Central Limit Theorem Answers
For each question, either give the correct answer accurate to 4 decimal places (ex:
23.52%) or if you feel that the question cannot be asked, explain why.
1. Credit card balances for young couples are roughly normally distributed and
have a mean of $750 and a standard deviation of $265.
a. What is the probability that a typical couple’s balance is more than $1,000?
z
x

z
1000  750
 .94
265
Px  1000  1  .8264  .1736
b. What is the probability that the average balance of an SRS of 5 couples is more
than $1,000?
z
x

z
n
1000  750
 2.11
265
5
Px  1000  1  .9826  .0174
c. What is the probability that the average balance of an SRS of 100 couples is
less than $700?
z
x

z
700  750
 1.89
265
Px  1000  .0294
100
n
2. The amount of snow that falls in Buffalo, NY over a winter is normally distributed
with a mean of 15 feet, 7 inches and a standard deviation of 6 feet, 2 inches.
a. What is the probability that Buffalo will have over 12 feet of snow next winter?
z
x

z
144  187
 .58
74
Px  144  1  .2810  .7190
b. What is the probability that Buffalo’s next 4 winters will average over 10 feet of
snow?
z
x

z
n
120  187
 1.81
74
4
Px  144  1  .0351  .9649
3. Tomato plants watered lightly for a month show an average growth of 27
centimeters with a standard deviation of 8.3 centimeters. Assume that tomato
plants grown according to a normal distribution.
a. What is the probability that a plant will grow over 30 centimeters?
z
x

z
30  27
 .36
8.3
Px  30  1  .6406  .3594
b. What is the probability that an SRS of 50 plants will have an average growth of
over 30 centimeters?
z
x

n
z
30  27
 2.56
8.3
50
Px  30  1  .9948  .0052
c. Compare answer b to answer a. Explain why answer b is (smaller/greater) than
answer a.
Answer b is smaller. This answer is smaller since a sample mean is usually
much less variable than a single measurement, and if select properly will
tend to be close in value to the population measurements. A single
measurement has a much higher chance of being unusual. (Think about the
height of teachers example)
d. What is the probability that a plant will grow between 25 and 30 centimeters?
z
x

z
30  27
 .36
8.3
z
25  27
 .  .24
8.3
Find the individual probabilities that x is less than 25 and less than 30.
Px  30  .6406
Px  25  .4052
The answer is the difference of the probabilities
.6406  .4052  .2354
e. What is the probability that 100 plants will have an average growth between 25
and 30 centimeters?
z
x

z
n
25  27
 2.41
8.3
100
z
30  27
 3.61
8.3
100
Find the individual probabilities that x is less than 25 and less than 30.
Px  30  Almost _1
Px  25  .0080
The answer is the difference of the probabilities
1  .0080  .9920
f. Compare answer e to answer d. Explain why answer e is (smaller/greater) than
answer d.
Answer d is smaller. This answer is smaller since a sample mean is usually
much less variable than a single measurement, and if select properly will
tend to be close in value to the population measurements. A single
measurement has a much higher chance of being unusual. (Think about the
height of teachers example)
4. The average amount of money spent at lunch in the Wissahickon cafeteria is
$3.00 with a standard deviation of 75 cents. Assume the distribution of money
spent is normal.
a. What is the probability that a student spends more than $3.50 for lunch?
z
x
350  300
 .67
75
z

Px  350  1  .7486  .2514
b. What is the probability that the average amount spent by an SRS of 10 students
will be greater than $4?
z
x
400  300
 4.22
75
10
z

n
Px  400  Almost _ 0
c. What is the probability that a student spends less than $2.75 for lunch?
z
x
z

275  300
 .33
75
Px  275  .3707
d. Find the probability that the average amount spent by an SRS of 25 students is
less than $2.75.
z
x
z

n
275  300
 1.67
75
25
Px  275  .0475
5. The typical 6 ounce bag of potato chips is normally distributed by weight with a
standard deviation of .15 ounces.
a. What is the probability that a bag contains less than 5.9 ounces?
z
x

z
5.9  6
 .67
.15
Px  5.9  .2514
b. What is the probability that the average weight of an SRS of 12 bags will be less
than 5.9 ounces?
z
x

n
z
5 .9  6
 2.31
.15
12
Px  5.9  .0104
c. What is the probability that a bag will have more than 6.2 ounces or less tan 5.8
ounces?
First, find the probability of more than 6.2 ounces and less than 5.8 ounces
z
x
z
5.8  6
 .1.33
.15

Px  5.8  .0918
Px  6.2  1  .9082  .0918
z
6.2  6
 1.33
.15
Since the problem asks us for “or” the answer is the sum of the probabilities
Px  6.2 _ or _ x  5.8  .0918  .0918  .1836
d. Find the probability that an SRS of 5 bags has more than 6.2 ounces or less
than 5.8 ounces.
First, find the probability of more than 6.2 ounces and less than 5.8 ounces
z
x

n
z
5.8  6
 2.98
.15
5
z
6 .2  6
 2.98
.15
5
Px  5.8  .0014
Px  6.2  1  .9986  .0014
Since the problem asks us for “or” the answer is the sum of the probabilities
Px  6.2 _ or _ x  5.8  .0014  .0014  .0028
3. The average number of years that a particular washing machine lasts is 7.45
years with a standard deviation of 2.71 years. Assume normality. The warranty for
this machine is two years.
a. What is the probability that a machine will last more than 8 years?
z
x

z
8  7.45
 .20
2.71
Px  8  1  .5793  .4207
b. What is the probability that an SRS of 100 washing machines will last more than
8 years?
z
x

z
n
8  7.45
 2.03
2.71
100
Px  5.9  .0104
c. What is the probability that a machine will fail within the warranty period?
z
x

z
2  7.45
 2.01
2.71
Px  2  .0222
d. What is the probability that an SRS of 15 will all fail within the warranty period?
z
x

z
n
2  7.45
 7.79
2.71
15
Px  2  Almost _ 0
6. The average amount of time that people spend going through airport security for
planes taking off between 8 AM and 10 AM at a busy airport is 21 minutes with a
standard deviation of 4.2 minutes.
a. What is the probability that a person has to wait more than 25 minutes?
z
x

z
25  21
 .95
4.2
Px  25  1  .8289  .1711
b. What is the probability that the average wait for the next 8 people in line have to
wait is more than 25 minutes?
z
x

z
n
25  21
 2.69
4 .2
8
Px  25  1  .9964  .0036
c. What is the probability that an SRS of 75 people will wait an average of 20
minutes or less?
z
x

20  21
 2.06
4.2
75
z
n
Px  25  .0197
7. An SAT review course claims that it can increase SAT scores with great
success. It reports that the average gain of a student is 50 points with a standard
deviation of 21.2. No other information is given.
a. If a student takes the course, what is the probability that her scores will increase
50 points or more?
z
x

50  50
0
21.2
z
Px  50  1  .5  .5
b) If a student takes the course, what is the probability that her scores will increase
55 points or more?
z
x

z
55  50
 .24
21.2
Px  55  1  .5948  .4052
c) If 10 students take the course, what is the probability that the average gain will
be 55 points or more?
z
x

n
z
55  50
 7.19
21.2
10
Px  55  1  almost _1  almost _ 0
d) If 100 students, take the course, what is the probability that the average gain will
be 55 points or more?
z
x

z
n
55  50
 22.73
21.2
100
Px  55  1  almost _1  almost _ 0
e) If 200 students, take the course, what is the probability that the average gain will
be 55 points or more?
z
x

z
n
55  50
 32.14
21.2
200
Px  55  1  almost _1  almost _ 0
f) Which of these answers are you most confident of? Why?
I’ll leave this one to you. All these answers make sense. Be sure to back up
your response in a logical, ordered way.
8. At 8 AM on a typical weekday morning, the average number of people in a 7-11
store is 24.5 with a standard deviation of 4.4.
a. What is the approximate distribution of the mean number of persons x in 500
randomly selected 7-11’s?
Since the number of people in the stores can be assumed to be normally
distributed, the sample can also be assumed to be normally distributed.
b. What is the probability that 500 selected stores will have more than 12,500
people in them at 8 AM?
z
x

n
z
25  24.5
 2.54
4.4
500
Px  25  1  .9945  .0055