Information from the American Institute of Insurance indicates the meanamount of life insurance per household in the United States is $110,000. This distribution is positively skewed. The standard deviationof the population is not known.

A. a random sampleof 50 household revealed a mean of $112,000 and a standard deviation of $40,000. What is the standard errorof the mean?

B. Suppose that you selected 50 samples of households. 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 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.

A. a random sample of 50 household revealed a mean of $112,000 and a standard deviation of $40,000. What is the standard error of the mean?

N = 50

M = 112000

SD = 40000

Standard error is SE = SD / SQRT(n) = 40000/SQRT(50)= 5,656.85

B. Suppose that you selected 50 samples of households. What is the expected …

This solution contains complete steps of calculations and brief explanations to determine standard error of the mean, expected shape of the distribution, and selecting sample with specific means.