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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
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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
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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
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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
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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