The mean amount of life insurance per household is $111 000. This distribution is positively skewed. The standard deviation of the population is $39 000. Use Appendix B.1 for the z values.

a. A random sample of 40 households revealed a mean of $115 000. What is the standard error of the mean? (Round the final answer to the nearest whole number.)

Standard error of the mean

b. Suppose that you selected 40 samples of households. What is the expected shape of the distribution of the sample mean?

Shape (Click to select) Not normal, the standard deviation is unknown Normal Uniform Unknown

c. What is the likelihood of selecting a sample with a mean of at least $115 000? (Round the z values to 2 decimal places and the final answers to 4 decimal places.)

Probability

d. What is the likelihood of selecting a sample with a mean of more than $96 000? (Round the z values to 2 decimal places and the final answers to 4 decimal places.)

Probability

e. Find the likelihood of selecting a sample with a mean of more than $96 000 but less than $115 000. (Round the z values to 2 decimal places and the final answers to 4 decimal places.)

Probability