SOLUTION: The mean amount of life insurance per household is 110,000 normal distribution a) with a standarf deviation of $40,000 what is the likelihood of selecting a sample with a mean of a

Question 219059: The mean amount of life insurance per household is 110,000 normal distribution a) with a standarf deviation of $40,000 what is the likelihood of selecting a sample with a mean of at least $112,000?
b) What is the expected shape of the distribution of the sample mean?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The mean amount of life insurance per household is 110,000 normal distribution a) with a standard deviation of $40,000 what is the likelihood of selecting a sample with a mean of at least $112,000?
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z(112,000) = (112,000-110,000)/[40,000] = 2/40
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P(z>(2/40) = normalcdf((2/40),10) = 0.48
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b) What is the expected shape of the distribution of the sample mean?
The sample means will have a normal distribution
as guaranteed by the Central Limit Theorem.
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Cheers,
Stan H.

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