Question 1192903: Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $115,000. This distribution follows the normal distribution with a standard deviation of $37,000.
a.What is the likelihood of selecting a sample with a mean of at least $120,000? (Round your z-value to 2 decimal places and final answer to 4 decimal places.)
b.What is the likelihood of selecting a sample with a mean of more than $106,000? (
c. Find the likelihood of selecting a sample with a mean of more than $106,000 but less than $120,000.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
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A. z>(120000-115000)/37000=5/37 so probability is 0.4463
B. this is z-9000/37000 or z >-9/37, probability is 0.5961
C.This is the difference between the two or probability of 0.1498
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