Question 468888: i really need some help with this problem, I am really struggling with it.
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?
b. What is the expected shape of the distribution of the sample mean?
c. What is the likelihood of selecting a sample with a mean of at least $112,000?
d. What is the likelihood of selecting a sample with a mean of more than $100,000?
e. 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 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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