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.)

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a Standard error of the mean SD Sqrtn 40000 sqrt50 565685 b No... View the full answer