SOLUTION: Information from the American Institute of Insurance indicates the mean amount of the insurance per household in the United States is $110,000. This distribution follows the normal

Question 212987: Information from the American Institute of Insurance indicates the mean amount of the insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000.
If we select a random sample of 50 households, what is the standard error of mean?
What is the expected shape of the distribution of the sample mean?
What is the likelihood of selecting a sample with a mean of at least $112,000?
What is the likelihood of selecting a sample with a mean of more than $100,000?
Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Information from the American Institute of Insurance indicates the mean amount of the insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000.
If we select a random sample of 50 households, what is the standard error of mean?
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E = 1.96[40000/sqrt(50}] = 11087.43
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What is the expected shape of the distribution of the sample mean?
mean of the sample means = 110,000
standard deviation of the sample means = 40,000/sqrt(50)
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What is the likelihood of selecting a sample with a mean of at least $112,000?
Find the z-score of 112,000 = (112,000-110,000)/[40,000/sqrt(50)] = 0.3536
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P(x-bar > 112,000) = P(z > 0.3536) = 0.3618..
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What is the likelihood of selecting a sample with a mean of more than $100,000?
Same procedure as above.
Find the z-score of 100,000
Find the probability that z is greater than that z-value.
Ans: 0.9615..
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Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000?
Ans: 0.5996...
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Cheers,
Stan H.

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