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The Normal Distribution and Other Continuous Distributions 153
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.
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,  .
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.
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.
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.
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.
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The Normal Distribution and Other Continuous Distributions
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.
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.
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.
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?
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?
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.
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
The Normal Distribution and Other Continuous Distributions 155
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
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
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
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
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
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The Normal Distribution and Other Continuous Distributions
b) 0.3085
c) 0.2674
d) 0.1915
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
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
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
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
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 _______?
The Normal Distribution and Other Continuous Distributions 157
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 _______?
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
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?
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 _______?
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
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
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
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The Normal Distribution and Other Continuous Distributions
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
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.
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.
33. True or False: The probability that a standard normal random variable, Z, falls between – 1.50
and 0.81 is 0.7242.
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.
35. True or False: The probability that a standard normal random variable, Z, is below 1.96 is
0.4750.
36. True or False: The probability that a standard normal random variable, Z, is between 1.00 and
3.00 is 0.1574.
37. True or False: The probability that a standard normal random variable, Z, falls between –2.00
and –0.44 is 0.6472.
38. True or False: The probability that a standard normal random variable, Z, is less than 50 is
approximately 0.
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.
The Normal Distribution and Other Continuous Distributions 159
40. True or False: Theoretically, the mean, median, and the mode are all equal for a normal
distribution.
41. True or False: Any set of normally distributed data can be transformed to its standardized form.
42. True or False: The "middle spread," that is the middle 50% of the normal distribution, is equal to
one standard deviation.
43. True or False: A normal probability plot may be used to assess the assumption of normality for a
particular batch of data.
44. True or False: If a data batch is approximately normally distributed, its normal probability plot
would be S-shaped.
45. The probability that a standard normal variable Z is positive is ________.
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?
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?
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?
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?
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?
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The Normal Distribution and Other Continuous Distributions
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?
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?
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?
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?
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?
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 __________.
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 __________.
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 __________.
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 __________.
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 __________.
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 __________.
The Normal Distribution and Other Continuous Distributions 161
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 __________.
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.
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.
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 __________.
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 __________.
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).
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).
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.
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.
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.
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.
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The Normal Distribution and Other Continuous Distributions
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.
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.
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.
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.
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.
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.
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).
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.
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.
The Normal Distribution and Other Continuous Distributions 163
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 ________.
83. Referring to Table 6-1, the fourth standard normal quantile is ________.
84. Referring to Table 6-1, the ninth standard normal quantile is ________.
85. Referring to Table 6-1, the fourteenth standard normal quantile is ________.
86. Referring to Table 6-1, the last standard normal quantile is ________.
87. Referring to Table 6-1, construct a normal probability plot for the data.
88. True or False: Referring to Table 6-1, the data appear reasonably normal but not perfectly normal.
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 ________.
90. Referring to Table 6-2, the fifth standard normal quantile is ________.
91. Referring to Table 6-2, the sixth standard normal quantile is ________.
92. Referring to Table 6-2, the second standard normal quantile is ________.
93. Referring to Table 6-2, the seventh standard normal quantile is ________.
164
The Normal Distribution and Other Continuous Distributions
94. Referring to Table 6-2, construct a normal probability plot.
95. True or False: Referring to Table 6-2, the data appear normal.
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
______.
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
______.
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 ______.
99. Patients arriving at an outpatient clinic follow an exponential distribution with mean 15 minutes.
What is the average number of arrivals per minute?
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?
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?
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?
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?
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?
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?
The Normal Distribution and Other Continuous Distributions 165
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?
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?
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?
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?
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?
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?
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?
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
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
166
The Normal Distribution and Other Continuous Distributions
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?
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?
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?
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.
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?
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?
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?
121. Referring to Table 6-5, what is the variance of the time interval?
122. Referring to Table 6-5, what is the standard deviation of the time interval?
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?
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?
The Normal Distribution and Other Continuous Distributions 167
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?
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?
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?
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?
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?
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?
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?
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?
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?
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?
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The Normal Distribution and Other Continuous Distributions
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?
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?
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.
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.
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.
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.
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.
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.
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.
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.
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.
The Normal Distribution and Other Continuous Distributions 169
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.
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.
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.
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.
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.
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.
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.
170
The Normal Distribution and Other Continuous Distributions
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.
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.
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.
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.
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.
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.
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.
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.
The Normal Distribution and Other Continuous Distributions 171
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.