Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the U.S is $115,000. This distribution follows the normal distribution of $$
c. What is the likelihood of selecting a sample with a mean of at least $120,000 (Round z values to 2 decimal places and final answer to 4 decimal places)
d) What is the likelihood of selecting a sample with a mean of more than $106,000? (Round z values to 2 decimal places and final answer to 4 decimal places)
Assuming the same standard deviation as from the other problem:
c. Z(120000) = 120000-115000 / 37000 = 0.14
P(Z=0.14) = 0.5557
But we want AT LEAST $120,000 which is the right portion of the normal curve
So, the correct P value is 1 - 0.5557 = 0.4443
d. Z(106000) = 106000 - 115000 / 37000 = - 0.24
P(Z= -0.24) = 0.4052
But we want MORE than $106,000 so again we are looking at the right portion of the normal curve
So, P = 1- 0.4052 = 0.5948
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?