34.Information from the American Institute of Insurance indicatesthe mean amount

Question

34.Information from the American Institute of Insurance indicatesthe mean amount of life insurance per household in the UnitedStates is $110,000. This distribution follows the normaldistribution with a standard deviation of $40,000.

1. If we select a random sample of 50 households, what is thestandard error of the mean?
2. What is the expected shape of the distribution of the samplemean?
3. What is the likelihood of selecting a sample with a mean of atleast $112,000?
4. What is the likelihood of selecting a sample with a mean ofmore than $100,000?
5. Find the likelihood of selecting a sample with a mean of morethan $100,000 but less than $112,000.

Explanation / Answer

standard error: 2(40000)/50 = 11313.7 normal distribution (bell-curve) P(x> 112,000) z = (x-u)/o = (112000-110000)/40000 = 0.05 P(x>112,000) = P(z>0.05) = normalcdf(0.05, 100) =0.48 P(x> 100,000) z = (x-u)/o = (112000-100000)/40000 = 0.3 P(x>100000) = P(z>0.3) = normalcdf(0.3, 100) =0.382 P(100,000<x<112,000) = P(0.05<z<0.3) =normalcdf(0.05, 0.3) = 0.09797

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