Practice Exam
Chapter 6 - The Normal Distribution
SECTION I: MULTIPLE - CHOICE
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. 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.
3. 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
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4. 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
5. 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.
i. 0.0919
ii. 0.2255
iii. 0.4938
iv. 0.7745
6. 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. 5.2 pounds
d. 7.36 pounds
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Section II: TRUE Or FALSE
1. The probability that a standard normal random variable, Z, falls
between – 1.50 and 0.81 is 0.7242.
True
2. The probability that a standard normal random variable, Z, is
below 1.96 is 0.4750.
False
3. The probability that a standard normal random variable, Z, falls
between –2.00 and –0.44 is 0.6472.
False
4. 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.
False
5. Any set of normally distributed data can be transformed to its
standardized form.
True
6. The "middle spread," that is the middle 50% of the normal
distribution, is equal to one standard deviation.
False
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7. If a data batch is approximately normally distributed, its normal
probability plot would be S-shaped.
False
SECTION III: FREE RESPONSE PROBLEMS
1. 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.
a. What proportion of the plan recipients would receive payments
beyond age 75?
0.0228
b. Find the age at which payments have ceased for approximately
86% of the plan participants.
71.78 years old
2. 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.
0.8944
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3. 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 _______?
0.1056
4. The amount of pyridoxine (in grams) in a multiple vitamin is
normally distributed with µ = 110 grams and σ = 25 grams.
a. What is the probability that a randomly selected vitamin will
contain between 100 and 110 grams of pyridoxine? 0.1554
b. What is the probability that a randomly selected vitamin will
contain between 82 and 100 grams of pyridoxine? 0.2132
c. What is the probability that a randomly selected vitamin will
contain at least 100 grams of pyridoxine? 0.6554
d. What is the probability that a randomly selected vitamin will
contain between 100 and 120 grams of pyridoxine?0.3108
e. What is the probability that a randomly selected vitamin will
contain less than 100 grams of pyridoxine? 0.3446
f. What is the probability that a randomly selected vitamin will
contain less than 100 grams or more than 120 grams of pyridoxine?
0.6892
g. Approximately 83% of the vitamins will have at least how many
grams of pyridoxine? 86.25 using Table E.2
5. 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
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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%.
a. What is the probability that the time lapsed between two
consecutive trades will be longer than 17 seconds? 7% or 0.07
b. What is the probability that the time lapsed between two
consecutive trades will be longer than 17 seconds?7% or 0.07
c. What is the probability that the time lapsed between two
consecutive trades will be between 13 and 14 seconds?13% or 0.13
d. What is the probability that the time lapsed between two
consecutive trades will be between 15 and 16 seconds? 30% or 0.30
e. What is the probability that the time lapsed between two
consecutive trades will be between 14 and 15 seconds?30% or 0.30
f. What is the probability that the time lapsed between two
consecutive trades will be between 13 and 16 seconds?73% or 0.73
g. What is the probability that the time lapsed between two
consecutive trades will be between 14 and 17 seconds? 73% or 0.73
h. The probability is 20% that the time lapsed will be shorter how
many seconds?14 seconds
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i. The probability is 80% that the time lapsed will be longer than
how many seconds? 14 seconds
j. The middle 60% of the time lapsed will fall between which two
numbers?14 seconds and 16 seconds
k. The middle 86% of the time lapsed will fall between which two
numbers? 13 seconds and 17 seconds
6. Suppose Z has a standard normal distribution with a mean of 0
and standard deviation of 1.
a. The probability that Z is less than 1.15 is __________. 0.8749
b. The probability that Z is less than 1.15 is __________. 0.8749
c. The probability that Z is more than 0.77 is __________. 0.2206
d. The probability that Z is less than -2.20 is __________.0.0139
e. The probability that Z is more than -0.98 is __________.0.8365
f. The probability that Z is between -2.33 and 2.33 is _______.0.9802
g. The probability that Z is between -2.89 and -1.03 is ______.0.1496
h. The probability that Z is between -0.88 and 2.29 is ______.0.7996
i. The probability that Z values are larger than ______ is 0.3483. 0.39
j. The probability that Z values are larger than ___ is 0.6985. -0.52
k. So 27% of the possible Z values are smaller than _______. -0.61
l. So 85% of the possible Z values are smaller than _______.1.04
m. So 96% of the possible Z values are between ______ and
__________ (symmetrically distributed about the mean). -2.05 and
2.05 or -2.06 and 2.06 6
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n. So 50% of the possible Z values are between ___ and ___ (
symmetrically distributed about the mean)
.-0.67 and 0.67 or -0.68 and 0.68
7. The owner of a fish market determined that the average 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%.
a. The probability that a randomly selected catfish will weigh less
than 3.6 pounds is ______.70% or 0.7
b. The probability that a randomly selected catfish will weigh
between 2.6 and 3.6 pounds is ______.50% or 0.5
c. The middle 40% of the catfish will weigh between ______ pounds
and ______ pounds.2.8 and 3.6
TABLE (A)
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.
a. Referring to Table (A), the first standard normal quantile is
________.-1.5341
b. Referring to Table (A), the fourth standard normal quantile is
________.-0.6745
c. Referring to Table (A), the ninth standard normal quantile is
________.+0.1573
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d. Referring to Table (A), the fourteenth standard normal quantile is
________.+1.1503
e. Referring to Table (A), the last standard normal quantile is
________.+1.5341
Table (B)
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.
a. Referring to Table (B), for a randomly chosen Monday, what is
the probability there will be less than 340 column inches of classified
advertisement?0.8413
b. Referring to Table (B), for a randomly chosen Monday, what is
the probability there will be between 280 and 360 column inches of
classified advertisement? 0.9544 using Table E.2
c. Referring to Table (B), for a randomly chosen Monday the
probability is 0.1 that there will be less than how many column inches
of classified advertisements?294.4
d. Referring to Table (B), 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:
i. 300 --320
ii. 310 --330
iii. 320 -- 340
iv. 330 -- 350
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