Question 210776: This is from the textbook Statistical Techniques in Business and Economics, 13th edition, exercise #34
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the Unites States is $110,00. 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 the 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 $100,00
find the likelihood of selecting a sample with a mean of no more than $100,000 but no 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 life insurance per household in the Unites States is $110,00. 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 the mean?
Ans: 40,000/sqrt(50)
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what is the expected shape of the distribution of the sample mean
The mean of the sample means is $110,000; standard deviation = 40,000/sqrt(50)
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what is the likelihood of selecting a sample with a mean of at least $112,000
z(112,000) = (112,000-110,000)/[40,000/sqrt(50)] = 0.3536
P(x-bar >=112000) = 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
Find the z value of 100,000
Find the P(z is greater than that value)
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find the likelihood of selecting a sample with a mean of no more than $100,000 but no less than $112,000.
Use the z-value of 100,000 and the z-value of 112,000
Find the probability z is between those values.
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Cheers,
Stan H.
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