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# Normal Distribution

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is \$110,000. This distribution follows the normal distribution with a standard deviation of \$40,000.
a. If we select a random sample of 50 households, what is the standard error of the mean?
b. What is the expected shape of the distribution of the sample?
c. What is the likelihood of selecting a sample with a mean of at least \$112,000?
d. What is the likelihood of selecting a sample with a mean of more than \$100,000?
e. Find the likelihood of selecting a sample with a mean of more than \$100,000 but less than \$112,000.

#### Solution Preview

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is \$110,000. This distribution follows the normal distribution with a standard deviation of \$40,000.

a. If we select a random sample of 50 households, what is the standard error of the mean?

Standard deviation =s= \$40,000
sample size=n= 50
sx=standard error of mean=s/square root of n= \$5,657 = ( 40000 /square root of 50)

Answer: standard error of the mean= \$5,657

b. What is the expected shape of the distribution of the sample?

Expected shape is normal distribution as sample size is large
(This ...

#### Solution Summary

Calculates probabilities using standard normal distribution

\$2.19