Question 468888
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. 
This distribution follows the normal distribution with a standard deviation of $40,000.
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a. If we select a random sample of 50 households, what is the standard error of the mean?
Ans: s = 40,000/sqrt(50) = 5656.85
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b. What is the expected shape of the distribution of the sample mean?
Ans: normal
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c. What is the likelihood of selecting a sample with a mean of at least $112,000?
t(112,000) = (112,000-110,000)/[5656.85 = 0.3536
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P(x-bar >= 112,000) = P(t >= 0.3536 when df = 49) = 0.3626
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d. What is the likelihood of selecting a sample with a mean of more than $100,000?
t(100,000) = (100,000-112,000)/5656.85 = -2.1213
P(x-bar > 100,000) = P(t > -2.1213 when df = 49) = 0.9805
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e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000.
P(100,000 < x-bar < 112,000) = P(-2.1213 < t < 0.3626 when df=49) = 0.6213
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Cheers,
Stan H.
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