CHAPTER 6: THE NORMAL DISTRIBUTION
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, μ, and standard deviation, σ.
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
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
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.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
14.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
15.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
ANSWER:
a
TYPE: MC DIFFICULTY: Moderate
KEYWORDS: normal distribution, value
16.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
17.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
18.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
19.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
20.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
21.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
22.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
23.True or False: The probability that a standard normal random variable, Z, falls between – 1.50
and 0.81 is 0.7242.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: standardized normal distribution, probability
24.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
25.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
26.True or False: The probability that a standard normal random variable, Z, is between 1.00 and
3.00 is 0.157
4.
ANSWER:
True
TYPE: TF DIFFICULTY: Easy
KEYWORDS: standardized normal distribution, probability
27.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
28.True or False: The probability that a standard normal random variable, Z, is less than 5.0 is
approximately 0.
ANSWER:
False
TYPE: TF DIFFICULTY: Easy
KEYWORDS: standardized normal distribution, probability
29.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
30.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
31.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
32.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
33.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
34.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
35.The probability that a standard normal variable Z is positive is ________.
ANSWER:
0.50
TYPE: FI DIFFICULTY: Easy
KEYWORDS: standardized normal distribution
36.The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with μ = 110
grams and σ = 25 grams. What is the probability that a randomly selected vitamin will contain between 100 and 110 grams of pyridoxine?
ANSWER:
0.1554
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
37.The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with μ = 110
grams and σ = 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
38.The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with μ = 110
grams and σ = 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
39.The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with μ = 110
grams and σ = 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
40.The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with μ = 110
grams and σ = 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
41.The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with μ = 110
grams and σ = 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
42.The amount of pyridoxine (in grams) in a multiple vitamin is normally distributed with μ = 110
grams and σ = 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
43.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
44.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
45.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
46.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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?
ANSWER:
7% or 0.07
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
47.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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?
ANSWER:
13% or 0.13
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
48.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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?
ANSWER:
30% or 0.30
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
49.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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?
ANSWER:
30% or 0.30
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
50.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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?
ANSWER:
73% or 0.73
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
51.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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?
ANSWER:
73% or 0.73
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
52.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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?
ANSWER:
14 seconds
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
53.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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?
ANSWER:
14 seconds
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
54.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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? ANSWER:
14 seconds and 16 seconds
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
55.You were told that the amount of time lapsed between consecutive trades on the New York Stock
Exchange 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? ANSWER:
13 seconds and 17 seconds
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
56.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?
ANSWER:
4.41% or 0.0441
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
57.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? ANSWER:
2.27% or 0.0227
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
58.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?
ANSWER:
2.27% or 0.0227
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
59.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?
ANSWER:
43.32% or 0.4332
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
60.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?
ANSWER:
43.32% or 0.4332
TYPE: PR DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
61.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?
ANSWER:
91.05% or 0.9105
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
62.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?
ANSWER:
91.05% or 0.9105
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
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 55 and 95?
ANSWER:
95.46% or 0.9546
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
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%. The middle 86.64% of the students will score between which two scores?
ANSWER:
60 and 90
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
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%. The middle 95.46% of the students will score between which two scores?
ANSWER:
55 and 95
TYPE: PR DIFFICULTY: Moderate
KEYWORDS: normal distribution, probability
66.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
67.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:
0.2206
TYPE: FI DIFFICULTY: Easy
KEYWORDS: standardized normal distribution, probability
68.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
69.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
70.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
71.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
72.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
73.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
74.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.
ANSWER:
-0.52
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: standardized normal distribution, probability
75.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
76.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
77.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
78.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
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. The
probability is __________ that a product is assembled in less than 12 minutes. ANSWER:
0.0668
TYPE: FI DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
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. 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
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. 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
82.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
83.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
84.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:
0.9772
TYPE: FI DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
85.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
TYPE: FI DIFFICULTY: Easy
KEYWORDS: normal distribution, probability
86.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
87.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
88.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
89.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
90.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
91.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.
92.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
93.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
94.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
95.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
96.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
97.Referring to Table 6-1, construct a normal probability plot for the data.
ANSWER:
11
12
13-2-1.5-1-0.500.51 1.52
Z -Score
TYPE: PR DIFFICULTY: Moderate KEYWORDS: normal probability plot 98. True or False: Referring to Table 6-1, the data appear reasonably normal but not perfectly normal.
ANSWER: True
TYPE: TF DIFFICULTY: Moderate 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.
99. 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
100. Referring to Table 6-2, the fifth standard normal quantile is ________.
ANSWER: 0
TYPE: FI DIFFICULTY: Moderate
KEYWORDS: standard normal quantile, normal probability plot
101. Referring to Table 6-2, the sixth standard normal quantile is ________.
ANSWER: +0.25
TYPE: FI DIFFICULTY: Moderate