Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $120,000. This distribution follows the normal distribution with a standard deviation of $38,000.

Question

(A) If we select a random sample of 58 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)

Standard error of the mean=?

(B) What is the expected shape of the distribution of the sample mean?

Not normal, the standard deviation is unknown.
Unknown.
Uniform
Normal.

(C) What is the likelihood of selecting a sample with a mean of at least $124,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability=?

(D) What is the likelihood of selecting a sample with a mean of more than $112,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability=?

(E) Find the likelihood of selecting a sample with a mean of more than $112,000 but less than $124,000. (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability=?

Answer 1

(A) SEm = SD/√n

(B) Normal

(C, D, E) Z = (score-mean)/SEm

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.

Answer 2

500