1) A normal population has a mean of 54 and a standard deviation of 22. You select a random sample of 40. Use Appendix B1 for the z values.

Compute the probability that the sample mean is ( final answer to 4 decimals)

a) Greater than 57.

b) Less than 50.

c) Between 50 and 57.

2) The mean amount of life insurance per household is $111 000. This distribution is positively skewed. The standard deviation of the population is $39000. Use Appendix B1. for the z values.

a) A random sample of 40 households reveled a mean of $115 000. What is the standard error of the mean ? ( final answer to the 4decimals)

b) What is the likelihood of selecting a sample with a mean of at least $115 000 ? ( final answer to 4 decimal points)

c) What is the likelihood of selectinglikelihood of selecting a sample with a mean of more than $96 000 ? ( final answer to 4 decimal points)

d) Find the likelihood of selecting a sample with a mean of more than $96 000 but less than $ 115 000? ( final answer to 4 decimal place)