The mean amount of life insurance per household is $110,000. This distribution is positively skewed. The standard deviation of the population is $40,000. Use Appendix B.1 for the z -values.
a. A random sample of 50 households revealed a mean of $112,000. What is the standard error of the mean? (Round the final answer to 2 decimal places.)
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? (Round the final answer to 4 decimal places.)
d. What is the likelihood of selecting a sample with a mean of more than $100,000? (Round the final answer to 4 decimal places.)
e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000. (Round the final answer to 4 decimal places.)
