Question 1192825: 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.
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.)
d. What is the likelihood of selecting a sample with a mean of more than $106,000? (Round your z-value to 2 decimal places and final answer to 4 decimal places.)
e. Find the likelihood of selecting a sample with a mean of more than $106,000 but less than $120,000. (Round your z-value to 2 decimal places and final answer to 4 decimal places.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Here, you mean a sample of 1, or probability of selecting a person/insurance..., because there is no other sample size given
=(x-mean)/sigma
>(120000-115000)/37000=5/37=0.14
probability is 0.4443
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this is z>-9/37=-9.24
probability is 0.5948
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If you subtract the first from the second, you arrive at the 106-120 interval and that has probability of 0.1505. If you use a calculator and use the fraction, you will get a more exact result, but those aren't the directions.
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