The mean amount of life insurance per household is $117 000. This distribution is positively skewed. The standard deviation of the population is $33 000. Use Appendix B.1 for the z values.

a. A random sample of 70 households revealed a mean of $121 000. 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 70 samples of households. What is the expected shape of the distribution of the sample mean?

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

c. What is the likelihood of selecting a sample with a mean of at least $121 000? (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 $109 000? (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 $109 000 but less than $121 000. (Round the z values to 2 decimal places and the final answers to 4 decimal places.)

Probability