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# Random sampling: Mean amount of life insurance per household in the US.

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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 mean?
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.

https://brainmass.com/statistics/normal-distribution/random-sampling-determining-mean-amount-life-insurance-295323

#### Solution Preview

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

Standard error of mean = 40000/sqrt(50) = 5656.85

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

#### Solution Summary

This solution provides formula and calculations for the probability problem set in an easy to understand, step by step format, including calculations of the standard error of the mean, probability in each condition, and whether the distribution is normal.

\$2.49