Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
The Normal Distribution and Other Continuous Distributions 153 CHAPTER 6: THE NORMAL DISTRIBUTION AND OTHER CONTINUOUS DISTRIBUTIONS 1. In its standardized form, the normal distribution a) has a mean of 0 and a standard deviation of 1. b) has a mean of 1 and a variance of 0. c) has an area equal to 0.5. d) cannot be used to approximate discrete probability distributions. ANSWER: a TYPE: MC DIFFICULTY: Easy KEYWORDS: standardized normal distribution, properties 2. Which of the following about the normal distribution is not true? a) Theoretically, the mean, median, and mode are the same. b) About 2/3 of the observations fall within 1 standard deviation from the mean. c) It is a discrete probability distribution. d) Its parameters are the mean, , and standard deviation, . ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties 3. If a particular batch of data is approximately normally distributed, we would find that approximately a) 2 of every 3 observations would fall between 1 standard deviation around the mean. b) 4 of every 5 observations would fall between 1.28 standard deviations around the mean. c) 19 of every 20 observations would fall between 2 standard deviations around the mean. d) All the above. ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, properties 4. For some positive value of Z, the probability that a standard normal variable is between 0 and Z is 0.3770. The value of Z is a) 0.18. b) 0.81. c) 1.16. d) 1.47. ANSWER: c TYPE: MC DIFFICULTY: Moderate 154 The Normal Distribution and Other Continuous Distributions KEYWORDS: standardized normal distribution, value 5. For some value of Z, the probability that a standard normal variable is below Z is 0.2090. The value of Z is a) – 0.81. b) – 0.31. c) 0.31. d) 1.96. ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 6. For some positive value of Z, the probability that a standard normal variable is between 0 and Z is 0.3340. The value of Z is a) 0.07. b) 0.37. c) 0.97. d) 1.06. ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 7. For some positive value of X, the probability that a standard normal variable is between 0 and +2X is 0.1255. The value of X is a) 0.99. b) 0.40. c) 0.32. d) 0.16. ANSWER: d TYPE: MC DIFFICULTY: Difficult KEYWORDS: normal distribution, value 8. For some positive value of X, the probability that a standard normal variable is between 0 and +1.5X is 0.4332. The value of X is a) 0.10. b) 0.50. c) 1.00. d) 1.50. ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: normal distribution, value The Normal Distribution and Other Continuous Distributions 155 9. Given that X is a normally distributed random variable with a mean of 50 and a standard deviation of 2, find the probability that X is between 47 and 54. ANSWER: 0.9104 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 10. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What proportion of the plan recipients would receive payments beyond age 75? ANSWER: 0.0228 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 11. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What proportion of the plan recipients die before they reach the standard retirement age of 65? ANSWER: 0.1957 using Excel or 0.1949 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 12. A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. Find the age at which payments have ceased for approximately 86% of the plan participants. ANSWER: 71.78 years old TYPE: PR DIFFICULTY: Difficult KEYWORDS: normal distribution, value 13. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order? a) 0.22313 b) 0.48658 c) 0.51342 d) 0.77687 156 The Normal Distribution and Other Continuous Distributions ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 14. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. What proportion of customers having to hold more than 1.5 minutes will hang up before placing an order? a) 0.86466 b) 0.60653 c) 0.39347 d) 0.13534 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 15. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. Find the waiting time at which only 10% of the customers will continue to hold. a) 2.3 minutes b) 3.3 minutes c) 6.9 minutes d) 13.8 minutes ANSWER: c TYPE: MC DIFFICULTY: Difficult KEYWORDS: exponential distribution, value 16. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What proportion of callers is put on hold longer than 2.8 minutes? a) 0.367879 b) 0.50 c) 0.60810 d) 0.632121 ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability The Normal Distribution and Other Continuous Distributions 157 17. A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What is the probability that a randomly selected caller is placed on hold fewer than 7 minutes? a) 0.0009119 b) 0.082085 c) 0.917915 d) 0.9990881 ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 18. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes. a) 0.3551 b) 0.3085 c) 0.2674 d) 0.1915 ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, probability 19. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot. a) 0.0919 b) 0.2255 c) 0.4938 d) 0.7745 ANSWER: d TYPE: MC DIFFICULTY: Easy KEYWORDS: normal distribution, probability 20. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the point in the distribution in which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot. a) 2.8 minutes b) 3.2 minutes c) 3.4 minutes d) 4.2 minutes 158 The Normal Distribution and Other Continuous Distributions ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, value 21. Let X represent the amount of time it takes a student to park in the library parking lot at the university. If we know that the distribution of parking times can be modeled using an exponential distribution with a mean of 4 minutes, find the probability that it will take a randomly selected student more than 10 minutes to park in the library lot. a) 0.917915 b) 0.670320 c) 0.329680 d) 0.082085 ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 22. Let X represent the amount of time it takes a student to park in the library parking lot at the university. If we know that the distribution of parking times can be modeled using an exponential distribution with a mean of 4 minutes, find the probability that it will take a randomly selected student between 2 and 12 minutes to park in the library lot. a) 0.049787 b) 0.556744 c) 0.606531 d) 0.656318 ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 23. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh more than 4.4 pounds is _______? ANSWER: 0.0668 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 24. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh between 3 and 5 pounds is _______? ANSWER: 0.5865 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability The Normal Distribution and Other Continuous Distributions 159 25. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight (in pounds) should the citation designation be established? a) 1.56 pounds b) 4.84 pounds c) 5.20 pounds d) 7.36 pounds ANSWER: b TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, value 26. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, above what weight (in pounds) do 89.80% of the weights occur? ANSWER: 2.184 pounds TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 27. The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh less than 2.2 pounds is _______? ANSWER: 0.1056 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 28. The Tampa International Airport (TIA) has been criticized for the waiting times associated with departing flights. While the critics acknowledge that many flights have little or no waiting times, their complaints deal more specifically with the longer waits attributed to some flights. The critics are interested in showing, mathematically, exactly what the problems are. Which type of distribution would best model the waiting times of the departing flights at TIA? a) Uniform distribution b) Binomial distribution c) Normal distribution d) Exponential distribution ANSWER: d TYPE: MC DIFFICULTY: Moderate KEYWORDS: exponential distribution, properties 29. Scientists in the Amazon are trying to find a cure for a deadly disease that is attacking the rain forests there. One of the variables that the scientists have been measuring involves the diameter of the trunk of the trees that have been affected by the disease. Scientists have calculated that the 160 The Normal Distribution and Other Continuous Distributions average diameter of the diseased trees is 42 centimeters. They also know that approximately 95% of the diameters fall between 32 and 52 centimeters and almost all of the diseased trees have diameters between 27 and 57 centimeters. When modeling the diameters of diseased trees, which distribution should the scientists use? a) uniform distribution b) binomial distribution c) normal distribution d) exponential distribution ANSWER: c TYPE: MC DIFFICULTY: Moderate KEYWORDS: normal distribution, properties 30. In the game Wheel of Fortune, which of the following distributions can best be used to compute the probability of winning the special vacation package in a single spin? a) uniform distribution b) binomial distribution c) normal distribution d) exponential distribution ANSWER: a TYPE: MC DIFFICULTY: Moderate KEYWORDS: uniform distribution, properties 31. A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall below 10.875 ounces. ANSWER: 0.8944 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 32. A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall above 10.95 ounces. ANSWER: 0.0668 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 33. True or False: The probability that a standard normal random variable, Z, falls between – 1.50 and 0.81 is 0.7242. ANSWER: True The Normal Distribution and Other Continuous Distributions 161 TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 34. True or False: The probability that a standard normal random variable, Z, is between 1.50 and 2.10 is the same as the probability Z is between – 2.10 and – 1.50. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 35. True or False: The probability that a standard normal random variable, Z, is below 1.96 is 0.4750. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 36. True or False: The probability that a standard normal random variable, Z, is between 1.00 and 3.00 is 0.1574. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 37. True or False: The probability that a standard normal random variable, Z, falls between –2.00 and –0.44 is 0.6472. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 38. True or False: The probability that a standard normal random variable, Z, is less than 50 is approximately 0. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 39. True or False: A worker earns $15 per hour at a plant and is told that only 2.5% of all workers make a higher wage. If the wage is assumed to be normally distributed and the standard deviation of wage rates is $5 per hour, the average wage for the plant is $7.50 per hour. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal distribution, mean 162 The Normal Distribution and Other Continuous Distributions 40. True or False: Theoretically, the mean, median, and the mode are all equal for a normal distribution. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal distribution, properties 41. True or False: Any set of normally distributed data can be transformed to its standardized form. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal distribution, properties 42. True or False: The "middle spread," that is the middle 50% of the normal distribution, is equal to one standard deviation. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal distribution, probability, value 43. True or False: A normal probability plot may be used to assess the assumption of normality for a particular batch of data. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: normal probability plot 44. True or False: If a data batch is approximately normally distributed, its normal probability plot would be S-shaped. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: normal probability plot 45. The probability that a standard normal variable Z is positive is ________. ANSWER: 0.50 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution 46. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 110 grams of pyridoxine? The Normal Distribution and Other Continuous Distributions 163 ANSWER: 0.1554 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 47. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain between 82 and 100 grams of pyridoxine? ANSWER: 0.2132 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 48. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain at least 100 grams of pyridoxine? ANSWER: 0.6554 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 49. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 120 grams of pyridoxine? ANSWER: 0.3108 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 50. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain less than 100 grams of pyridoxine? ANSWER: 0.3446 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 51. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. What is the probability that a randomly selected vitamin will contain less than 100 grams or more than 120 grams of pyridoxine? ANSWER: 0.6892 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 164 The Normal Distribution and Other Continuous Distributions 52. The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with = 110 grams and = 25 grams. Approximately 83% of the vitamins will have at least how many grams of pyridoxine? ANSWER: 86.15 using Excel or 86.25 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 53. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be between 121 and 124 inches? ANSWER: 0.8186 using Excel or 0.8185 using Table E.2 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 54. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be over 125 inches in length? ANSWER: 0.0228 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 55. The true length of boards cut at a mill with a listed length of 10 feet is normally distributed with a mean of 123 inches and a standard deviation of 1 inch. What proportion of the boards will be less than 124 inches? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 56. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is less than 1.15 is __________. ANSWER: 0.8749 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 57. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is more than 0.77 is __________. ANSWER: 0.2207 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability The Normal Distribution and Other Continuous Distributions 165 58. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is less than -2.20 is __________. ANSWER: 0.0139 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 59. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is more than -0.98 is __________. ANSWER: 0.8365 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 60. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -2.33 and 2.33 is __________. ANSWER: 0.9802 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 61. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -2.89 and -1.03 is __________. ANSWER: 0.1496 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 62. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -0.88 and 2.29 is __________. ANSWER: 0.7996 TYPE: FI DIFFICULTY: Easy KEYWORDS: standardized normal distribution, probability 63. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z values are larger than __________ is 0.3483. ANSWER: 0.39 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, probability 64. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z values are larger than __________ is 0.6985. 166 The Normal Distribution and Other Continuous Distributions ANSWER: -0.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, probability 65. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 27% of the possible Z values are smaller than __________. ANSWER: -0.61 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 66. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 85% of the possible Z values are smaller than __________. ANSWER: 1.04 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 67. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 96% of the possible Z values are between __________ and __________ (symmetrically distributed about the mean). ANSWER: -2.05 and 2.05 or -2.06 and 2.06 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 68. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. So 50% of the possible Z values are between __________ and __________ (symmetrically distributed about the mean). ANSWER: -0.67 and 0.67 or -0.68 and 0.68 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standardized normal distribution, value 69. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in less than 12 minutes. ANSWER: 0.0668 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability The Normal Distribution and Other Continuous Distributions 167 70. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 14 and 16 minutes. ANSWER: 0.3829 using Excel or 0.3830 using Table E.2 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 71. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 10 and 12 minutes. ANSWER: 0.0606 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 72. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 15 and 21 minutes. ANSWER: 0.49865 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 73. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in between 16 and 21 minutes. ANSWER: 0.30719 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 74. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in more than 11 minutes. ANSWER: 0.97725 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 75. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in more than 19 minutes. ANSWER: 0.0228 168 The Normal Distribution and Other Continuous Distributions TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 76. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. The probability is __________ that a product is assembled in less than 20 minutes. ANSWER: 0.9938 TYPE: FI DIFFICULTY: Easy KEYWORDS: normal distribution, probability 77. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 15% of the products require more than __________ minutes for assembly. ANSWER: 17.0729 using Excel or 17.08 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 78. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 90% of the products require more than __________ minutes for assembly. ANSWER: 12.44 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value 79. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 60% of the products would be assembled within __________ and __________ minutes (symmetrically distributed about the mean). ANSWER: 13.32 and 16.68 or 13.31 and 16.69 TYPE: FI DIFFICULTY: Difficult KEYWORDS: normal distribution, value 80. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 17% of the products would be assembled within __________ minutes. ANSWER: 13.1 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value The Normal Distribution and Other Continuous Distributions 169 81. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 70% of the products would be assembled within __________ minutes. ANSWER: 16.0488 using Excel or 16.04 using Table E.2 TYPE: FI DIFFICULTY: Moderate KEYWORDS: normal distribution, value TABLE 6-1 The manager of a surveying company believes that the average number of phone surveys completed per hour by her employees has a normal distribution. She takes a sample of 15 days output from her employees and determines the average number of surveys per hour on these days. The ordered array for this data is: 10.0, 10.1, 10.3, 10.5, 10.7, 11.2, 11.4, 11.5, 11.7, 11.8, 11.8, 12.0, 12.2, 12.2, 12.5. 82. Referring to Table 6-1, the first standard normal quantile is ________. ANSWER: -1.5341 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 83. Referring to Table 6-1, the fourth standard normal quantile is ________. ANSWER: -0.6745 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 84. Referring to Table 6-1, the ninth standard normal quantile is ________. ANSWER: +0.1573 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 85. Referring to Table 6-1, the fourteenth standard normal quantile is ________. ANSWER: +1.1503 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 86. Referring to Table 6-1, the last standard normal quantile is ________. ANSWER: +1.5341 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 170 The Normal Distribution and Other Continuous Distributions 87. Referring to Table 6-1, construct a normal probability plot for the data. ANSWER: 13 Surveys 12 11 10 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Z -Score TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal probability plot 88. True or False: Referring to Table 6-1, the data appear reasonably normal but not perfectly normal. ANSWER: True TYPE: TF DIFFICULTY: Difficult KEYWORDS: normal probability plot TABLE 6-2 The city manager of a large city believes that the number of reported accidents on any weekend has a normal distribution. She takes a sample of nine weekends and determines the number of reported accidents during each. The ordered array for this data is: 15, 46, 53, 54, 55, 76, 82, 256, 407. 89. Referring to Table 6-2, the first standard normal quantile is ________. ANSWER: -1.28 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 90. Referring to Table 6-2, the fifth standard normal quantile is ________. ANSWER: 0 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot The Normal Distribution and Other Continuous Distributions 171 91. Referring to Table 6-2, the sixth standard normal quantile is ________. ANSWER: +0.25 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 92. Referring to Table 6-2, the second standard normal quantile is ________. ANSWER: -0.84 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 93. Referring to Table 6-2, the seventh standard normal quantile is ________. ANSWER: +0.52 TYPE: FI DIFFICULTY: Moderate KEYWORDS: standard normal quantile, normal probability plot 94. Referring to Table 6-2, construct a normal probability plot. ANSWER: 450 Accidents 400 350 300 250 200 150 100 50 0 -1.5 -1 -0.5 0 0.5 1 Z -Scores TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal probability plot 95. True or False: Referring to Table 6-2, the data appear normal. ANSWER: False TYPE: TF DIFFICULTY: Difficult 1.5 172 The Normal Distribution and Other Continuous Distributions KEYWORDS: normal probability plot 96. Times spent watching TV every week by first graders follow an exponential distribution with mean 10 hours. The probability that a given first grader spends less than 20 hours watching TV is ______. ANSWER: 0.8647 TYPE: FI DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 97. Times spent watching TV every week by first graders follow an exponential distribution with mean 10 hours. The probability that a given first grader spends more than 5 hours watching TV is ______. ANSWER: 0.6065 TYPE: FI DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 98. Times spent watching TV every week by first graders follow an exponential distribution with mean 10 hours. The probability that a given first grader spends between 10 and 15 hours watching TV is ______. ANSWER: 0.1447 TYPE: FI DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 99. Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes. What is the average number of arrivals per minute? ANSWER: 0.06667 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, mean 100. Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes. What is the probability that a randomly chosen arrival to be more than 18 minutes? ANSWER: 0.3012 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 101. Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes. What is the probability that a randomly chosen arrival to be less than 15 minutes? ANSWER: 0.6321 TYPE: PR DIFFICULTY: Easy The Normal Distribution and Other Continuous Distributions 173 KEYWORDS: exponential distribution, probability 102. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour. What is the probability that a randomly chosen arrival to be less than 15 minutes? ANSWER: 0.9765 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 103. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour. What is the probability that a randomly chosen arrival to be more than 5 minutes? ANSWER: 0.2865 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 104. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 15 patients per hour. What is the probability that a randomly chosen arrival to be between 5 minutes and 15 minutes? ANSWER: 0.2630 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 105. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival to be more than 1 hour? ANSWER: 0.3679 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 106. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival to be more than 2.5 hours? ANSWER: 0.0821 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 107. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1 patient per hour. What is the probability that a randomly chosen arrival to be less than 20 minutes? ANSWER: 0.2835 TYPE: PR DIFFICULTY: Easy 174 The Normal Distribution and Other Continuous Distributions KEYWORDS: exponential distribution, probability 108. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1.5 patients per hour. What is the probability that a randomly chosen arrival to be less than 10 minutes? ANSWER: 0.2212 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 109. Patients arriving at an outpatient clinic follow an exponential distribution at a rate of 1.5 patients per hour. What is the probability that a randomly chosen arrival to be between 10 and 15 minutes? ANSWER: 0.0915 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability TABLE 6-3 The number of column inches of classified advertisements appearing on Mondays in a certain daily newspaper is normally distributed with population mean 320 and population standard deviation 20 inches. 110. Referring to Table 6-3, for a randomly chosen Monday, what is the probability there will be less than 340 column inches of classified advertisement? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 111. Referring to Table 6-3, for a randomly chosen Monday, what is the probability there will be between 280 and 360 column inches of classified advertisement? ANSWER: 0.9545 using Excel or 0.9544 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, probability 112. Referring to Table 6-3, for a randomly chosen Monday the probability is 0.1 that there will be less than how many column inches of classified advertisements? ANSWER: 294.4 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value The Normal Distribution and Other Continuous Distributions 175 113. Referring to Table 6-3, a single Monday is chosen at random. State in which of the following ranges the number of column inches of classified advertisement is most likely to be: a) 300 --320 b) 310 --330 c) 320 -- 340 d) 330 -- 350 ANSWER: b TYPE: MC Difficulty: Moderate KEYWORDS: normal distribution, probability TABLE 6-4 John has two jobs. For daytime work at a jewelry store he is paid $200 per month, plus a commission. His monthly commission is normally distributed with mean $600 and standard deviation $40. At night he works as a waiter, for which his monthly income is normally distributed with mean $100 and standard deviation $30. John's income levels from these two sources are independent of each other. 114. Referring to Table 6-4, for a given month, what is the probability that John's commission from the jewelry store is less than $640? ANSWER: 0.8413 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 115. Referring to Table 6-4, for a given month, what is the probability that John's income as a waiter is between $70 and $160? ANSWER: 0.8186 using Excel or 0.8185 using Table E.2 TYPE: PR DIFFICULTY: Easy KEYWORDS: normal distribution, probability 116. Referring to Table 6-4, the probability is 0.9 that John's income as a waiter is less than how much in a given month? ANSWER: $138.45 using Excel or $138.40 using Table E.2 TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal distribution, value 117. Referring to Table 6-4, find the mean and standard deviation of John's total income from these two jobs for a given month. ANSWER: $900; $50 TYPE: PR DIFFICULTY: Difficult EXPLANATION: Total mean = $200 + $600 + $100 = $900; Total variance = 402 + 302 = 2,500 176 The Normal Distribution and Other Continuous Distributions KEYWORDS: normal distribution, mean, standard deviation 118. Referring to Table 6-4, for a given month, what is the probability that John's total income from these two jobs is less than $825? ANSWER: 0.0668 TYPE: PR DIFFICULTY: Difficult EXPLANATION: Total mean = $900, Total standard deviation = $50 KEYWORDS: normal distribution, probability 119. Referring to Table 6-4, the probability is 0.2 that John's total income from these two jobs in a given month is less than how much? ANSWER: $857.92 using Excel or $858 using Table E.2 TYPE: PR DIFFICULTY: Difficult KEYWORDS: normal distribution, value TABLE 6-5 Suppose the time interval between two consecutive defective light bulbs from a production line has a uniform distribution over an interval from 0 to 90 minutes. 120. Referring to Table 6-5, what is the mean of the time interval? ANSWER: 45 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, mean 121. Referring to Table 6-5, what is the variance of the time interval? ANSWER: 675 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, variance 122. Referring to Table 6-5, what is the standard deviation of the time interval? ANSWER: 25.9808 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, standard deviation 123. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be exactly 10 minutes? ANSWER: 0.0 TYPE: PR DIFFICULTY: Moderate The Normal Distribution and Other Continuous Distributions 177 KEYWORDS: uniform distribution, probability 124. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be less than 10 minutes? ANSWER: 0.1111 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 125. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be between 10 and 20 minutes? ANSWER: 0.1111 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 126. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be between 10 and 35 minutes? ANSWER: 0.2778 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 127. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be at least 50 minutes? ANSWER: 0.4444 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 128. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be at least 80 minutes? ANSWER: 0.1111 TYPE: PR DIFFICULTY: Easy KEYWORDS: uniform distribution, probability 129. Referring to Table 6-5, what is the probability that the time interval between two consecutive defective light bulbs will be at least 90 minutes? ANSWER: 0.0 TYPE: PR DIFFICULTY: Moderate KEYWORDS: uniform distribution, probability 178 The Normal Distribution and Other Continuous Distributions 130. Referring to Table 6-5, the probability is 50% that the time interval between two consecutive defective light bulbs will fall between which two values that are the same distance from the mean? ANSWER: 22.5 and 67.5 TYPE: PR DIFFICULTY: Difficult KEYWORDS: uniform distribution, value 131. Referring to Table 6-5, the probability is 75% that the time interval between two consecutive defective light bulbs will fall between which two values that are the same distance from the mean? ANSWER: 11.25 and 78.75 TYPE: PR DIFFICULTY: Difficult KEYWORDS: uniform distribution, value 132. Referring to Table 6-5, the probability is 90% that the time interval between two consecutive defective light bulbs will fall between which two values that are the same distance from the mean? ANSWER: 4.5 and 85.5 TYPE: PR DIFFICULTY: Difficult KEYWORDS: uniform distribution, value TABLE 6-6 The interval between consecutive hits at a web site is assumed to follow an exponential distribution with an average of 40 hits per minute. 133. Referring to Table 6-6, what is the average time between consecutive hits? ANSWER: 0.025 minutes TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, mean 134. Referring to Table 6-6, what is the probability that the next hit at the web site will occur within 10 seconds after just being hit by a visitor? ANSWER: 0.9987 TYPE: PR DIFFICULTY: Easy KEYWORDS: exponential distribution, probability 135. Referring to Table 6-6, what is the probability that the next hit at the web site will occur within no sooner than 5 seconds after just being hit by a visitor? ANSWER: The Normal Distribution and Other Continuous Distributions 179 0.0357 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 136. Referring to Table 6-6, what is the probability that the next hit at the web site will occur between the next 1.2 and 1.5 seconds after just being hit by a visitor? ANSWER: 0.08145 TYPE: PR DIFFICULTY: Moderate KEYWORDS: exponential distribution, probability 137. True or False: One of the reasons that a correction for continuity adjustment is needed when approximating the binomial distribution with a normal distribution is because the normal distribution is used for a discrete random variable while the binomial distribution is used for a continuous random variable. ANSWER: False TYPE: TF DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 138. True or False: One of the reasons that a correction for continuity adjustment is needed when approximating the binomial distribution with a normal distribution is because the probability of getting a specific value of a random variable is zero with the normal distribution. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 139. True or False: One of the reasons that a correction for continuity adjustment is needed when approximating the binomial distribution with a normal distribution is because a random variable having a binomial distribution can have only a specified value while a random variable having a normal distribution can take on any values within an interval around that specified value. ANSWER: True TYPE: TF DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 140. True or False: To determine the probability of getting fewer than 3 successes in a binomial distribution, you will find the area under the normal curve for X = 3.5 and below. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 141. True or False: To determine the probability of getting more than 3 successes in a binomial distribution, you will find the area under the normal curve for X = 3.5 and above. 180 The Normal Distribution and Other Continuous Distributions ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 142. True or False: To determine the probability of getting at least 3 successes in a binomial distribution, you will find the area under the normal curve for X = 2.5 and above. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 143. True or False: To determine the probability of getting no more than 3 successes in a binomial distribution, you will find the area under the normal curve for X = 2.5 and below. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 144. True or False: To determine the probability of getting between 3 and 4 successes in a binomial distribution, you will find the area under the normal curve between X = 3.5 and 4.5. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 145. True or False: To determine the probability of getting between 2 and 4 successes in a binomial distribution, you will find the area under the normal curve between X = 1.5 and 4.5. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 146. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever the sample size is at least 30. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment The Normal Distribution and Other Continuous Distributions 181 147. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever the sample size is at least 15. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 148. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever np is at least 5. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 149. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever n(p-1) is at least 5. ANSWER: False TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 150. True or False: As a general rule, one can use the normal distribution to approximate a binomial distribution whenever n and n(p-1) are at least 5. ANSWER: True TYPE: TF DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment TABLE 6-7 A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.15. 151. Referring to Table 6-7, which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than 2 computers will require repair on a given day? a) The probability that any one of the computers will require repair on a given day is constant. b) The probability that a computer that will require repair in the morning is the same as that in the afternoon. c) The number of computers that will require repair in the morning is independent of the number of computers that will require repair in the afternoon. d) The probability that two or more computers that will require repair in a given day approaches zero. ANSWER: a 182 The Normal Distribution and Other Continuous Distributions TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 152. Referring to Table 6-7, which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than 2 computers will require repair on a given day? a) The probability that a computer that will require repair in the morning is the same as that in the afternoon. b) A randomly selected computer on a given day will either require a repair or will not. c) The number of computers that will require repair in the morning is independent of the number of computers that will require repair in the afternoon. d) The probability that two or more computers that will require repair in a given day approaches zero. ANSWER: b TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 153. Referring to Table 6-7, which of the following is one of the properties required so that the binomial distribution can be used to compute the probability that no more than 2 computers will require repair on a given day? a) The probability that a computer that will require repair in the morning is the same as that in the afternoon. b) The number of computers that will require repair in the morning is independent of the number of computers that will require repair in the afternoon. c) The probability that any one of the computers that will require repair on a given day will not affect or change the probability that any other computers that will require repair on the same day. d) The probability that two or more computers that will require repair in a given day approaches zero. ANSWER: c TYPE: MC DIFFICULTY: Easy KEYWORDS: binomial distribution, properties 154. Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be no more than 8 computers that require repair on a given day using a normal approximation. ANSWER: 0.0051 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 155. Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be less than 8 computers that require repair on a given day using a normal approximation. ANSWER: The Normal Distribution and Other Continuous Distributions 183 0.0024 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 156. Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be exactly 10 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0091 using Excel or 0.0090 using Table E.2 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 157. Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be at least 25 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0749 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 158. Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be more than 25 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0454 TYPE: PR DIFFICULTY: Easy KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 159. Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be between 25 and 30 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0733 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 160. Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be more than 25 but less than 30 computers that requires repair on a given day using a normal approximation. ANSWER: 0.0419 TYPE: PR DIFFICULTY: Moderate KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment 184 The Normal Distribution and Other Continuous Distributions 161. Referring to Table 6-7 and assuming that the number of computers that requires repair on a given day follows a binomial distribution, compute the probability that there will be less than 25 or more than 30 computers that requires repair on a given day using a normal approximation. ANSWER: 0.9267 TYPE: PR DIFFICULTY: Difficult KEYWORDS: approximation, normal distribution, binomial distribution, continuity adjustment