Question 240859
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.
a. If we select a random sample of 50 households, what is the standard error of the mean?
The standard error depends on the level of significance.

If alpha is 5%, SE = 2.0096*40,000/sqrt(50) = 11369.0143
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b. What is the expected shape of the distribution of the sample mean?
Normal with mean = 110,000 and std = 40,000/sqrt(25) = 5656.85
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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)/[40,000/sqrt(50)] = 0.3536..
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P(x >= 112000) = P(t >= 0.3536 with 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
Same process as "c"
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e. What is the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000?
Combine the t-values from "c" and "d".
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Cheers,
Stan H.