000, and σX = 12,000, what are and σY? A) 6,000,000 and 12,000 B) 6,000,000 and 720,000 C) 4,500,000 and 518,400,000,000 D) 4,500,000 and 720,000 Answer: D Difficulty: Medium


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18. The number of ways to arrange x successes among n trials is equal to n! A) x !(n − x)! B) C)

n! (n − x)! n

x

n! x! D) Answer: A Difficulty: Medium

19. Which of the following is a valid probability value for a discrete random variable? A) .2 B) 1.01 C) -.7 D) All of the above Answer: A Difficulty: Medium 20. A total of 50 raffle tickets are sold for a contest to win a car. If you purchase one ticket, what are your odds against winning? A) 49 to 1 B) 50 to 1 C) .05 D) .01 Answer: A Difficulty: Medium 21. Which one of the following statements is not an assumption of the binomial distribution? A) Sampling is with replacement. B) The experiment consists of n identical trials. C) The probability of success remains constant from trial to trial. D) Trials are independent of each other. E) Each trial results in one of two mutually exclusive outcomes. Answer: A Difficulty: Medium (REF) 22. The binomial distribution is characterized by situations that are analogous to A) Drawing balls from and urn B) Coin tossing C) Counting defects on an item D) Measuring the length of an item. Answer: B Difficulty: Medium

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23. A random variable is said to be discrete if A) Its outcomes are countable B) It can assume any real number within a given interval C) The rules of probability apply D) It can be represented graphically Answer: A Difficulty: Easy 24. Two characteristics/assumptions of the Poisson distribution include: A) Probability of success remains constant from trial to trial and the random variable of interest is continuous. B) The event occurring in one interval is independent of the event occurring in any other nonoverlapping interval, and the random variable of interest is continuous. C) The event occurring in one interval is independent of the event occurring in any other nonoverlapping interval, and the random variable of interest is discrete. D) The event occurring in one interval is dependent of the event occurring in any other nonoverlapping interval, and the random variable of interest is continuous. Answer: C Difficulty: Medium (REF) Use the following information to answer questions 25-27: The Securities and Exchange Commission has determined that the number of companies listed in NYSE declaring bankruptcy is approximately a Poisson distribution with a mean of 2.6 per month. 25. Find the probability that exactly 4 bankruptcies occur next month. A) .8774 B) .1414 C) .1557 D) .2176 Answer: B Difficulty: Medium 26. Find the probability that more than 1 bankruptcy occur next month. A) .1931 B) .9257 C) .7326 D) .4816 E) .2674 Answer: C Difficulty: Medium

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27. Find the probability that no more than one bankruptcy occurs next month. A) .1931 B) .9257 C) .7326 D) .4816 E) .2674 Answer: E Difficulty: Medium 28. A fair die is rolled 10 times. What is the probability that an odd number (1, 3, or 5) will occur less than 3 times? A) .0547 B) .1172 C) .1550 D) .7752 E) .8450 Answer: A Difficulty: Hard 29. A fair die is rolled 10 times. What is the probability that an even number (2, 4, or 6) will occur between 2 and 4 times? A) .6123 B) .1709 C) .1611 D) .3662 E) .3223 Answer: D Difficulty: Hard 30. A fair die is rolled 10 times. What is the average number of even number (2, 4, or 6) outcomes? A) 3 B) 4 C) 5 D) 6 E) 7 Answer: C Difficulty: Easy


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31. A fair die is rolled 36 times. What is the standard deviation of the even number (2, 4, or 6) outcomes? A) 18 B) 9 C) 5 D) 3 E) 1.732 Answer: D Difficulty: Medium 32. Whenever p = .5, the binomial distribution will _________ be symmetric. A) Always B) Sometimes C) Never Answer: A Difficulty: Medium 33. Which of the following statements about the binomial distribution is not correct? A) Each trial results in a success or failure. B) Trials are independent of each other. C) The probability of success remains constant from trial to trial. D) The random variable of interest is continuous. E) The experiment consists of n identical trials. Answer: D Difficulty: Easy 34. If n = 20 and p = .4, then the mean of the binomial distribution is A) .4 B) 4.8 C) 8 D) 12 Answer: C Difficulty: Easy 35. If n = 15 and p = .4, then the standard deviation of the binomial distribution is A) 9 B) 6 C) 3.6 D) 1.8974 E) .4 Answer: D Difficulty: Easy

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36. The equation for the variance of the binomial distribution is given by: A) px(1 – p)n-x B) np C) np(1-p) np (1 − p ) D) Answer: C Difficulty: Easy Use the following information to answer questions 37-38: The manager of the local grocery store has determined that, on the average, 4 customers use the service desk every half-hour. Assume that the number of customers using the service desk has a Poisson distribution. 37. What is the probability that during a randomly selected half-hour period exactly 2 customers use the service desk? A) .0183 B) .0733 C) .1465 D) .9084 E) .7619 Answer: C Difficulty: Medium 38. What is the probability that during a randomly selected half-hour period no more than 2 customers use the service desk? A) .2381 B) .1465 C) .7619 D) .8535 E) .0916 Answer: A Difficulty: Medium Fill-in-the-Blank 39. The variable "employment status" which can take on one of two values: 1 for "employed" and 0 for "unemployed" is and example of a(n) _____________ random variable. Answer: Discrete Difficulty: Medium 40. If x is a binomial random variable, then the standard deviation of x is given by Answer: square root (npq) Difficulty: Medium


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41. A random variable that is defined to be the total number of successes in n trials is a(n) _____ random variable. Answer: Binomial Difficulty: Medium 42. A discrete variable that can often be used to describe the number of occurrences of an event over a specified interval of time or space is a(n) _____ random variable. Answer: Poisson Difficulty: Medium 43. The requirement that the probability of success remains constant from trial to trial is a property of the _______________ distribution. Answer: Binomial Difficulty: Medium 44. The hypergeometric distribution is used when sampling _____ replacement from small populations Answer: Without Difficulty: Medium 45. The distribution whose mean is equal to its variance is the _________ distribution. Answer: Poisson Difficulty: Easy 46. For a random variable X, the mean value of the squared deviations of its values from their expected value is called its Answer: Variance Difficulty: Hard Essay Use the following information to answer questions 47-50: The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 10%. 47. If 3 calculators are selected at random, what is the probability that one of the calculators will be defective? Answer: .243 3! (.1)1 (.9) 2 = .243 1!(3 −1)! Difficulty: Medium

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48. If 10 calculators are selected at random, what is the probability that 3 or more of the calculators will be defective? Answer: .0702 P(X ≥ 3) = 1 – P(X ≤ 2) = 1 - .9298 = .0702 Difficulty: Medium 49. If 100 calculators are selected at random, what is the expected number of defectives? Answer: 10 E[X] = µ x = (.10)(100) = 10 Difficulty: Medium 50. If 100 calculators are selected at random, what is the standard deviation of the number of defectives? Answer: 3 σ X = (.1)(.9)(100) = 3 Difficulty: Medium Use the following information to answer questions 51-52: Historical data shows that the average number of patient arrivals at the intensive care unit of General Hospital is 3 patients every 2 hours. Assume that the patient arrivals are distributed according to a Poisson distribution. 51. Determine the probability of 6 patients arriving in a five-hour period. Answer: .136 (e −7.5 )(7.5) 6 P( X = 6) = = .1359 6! Difficulty: Medium 52. Determine the probability of at least 4 but no more than 8 patients arriving in a three-hour period. Answer: .6174 P(4 ≤ X ≤ 8) = P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) P(4 ≤ X ≤ 8) = (.0463) + (.0824) + (.1281) + (.1708) + (.1898) = .6174 Difficulty: Medium


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Use the following information to answer questions 53-55: If the probability distribution of X is: X 3 4 5 6

P(X) 1/8 1/8 3/8 3/8

53. What is the expected value of X? Answer: 5.0 E [ X ] = (3) 1 + (4) 1 + 5 3 + 6 3 = 40 = 5.0 8 8 8 8 8 Difficulty: Medium

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( ) ( ) ( )

54. What is the variance of X? Answer: 1.0 E [ X ] = (3) 1 + (4) 1 + 5 3 + 6 3 = 40 = 5.0 8 8 8 8 8

( )

( ) ( ) ( )

1  1  3  3 σ X2 = (3 − 5) 2   + (4 − 5) 2   + (5− 5) 2  + (6− 5) 2  = 1 8  8  8  8

Difficulty: Medium 55. Assume the number of trucks passing an intersection has a Poisson distribution with mean of 5 trucks per minute. What is the probability of 0 or 1 trucks in one minute? Answer: .0404 e −5 50 e− 5 51 + = .0404 0! 1! Difficulty: Medium

Use the following information to answer questions 56-57: A vaccine is 95 percent effective. What is the probability that it is not effective for: 56. One and only one individual out of 20 individuals? Answer: .3774 (20) ( .95 ) 19 ( .05) 1 = .3774 Difficulty: Medium

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57. More than one out of 20 individuals? Answer: .2641 P(X ≥ 2) = 1 – [P(X = 0) + p(X =1)] P(X ≥ 2) = 1 – [(.3585) + (.3774)] = .2641 Difficulty: Medium 58. If the probability of a success on a single trial is .2, what is the probability of obtaining 3 successes in 10 trials if the number of successes is binomial? Answer: .2013 10! (.2)3 (.8) 7 = .2013 3!(10 − 3)! Difficulty: Medium

59. The number of calls coming into a PBX follows a Poisson process with a mean of 120 calls per hour. What is the probability of no calls in a one-minute interval? Answer: .1353 e −2 20 P( X = 0) = = e −2 = .1353 0! Difficulty: Medium (AS) 60. If x is a Poisson random variable with a mean of 10, what is the probability that x is greater than 6? Answer: .8698 P(X ≥ 7) = 1 – P(X ≤ 6) = 1 – (.1302) = .8698 Difficulty: Medium 61. An archer hits a target 95% of the time. What is the probability he/she will miss the target for the first time on the 15th shot? Answer: .0244 ( .95) 14 ( .05) 1 =.0244 Difficulty: Easy


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62. Three candidates run for different offices in different counties. Each has a one in three chance of being elected in his/her county. What is the probability that at least one of them will be elected? Answer: .7037 P ( X ≥ 1) = 1 − P ( X = 0) 0

3

8  3!  1   2  P ( X = 0) =      = 27  3!  3   3  19 P ( X ≥ 1) = 1 − 8 = = .7037 27 27 Difficulty: Hard

63. A test has 6 multiple choice questions, each with 4 alternatives. What is the probability of guessing 5 or more questions correctly? Answer: .0046 P(X ≥ 5) = P(X = 5) + P(X = 6) P(X ≥ 5) = (.0044) + (.0002) = .0046 Difficulty: Hard 64. If x is a Poisson random variable with a mean of 10, what is the probability that x is greater than or equal to 2? Answer: .9995 P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)] P(X ≥ 2) = 1 – (0 + .0005) = .9995 Difficulty: Medium (AS) 65. If x is a Poisson random variable with a mean of 10, what is the probability that x is equal to 8? Answer: .1126 e−10 108 P ( X = 8) = = .1126 8! Difficulty: Medium 66. Twenty coins are tossed. What is the probability of getting exactly 10 heads? Answer: .1762 20! P ( X = 10) = (.5) 10(.5) 10 = .1762 10!(20 −10)! Difficulty: Medium

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67. Determine the probability that a 3 will appear twice, if a fair die is rolled 10 times. Answer: .2907 2 8  10!   1   5  P( X = 2) =       = .2907  2!8!  6   6  Difficulty: Hard Use the following information to answer questions 68-69: During off hours, cars arrive at a tollbooth on the East-West toll road at an average rate of 0.5 cars per minute. The arrivals are distributed according to the Poisson distribution. 68. What is the probability that during the next minute three cars will arrive? Answer: .0126 e −.5 .53 (.6065)(.125) P ( X = 3) = = = .0126 3! 6 Difficulty: Medium 69. What is the probability that during the next five minutes three cars will arrive? Answer: .2138 e −2.5 2.53 (.0821)(15.625) P ( X = 3) = = = .2138 3! 6 Difficulty: Medium Use the following information to answer questions 70-75: For a binomial process, the probability of success is 40% and the number of trials is 5. 70. Find the expected value. Answer: 2 E[X] = (5)(.40) = 2 Difficulty: Medium 71. Find the variance. Answer: 1.2 σ 2 x = (5)(.4)(.6) = 1.2 Difficulty: Medium


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72. Find the standard deviation. Answer: 1.0954

σ X = (5)(.4)(.6) = 1.2= 1.0954 Difficulty: Medium 73. Find P(X ≤ 1). Answer: .3370 P(X ≤ 1) = [P(X = 0) + P(X = 1) P(X ≤ 1) = (.0778) + (.2592) = .337 Difficulty: Medium 74. Find P(X > 4). Answer: .0102 P(X = 5) = (.4)5 = .0102 Difficulty: Medium 75. Find the P(X = 2). Answer: .3456  5!  2 3 P ( X = 2) =   (.4) (.6) = .3456  2!3! Difficulty: Medium 76. Consider a Poisson distribution with an average of 3 customers per minute at the local grocery store. Determine the expected number of customer arrivals for a five-minute period Answer: 15 µ = (3)(5) = 15 Difficulty: Easy (AS) 77. Consider a Poisson distribution with an average of 3 customers per minute at the local grocery store. Write a formula for the probability of X, where X = the number of arrivals per minute. Answer: e −3 3 X P( X ) = X! Difficulty: Medium Use the following information to answer questions 78-83: Consider a Poisson distribution with an average of 3 customers per minute at the local grocery store. If X = the number of arrivals per minute:

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78. Find the expected value of X. Answer: 3 Difficulty: Easy 79. Consider a Poisson distribution with an average of 3 customers per minute at the local grocery store. If X = the number of arrivals per minute, find the variance of X. Answer: 3 σ 2 = µ = 3 Difficulty: Easy 80. Consider a Poisson distribution with an average of 3 customers per minute at the local grocery store. If X = the number of arrivals per minute, find the standard deviation of X. Answer: 1.732 σ X = 3 = 1.7321 Difficulty: Easy 81. Find the probability of 10 customers or less arriving within a minute. Answer: .9997 P(X ≤ 10) = 1 – P(X ≥ 11) = 1 – (.0002 + .0001) = .9997 Difficulty: Medium 82. Find the probability of more than 7 customers arriving within a minute. Answer: .0119 P(X ≥ 8) = .0081 + .0027 + .0008 + .0002 + .0001 = .0119 Difficulty: Medium 83. Find the probability of 3 customers arriving within a minute. Answer: .224 e −3 33 (.0498)(27) P ( X = 3) = = = .224 3! 6 Difficulty: Medium 84. One die is thrown. What is the expected value of the number of dots on the top face of the die? Answer: 3.5 1 1 1 1 1 1 E [ X ] = 1  + 2   + 3   + 4   + 5   + 6   6 6 6 6 6 6 Difficulty: Medium


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85. If X has the probability distribution X -1 0 1

P(X) .2 .3 .5

compute the expected value of X. Answer: .3 E[X] = -1(.2) + 0(.3) + 1(.5) = .3 Difficulty: Medium 86. If X has the probability distribution X P(X )

-2 .2

-1 .2

9 .2

1 .2

2 .2

compute the expected value of X. Answer: 1.8 E[X] = (-2)(.2) + (-1)(.2) + (1)(.2) + (2)(.2) + (9)(.2) = 1.8 Difficulty: Medium (AS)

Use the following information to answer questions 87-88: X has the following probability distribution P(X): X P(X)

1 .1

2 .5

3 .2

4 .2

87. Compute the expected value of X. Answer: 2.5 E[X] = (1)(.1) + (2)(.5) + (3)(.2) + (4)(.2) = 2.5 Difficulty: Medium 88. Compute the variance value of X. Answer: .9125 E[X] = (1)(.1) + (2)(.5) + (3)(.2) + (4)(.2) = 2.5 σ 2 x = = (1 – 2.5) 2 (.1) + (2 – 2.5) 2 (.5) + (3 – 2.5) 2 (.2) + (4 – 2.5) 2 (.2) = .9125 Difficulty: Medium

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Use the following information to answer questions 89-91: Historical data for a local steel manufacturing company shows that the average number of defects per standard sheet of steel is 2. In addition, the number of defects per unit is distributed according to Poisson distribution. 89. What is the probability that there will be a total of 7 defects on four standard sheets of steel? Answer: 0.1396 e −8 87 (.0003)(2, 097,152) P ( X = 7) = = = .1396 7! 5040 Difficulty: Medium 90. A batch has just been completed. What is the probability that the first three units manufactured in this batch will contain at least a total of 4 defects? Answer: 0.8488 P(X ≥ 4) = 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)] P(X ≥ 4) = 1 – (.0025 + .0149 + .0446 + .0892) = .8488 Difficulty: Medium 91. Determine the standard deviation of the number of defective units for 32 sheets of metal. Answer: 8 (2)(32) = 64 = 8 Difficulty: Easy 92. Consider the experiment of tossing a fair coin three times and observing the number of heads that result (X = number of heads). Construct the probability distribution for the random variable X. Answer: X P(X) 0 1/8 1 3/8 2 3/8 3 1/8 Difficulty: Hard (AS) Use the following information to answer questions 93-97: Consider the experiment of tossing a fair coin three times and observing the number of heads that result (X = number of heads).


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93. Determine the expected number of heads. Answer: 1.5 E [ X ] = µ X = (0) 1 + (1) 3 + (2) 3 + 3 1 = 1.5 8 8 8 8 Difficulty: Hard

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94. What is the variance for this distribution? Answer: .75 E [ X ] = µ X = (0) 1 + (1) 3 + (2) 3 + 3 1 = 1.5 8 8 8 8 1 3 3 1 σ X2 = (0 − 1.5)2   + (1 − 1.5)2   + (2 − 1.5)2   + (3 − 1.5)2   = .75 8 8 8 8 Difficulty: Hard

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( )

( ) ( )

95. What is the standard deviation for this distribution? Answer: .866 E [ X ] = µ X = (0) 1 + (1) 3 + (2) 3 + 3 1 = 1.5 8 8 8 8 1  3  3  σ X2 = (0 − 1.5) 2   + (1− 1.5) 2   + (2 − 1.5) 2  + (3− 1.5) 2 8  8  8 

( )

( )

( ) ( )

1  = .75 8

σ X = .75 = .866 Difficulty: Hard 96. If you were asked to play a game in which you tossed a fair coin three times and were given $2 for every head you threw, how much would you expect to win on average? Answer: $3 Expected return = (0) 1 + ($2)(1) 3 + ($2)(2) 3 + ($2)(3) 1 = 3.0 8 8 8 8 Difficulty: Hard

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97. A pharmaceutical company has determined that if a new cholesterol-reducing drug is manufactured (introduced to the market), the following probability distribution will describe this drug's contribution to the company's profits during the next six months. Profit Contribution -$ 30,000 $ 50,000 $200,000

Probability of profit contribution .20 .50 .30

The company management has decided to market this product if the expected contribution to profit for the next six months is more than $90,000. Based on the information given above, should the company begin manufacturing the new drug? Answer: No (79,000 < 90,000) µ x = .2(-30,000) + .5(50,000) + .3(200,000) = 79,000 Difficulty: Medium Use the following information to answer questions 98-102: According to data from the state blood program, 40% of all individuals have group A blood. If six (6) individuals give blood, find the probability 98. None of the individuals has group A blood? Answer: .0467 Difficulty: Easy 99. Exactly three of the individuals has group A blood? Answer: .2765 Difficulty: Easy 100. At least 3 of the individuals have group A blood Answer: .4557 P(x≥3) = P(x=3) + p(x=4) + p(x=5) + p(x=6)=.4557 Difficulty: Medium 101. Find the mean number of individuals having group A blood. Answer: 2.4 µ x = np = (.6)(.4)=2.4 Difficult: Easy


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102. Suppose that of the six randomly selected individuals, 3 have group A blood. Based on the probability in question 100, would you believe the data from the state blood program? Why or why not? Answer: Yes because probability > .05 Difficulty: Medium 103. A lawyer believes that the probability is .3 that she can win a discrimination suit. If she wins the case she will make $40,000, but if she loses she gets nothing. Assume that she has to spend $5000 preparing the case. What is her expected gain? Answer: $7000 µ x = (.7)(-5000) + (.3)(35000)=7000 Difficulty: Hard Use the following information to answer questions 104-36: Your company’s internal auditor believes that 10% of the company’s invoices contain errors. To check this theory, 20 invoices are randomly selected and 5 are found to have errors. 104. What is the probability that of the 20 invoices written, five or more would contain errors if the theory is valid? Answer: .0433 P(x≥5)= (.0319)+(.0089)+(.0020)+(.0004)+(0001)=.0433 Difficulty: Easy 105. Would you accept or reject the claim? Explain. Answer: Reject claim because P < .05 Use the following information to answer questions 106-109: An important part of the customer service responsibilities of a cable company relates to the speed with which trouble in service can be repaired. Historically, the data show that the likelihood is 0.75 that troubles in a residential service can be repaired on the same day. For the first five troubles reported on a given day, what is the probability that: 106. All five will be repaired on the same day Answer: .2373 P(x = 5) = .2373 Difficulty: Easy

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107. Fewer than two troubles will be repaired on the same day? Answer: .0046 P(x<2) = P(x=0) + P(x=1) = .0156 Difficult: Medium 108. At least 3 troubles will be repaired on the same day. Answer: .9624 P(x≥3) = 1-(P≤2) = .8965Difficulty: Medium 109. Find the mean number of troubles repaired on the same day Answer: 3.75 µ x = np = (5)(.75) = 3.75Difficulty: Easy Use the following information to answer questions 110-113: The Post Office has established a record in a major Midwestern city for delivering 90% of its local mail the next working day. 110. If you mail eight local letters, what is the probability that all of then will be delivered the next day. Answer: .4305 P(x=8) = .4305 Difficulty Easy (AS) 111. Of the eight, what is the average number you expect to be delivered the next day? Answer: 7.2 µ x = np =(8)(.9) = 7.2 Difficulty: Easy (AS) 112. Calculate the standard deviation of the number delivered. Answer: .85 Σ= npq = (7.2)(.1) = .72 = .85 Difficulty: Medium (AS)


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113. What is the probability that the number of delivered will be within 2 standard deviations of the mean Answer: .9619 P(7.2 ±2(.85)) = p(7.2 ±1.7) = P(5.5≤x≤8)= P(6≤x≤8)=(.4305)+(.3826)+(.1488)=9616 Difficulty: Hard (AS) 114. A car wash loses $30 on rainy days and makes $120 on days when is does not rain. If the probability of rain is 0.15, calculate the car wash’s expected profit. Answer: $97.50 µ x = (-30)(.15) + (120)(.85)= -4.50+102= 97.50 Difficulty: Medium Use the following information to answer questions 115-116: An insurance company will insure a $75,000 Hummer for its full value against theft at a premium of $1500 per year. Suppose that the probability that the Hummer will be stolen is 0.0075. 115. Calculate the insurance company’s expected net profit. Answer:$937.50 µ x = (-73500)(.0075) + (1500)(.9925)=-551.25+1488.75=937.50 Difficulty: Hard 116. Find the premium that the insurance company should charge if its wants its expected net profit to be $2000? Answer: $2,562.50 2000 = (x-75000)(.0075) + x(.9925)=.0075x – 562.5 + .9925x=2562.5 Difficulty: Hard Use the following information to answer questions 117-119: A large disaster cleaning company estimates that 30% of the jobs it bids on are finished within the bid time. Looking at a random sample of 8 jobs that is has contracted: 117. Calculate the probability that exactly 4 of the jobs were not completed within the bid time. Answer: .1361 P(x=4) = .1361 Difficulty: Medium

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118. Calculate the mean number of jobs completed within the bid time. Answer: 2.4 µ x = np = 8(.3) = 2.4 Difficulty: Medium 119. Find the probability that x (number of jobs finished on time) is within one standard deviation of the mean. Answer: .5506 σ = npq = (2.4)(.7) = 1.68 =1.3, 2.4±1.3= (1.1,3.7) – 2,3 (.2965)+(.2541)=.5506 Difficulty: Hard Multiple Choice 120. The probability that a given computer chip will fail is 0.02. Find the probability that of 5 delivered chips, exactly 2 will fail. A) .9039 B) .0922 C) .0038 D) .0000 Answer: C Difficulty: Medium 121. According to a survey of adults, 64% have money in a regular savings account. If we plan on surveying 50 randomly selected adults, find the mean number of adults who would have regular savings accounts. A) 12 B) 22 C) 32 D) 42 Answer: C Difficult: Medium Use the following information to answer questions 122-123: In the most recent election, 19% of all eligible college students voted. If a random sample of 20 students were surveyed: 122. Find the probability that exactly half voted in the election. A) .0000 B) .0014 C) .0148 D) .4997 Answer: B Difficulty: Medium Bowerman, Essentials of Business Statistics, 2/e

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123. Find the probability that none of the students voted A) .0000 B) .0014 C) .0148 D) .4997 Answer: C Difficulty: Medium 124. Of all individual tax returns, 37% include errors made by the taxpayer. If IRS examiners are assigned randomly selected returns in batches of 12, find the mean and standard deviation for the number of erroneous returns per batch. A) µ = 2.80, σ = 1.67 B) µ = 4.44, σ = 1.67 C) µ = 4.44, σ = 2.80 D) µ = 7.56, σ = 2.80 Answer: B Difficulty: Medium 125. In a study conducted for the State Department of Education, 30% of the teachers who left teaching did so because they were laid off. Assume that we randomly select 10 teachers who have recently left their profession. Find the probability that exactly 4 of them were laid off. A) .3000 B) .2668 C) .2001 D) .0090 Answer: C Difficulty: Medium 126. A company manufactures an appliance, gives a warranty and 95% of its appliances do not require repair before the warranty expires. If an organization buys 10 of these appliances, calculate an interval that contains 95.44% of all the appliances that won’t require repair. A) [8.12 10.88] B) [7.43 11.57] C) [8.81 10.19] D) [8.55 10.45] Answer: A Difficulty: Medium

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127. A manufacturer tested a sample of semiconductor chips and found that 35 were defective and 190 were good. If additional tests are to be conducted with random samples of 160 semiconductor chips, find the mean for the number of defects in these groups of 160 (rounded to nearest whole number). A) 56 B) 35 C) 29 D) 25 Answer: D Difficulty: Medium 128. In a study conducted by UCLA, it was found that 25% of college freshmen support increased military spending. If 6 college freshmen are randomly selected, find the probability that: Fewer than 4 support increased military spending A) .0330 B) .7844 C) .9624 D) .9954 Answer: C Difficulty: Medium (AS) 129. Exactly 3 support increased military spending A) .0330 B) .1318 C) .7844 D) .9624 Answer: B Difficulty: Medium (AS) 130. Only 1 supports increased military spending A) .0330 B) .1318 C) .3560 D) .7844 Answer: C Difficulty: Medium (AS) 131. A multiple-choice test has 30 questions and each one has five possible answers, of which one is correct. If all answers were guesses, find the probability of getting exactly four correct answers. A) .0604 B) .1325 C) .2552 D) .8000 Answer: B Difficulty: Medium


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Chapter 04

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