The Normal Distribution and Other Continuous Distributions 6-1
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.
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 set 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.
6-2 The Normal Distribution and Other Continuous Distributions
4. The value of the cumulative standardized normal distribution at Z is 0.8770. The value of Z is
a) 0.18
b) 0.81
c) 1.16
d) 1.47
5. For some value of Z, the value of the cumulative standardized normal distribution is 0.2090. The
value of Z is
a) – 0.81
b) – 0.31
c) 0.31
d) 1.96
6. For some value of Z, the value of the cumulative standardized normal distribution is 0.8340. The
value of Z is
a) 0.07
b) 0.37
c) 0.97
d) 1.06
7. The value of the cumulative standardized normal distribution at Z is 0.6255. The value of Z is
a) 0.99
b) 0.40
c) 0.32
d) 0.16
The Normal Distribution and Other Continuous Distributions 6-3
8. The value of the cumulative standardized normal distribution at 1.5X is 0.9332. 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 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?
6-4 The Normal Distribution and Other Continuous Distributions
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 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
waiting time was found to be a variable best approximated by an exponential distribution with a
mean length of waiting time equal to 3 minutes (i.e. the mean number of calls answered in a
minute is 1/3). 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
14. A 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
waiting time was found to be a variable best approximated by an exponential distribution with a
mean length of waiting time equal to 3 minutes (i.e. the mean number of calls answered in a
minute is 1/3). 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
The Normal Distribution and Other Continuous Distributions 6-5
15. A 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
waiting time was found to be a variable best approximated by an exponential distribution with a
mean length of waiting time equal to 3 minutes (i.e. the mean number of calls answered in a
minute is 1/3). 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 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
waiting time was found to be a variable best approximated by an exponential distribution with a
mean length of waiting time equal to 2.8 minutes (i.e. the mean number of calls answered in a
minute is 1/2.8). What proportion of callers is put on hold longer than 2.8 minutes?
a) 0.3679
b) 0.50
c) 0.60810
d) 0.6321
17. A 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
waiting time was found to be a variable best approximated by an exponential distribution with a
mean length of waiting time equal to 2.8 minutes (i.e. the mean number of calls answered in a
minute is 1/2.8). 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
6-6 The Normal Distribution and Other Continuous Distributions
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
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, 75.8% of the college students will take more than how many minutes 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
The Normal Distribution and Other Continuous Distributions 6-7
21. Let X represent the amount of time until the next student will arrive in the library parking lot at
the university. If we know that the distribution of arrival time can be modeled using an
exponential distribution with a mean of 4 minutes (i.e. the mean number of arrivals is 1/4 per
minute), find the probability that it will take more than 10 minutes for the next student to arrive
at the library parking lot.
a) 0.917915
b) 0.670320
c) 0.329680
d) 0.082085
22. Let X represent the amount of time till the next student will arrive in the library parking lot at the
university. If we know that the distribution of arrival time can be modeled using an exponential
distribution with a mean of 4 minutes (i.e. the mean number of arrivals is 1/4 per minute), find
the probability that it will take between 2 and 12 minutes for the next student to arrive at the
library parking lot.
a) 0.049787
b) 0.556744
c) 0.606531
d) 0.656318
23. The owner of a fish market determined that the mean 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 _______?
24. The owner of a fish market determined that the mean 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 _______?
6-8 The Normal Distribution and Other Continuous Distributions
25. The owner of a fish market determined that the mean 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 mean 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 mean 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. 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.
The Normal Distribution and Other Continuous Distributions 6-9
29. 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.
30. True or False: The probability that a standard normal variable, Z, falls between – 1.50 and 0.81
is 0.7242.
31. True or False: The probability that a standard normal variable, Z, is between 1.50 and 2.10 is the
same as the probability Z is between – 2.10 and – 1.50.
32. True or False: The probability that a standard normal variable, Z, is below 1.96 is 0.4750.
33. True or False: The probability that a standard normal variable, Z, is between 1.00 and 3.00 is
0.1574.
34. True or False: The probability that a standard normal variable, Z, falls between –2.00 and –0.44
is 0.6472.
6-10 The Normal Distribution and Other Continuous Distributions
35. True or False: The probability that a standard normal variable, Z, is less than 5.0 is
approximately 0.
36. True or False: A worker earns $15 per hour at a plant in China 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 mean wage for the plant is $7.50 per hour.
37. True or False: Theoretically, the mean, median, and the mode are all equal for a normal
distribution.
38. True or False: Any set of normally distributed data can be transformed to its standardized form.
39. True or False: The “middle spread,” that is the middle 50% of the normal distribution, is equal to
one standard deviation.
40. True or False: A normal probability plot may be used to assess the assumption of normality for a
particular set of data.
The Normal Distribution and Other Continuous Distributions 6-11
41. True or False: If a data set is approximately normally distributed, its normal probability plot
would be S-shaped.
42. The probability that a standard normal variable Z is positive is ________.
43. The amount of tea leaves in a can from a particular production line is normally distributed with
= 110 grams and
= 25 grams. What is the probability that a randomly selected can will
contain between 100 and 110 grams of tea leaves?
44. The amount of tea leaves in a can from a particular production line is normally distributed with
= 110 grams and
= 25 grams. What is the probability that a randomly selected can will
contain between 82 and 100 grams of tea leaves?
45. The amount of tea leaves in a can from a particular production line is normally distributed with
= 110 grams and
= 25 grams. What is the probability that a randomly selected can will
contain at least 100 grams of tea leaves?
46. The amount of tea leaves in a can from a particular production line is normally distributed with
= 110 grams and
= 25 grams. What is the probability that a randomly selected can will
contain between 100 and 120 grams of tea leaves?
6-12 The Normal Distribution and Other Continuous Distributions
47. The amount of tea leaves in a can from a particular production line is normally distributed with
= 110 grams and
= 25 grams. What is the probability that a randomly selected can will
contain less than 100 grams of tea leaves?
48. The amount of tea leaves in a can from a particular production line is normally distributed with
= 110 grams and
= 25 grams. What is the probability that a randomly selected can will
contain less than 100 grams or more than 120 grams of tea leaves?
49. The amount of tea leaves in a can from a particular production line is normally distributed with
= 110 grams and
= 25 grams. Approximately 83% of the can will have at least how many
grams of tea leaves?
50. 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?
51. 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?
The Normal Distribution and Other Continuous Distributions 6-13
52. 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?
53. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. What is the probability that the time lapsed between two
consecutive trades will be longer than 17 seconds?
54. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. What is the probability that the time lapsed between two
consecutive trades will be between 13 and 14 seconds?
55. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. What is the probability that the time lapsed between two
consecutive trades will be between 15 and 16 seconds?
6-14 The Normal Distribution and Other Continuous Distributions
56. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. What is the probability that the time lapsed between two
consecutive trades will be between 14 and 15 seconds?
57. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. What is the probability that the time lapsed between two
consecutive trades will be between 13 and 16 seconds?
58. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. What is the probability that the time lapsed between two
consecutive trades will be between 14 and 17 seconds?
59. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. The probability is 20% that the time lapsed will be shorter how
many seconds?
The Normal Distribution and Other Continuous Distributions 6-15
60. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. The probability is 80% that the time lapsed will be longer than
how many seconds?
61. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. The middle 60% of the time lapsed will fall between which two
numbers?
62. You were told that the amount of time lapsed between consecutive trades on a foreign stock
exchange market followed a normal distribution with a mean of 15 seconds. You were also told
that the probability that the time lapsed between two consecutive trades to fall between 16 to 17
seconds was 13%. The probability that the time lapsed between two consecutive trades would
fall below 13 seconds was 7%. The middle 86% of the time lapsed will fall between which two
numbers?
63. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. What is the probability of a score between 90
and 95?
6-16 The Normal Distribution and Other Continuous Distributions
64. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. What is the probability of a score greater than
95?
65. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. What is the probability of a score lower than 55?
66. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. What is the probability of a score between 75
and 90?
67. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. What is the probability of a score between 60
and 75?
68. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. What is the probability of a score between 60
and 95?
The Normal Distribution and Other Continuous Distributions 6-17
69. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. What is the probability of a score between 55
and 90?
70. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. What is the probability of a score between 55
and 95?
71. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. The middle 86.64% of the students will score
between which two scores?
72. You were told that the mean score on a statistics exam is 75 with the scores normally distributed.
In addition, you know the probability of a score between 55 and 60 is 4.41% and that the
probability of a score greater than 90 is 6.68%. The middle 95.46% of the students will score
between which two scores?
73. 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 __________.
6-18 The Normal Distribution and Other Continuous Distributions
74. 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 __________.
75. 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 __________.
76. 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 __________.
77. 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 __________.
78. 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 __________.
79. 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 __________.
The Normal Distribution and Other Continuous Distributions 6-19
80. 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.
81. 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.
82. 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 __________.
83. 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 __________.
84. 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).
85. 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).
6-20 The Normal Distribution and Other Continuous Distributions
86. The amount of time necessary for assembly line workers to complete a product is a normal
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.
87. The amount of time necessary for assembly line workers to complete a product is a normal
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.
88. The amount of time necessary for assembly line workers to complete a product is a normal
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.
89. The amount of time necessary for assembly line workers to complete a product is a normal
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.
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.
The Normal Distribution and Other Continuous Distributions 6-21
91. The amount of time necessary for assembly line workers to complete a product is a normal
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.
92. The amount of time necessary for assembly line workers to complete a product is a normal
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.
93. The amount of time necessary for assembly line workers to complete a product is a normal
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.
94. The amount of time necessary for assembly line workers to complete a product is a normal
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.
95. The amount of time necessary for assembly line workers to complete a product is a normal
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.
6-22 The Normal Distribution and Other Continuous Distributions
96. The amount of time necessary for assembly line workers to complete a product is a normal
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).
97. The amount of time necessary for assembly line workers to complete a product is a normal
variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 17% of the
products would be assembled within __________ minutes.
98. The amount of time necessary for assembly line workers to complete a product is a normal
variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 70% of the
products would be assembled within __________ minutes.
99. The amount of time between successive TV watching by first graders follows an exponential
distribution with a mean of 10 hours. The probability that a given first grader spends less than 20
hours between successive TV watching is ______.
100. The amount of time between successive TV watching by first graders follows an exponential
distribution with a mean of 10 hours. The probability that a given first grader spends more than 5
hours between successive TV watching is ______.
The Normal Distribution and Other Continuous Distributions 6-23
101. The amount of time between successive TV watching by first graders follows an exponential
distribution with a mean of 10 hours. The probability that a given first grader spends between 10
and 15 hours between successive TV watching is ______.
102. The interval between patients arriving at an outpatient clinic follows an exponential
distribution with mean 15 minutes. What is the mean number of arrivals per minute?
103. The interval between patients arriving at an outpatient clinic follows an exponential
distribution with mean 15 minutes. What is the probability that a randomly chosen arrival to be
more than 18 minutes?
104. The interval between patients arriving at an outpatient clinic follows an exponential
distribution with mean 15 minutes. What is the probability that a randomly chosen arrival to be
less than 15 minutes?
105. The interval between patients arriving at an outpatient clinic follows 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?
6-24 The Normal Distribution and Other Continuous Distributions
106. The interval between patients arriving at an outpatient clinic follows 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?
107. The interval between patients arriving at an outpatient clinic follows 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?
108. The interval between patients arriving at an outpatient clinic follows 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?
109. The interval between patients arriving at an outpatient clinic follows 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?
110. The interval between patients arriving at an outpatient clinic follows 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?
The Normal Distribution and Other Continuous Distributions 6-25
111. The interval between patients arriving at an outpatient clinic follows 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?
112. The interval between patients arriving at an outpatient clinic follows 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?
SCENARIO 6-1
The number of column inches of classified advertisements appearing on Mondays in a certain daily
newspaper is normally distributed with population mean of 320 and population standard deviation of
20 inches.
113. Referring to Scenario 6-1, for a randomly chosen Monday, what is the probability there will be
less than 340 column inches of classified advertisement?
114. Referring to Scenario 6-1, for a randomly chosen Monday, what is the probability there will be
between 280 and 360 column inches of classified advertisement?
115. Referring to Scenario 6-1, for a randomly chosen Monday the probability is 0.1 that there will
be less than how many column inches of classified advertisements?
6-26 The Normal Distribution and Other Continuous Distributions
116. Referring to Scenario 6-1, 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
John has two jobs. For daytime work at a jewelry store he is paid $15,000 per month, plus a
commission. His monthly commission is normally distributed with mean $10,000 and standard
deviation $2000. At night he works occasionally as a waiter, for which his monthly income is
normally distributed with mean $1,000 and standard deviation $300. John’s income levels from these
two sources are independent of each other.
117. Referring to Scenario 6-2, for a given month, what is the probability that John’s commission
from the jewelry store is less than $13,000?
118. Referring to Scenario 6-2, for a given month, what is the probability that John’s commission
from the jewelry store is no more than $8,000?
119. Referring to Scenario 6-2, for a given month, what is the probability that John’s commission
from the jewelry store is at least than $12,000?
120. Referring to Scenario 6-2, for a given month, what is the probability that John’s commission
from the jewelry store is more than $9,500?
The Normal Distribution and Other Continuous Distributions 6-27
121. Referring to Scenario 6-2, for a given month, what is the probability that John’s commission
from the jewelry store is between $11,000 and $12,000?
122. Referring to Scenario 6-2, for a given month, what is the probability that John’s commission
from the jewelry store is between $5,000 and $7,000?
123. Referring to Scenario 6-2, for a given month, what is the probability that John’s commission
from the jewelry store is between $9,000 and $11,000?
124. Referring to Scenario 6-2, the probability is 0.75 that John’s commission from the jewelry store
is less than how much in a given month?
125. Referring to Scenario 6-2, the probability is 0.95 that John’s commission from the jewelry store
is at least how much in a given month?
126. Referring to Scenario 6-2, John’s commission from the jewelry store will be between what two
values symmetrically distributed around the population mean 80% of the time?
6-28 The Normal Distribution and Other Continuous Distributions
127. Referring to Scenario 6-2, John’s commission from the jewelry store will be between what two
values symmetrically distributed around the population mean 90% of the time?
128. Referring to Scenario 6-2, the probability is 0.10 that John’s commission from the jewelry store
is more than how much in a given month?
129. Referring to Scenario 6-2, the probability is 0.30 that John’s commission from the jewelry store
is no more than how much in a given month?
130. Referring to Scenario 6-2, for a given month, what is the probability that John’s income as a
waiter is between $700 and $1600?
131. Referring to Scenario 6-2, for a given month, what is the probability that John’s income as a
waiter is between $1,200 and $1,600?
132. Referring to Scenario 6-2, for a given month, what is the probability that John’s income as a
waiter is between $800 and $900?
The Normal Distribution and Other Continuous Distributions 6-29
133. Referring to Scenario 6-2, for a given month, what is the probability that John’s income as a
waiter is no more than $300?
134. Referring to Scenario 6-2, for a given month, what is the probability that John’s income as a
waiter is less than $1300?
135. Referring to Scenario 6-2, for a given month, what is the probability that John’s income as a
waiter is at least $1400?
136. Referring to Scenario 6-2, for a given month, what is the probability that John’s income as a
waiter is more than $900?
137. Referring to Scenario 6-2, the probability is 0.25 that John’s income as a waiter is no more than
how much in a given month?
138. Referring to Scenario 6-2, the probability is 0.45 that John’s income as a waiter is more than
how much in a given month?
6-30 The Normal Distribution and Other Continuous Distributions
139. Referring to Scenario 6-2, the probability is 0.35 that John’s income as a waiter is no less than
how much in a given month?
140. Referring to Scenario 6-2, John’s income as a waiter will be between what two values
symmetrically distributed around the population mean 80% of the time?
141. Referring to Scenario 6-2, John’s income as a waiter will be between what two values
symmetrically distributed around the population mean 90% of the time?
142. Referring to Scenario 6-2, the probability is 0.9 that John’s income as a waiter is less than how
much in a given month?
SCENARIO 6-3
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.
143. Referring to Scenario 6-3, what is the mean of the time interval?
144. Referring to Scenario 6-3, what is the variance of the time interval?
The Normal Distribution and Other Continuous Distributions 6-31
145. Referring to Scenario 6-3, what is the standard deviation of the time interval?
146. Referring to Scenario 6-3, what is the probability that the time interval between two
consecutive defective light bulbs will be exactly 10 minutes?
147. Referring to Scenario 6-3, what is the probability that the time interval between two
consecutive defective light bulbs will be less than 10 minutes?
148. Referring to Scenario 6-3, what is the probability that the time interval between two
consecutive defective light bulbs will be between 10 and 20 minutes?
149. Referring to Scenario 6-3, what is the probability that the time interval between two
consecutive defective light bulbs will be between 10 and 35 minutes?
150. Referring to Scenario 6-3, what is the probability that the time interval between two
consecutive defective light bulbs will be at least 50 minutes?
6-32 The Normal Distribution and Other Continuous Distributions
151. Referring to Scenario 6-3, what is the probability that the time interval between two
consecutive defective light bulbs will be at least 80 minutes?
152. Referring to Scenario 6-3, what is the probability that the time interval between two
consecutive defective light bulbs will be at least 90 minutes?
153. Referring to Scenario 6-3, 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?
154. Referring to Scenario 6-3, 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?
155. Referring to Scenario 6-3, 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?
The Normal Distribution and Other Continuous Distributions 6-33
SCENARIO 6-4
The interval between consecutive hits at a web site is assumed to follow an exponential distribution
with a mean of 40 hits per minute.
156. Referring to Scenario 6-4, what is the mean time between consecutive hits?
157. Referring to Scenario 6-4, what is the probability that the next hit at the web site will occur
within 10 seconds after just being hit by a visitor?
158. Referring to Scenario 6-4, 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?
159. Referring to Scenario 6-4, 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?
160. The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds. He
also knew that the probability of a randomly selected catfish that would weigh more than 3.8
pounds is 20% and the probability that a randomly selected catfish that would weigh less than 2.8
pounds is 30%. The probability that a randomly selected catfish will weigh between 2.6 and 3.6
pounds is ______.
6-34 The Normal Distribution and Other Continuous Distributions
161. The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds. He
also knew that the probability of a randomly selected catfish that would weigh more than 3.8
pounds is 20% and the probability that a randomly selected catfish that would weigh less than 2.8
pounds is 30%. The middle 40% of the catfish will weigh between ______ pounds and ______
pounds.
SCENARIO 6-5
A company producing orange juice buys all its oranges from a large orange orchard. The amount of
juice that can be squeezed from each of these oranges is approximately normally distributed with a
mean of 4.7 ounces and some unknown standard deviation. The company’s production manager
knows that the probability is 30.85% that a randomly selected orange will contain less than 4.5
ounces of juice. Also the probability is 10.56% that a randomly selected orange will contain more
than 5.2 ounces of juice. Answer the following questions without the help of a calculator, statistical
software or statistical table.
162. Referring to Scenario 6-5, what is the probability that a randomly selected orange will contain
between 4.5 and 5.2 ounces of juices?
163. Referring to Scenario 6-5, what is the probability that a randomly selected orange will contain
between 4.2 and 4.9 ounces of juices?
164. Referring to Scenario 6-5, what is the probability that a randomly selected orange will contain
at least 4.9 ounces of juices?
The Normal Distribution and Other Continuous Distributions 6-35
165. Referring to Scenario 6-5, what is the probability that a randomly selected orange will contain
no more than 4.9 ounces of juices?
166. Referring to Scenario 6-5, what is the probability that a randomly selected orange will contain
no more than 4.2 ounces of juices?
167. Referring to Scenario 6-5, what is the probability that a randomly selected orange will contain
more than 4.2 ounces of juices?
SCENARIO 6-6
According to Investment Digest, the arithmetic mean of the annual return for common stocks over an
85-year period was 9.5% but the value of the variance was not mentioned. Also 25% of the annual
returns were below 8% while 65% of the annual returns were between 8% and 11.5%. The article
claimed that the distribution of annual return for common stocks was bell-shaped and approximately
symmetric. Assume that this distribution is normal with the mean given above. Answer the following
questions without the help of a calculator, statistical software or statistical table.
168. Referring to Scenario 6-6, find the probability that the annual return of a random year will be
less than 11.5%.
169. Referring to Scenario 6-6, find the probability that the annual return of a random year will be
more than 11.5%.
6-36 The Normal Distribution and Other Continuous Distributions
170. Referring to Scenario 6-6, find the probability that the annual return of a random year will be
between 7.5% and 11%.
171. Referring to Scenario 6-6, find the probability that the annual return of a random year will be
less than 7.5%.
172. Referring to Scenario 6-6, find the probability that the annual return of a random year will be
more than 7.5%.
173. Referring to Scenario 6-6, what is the value above which will account for the highest 25% of
the possible annual returns?
174. Referring to Scenario 6-6, 75% of the annual returns will be lower than what value?
175. Referring to Scenario 6-6, find the two values that will bound the middle 50% of the annual
returns?
The Normal Distribution and Other Continuous Distributions 6-37
176. Referring to Scenario 6-6, find the two values that will bound the middle 80% of the annual
returns?
177. Referring to Scenario 6-6, 10% of the annual returns will be less than what amount?
178. Referring to Scenario 6-6, 10% of the annual returns will be at least what amount?