In: Statistics and Probability
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $128,000. This distribution follows the normal distribution with a standard deviation of $33,000.
If we select a random sample of 66 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)
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 $130,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)
What is the likelihood of selecting a sample with a mean of more than $120,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)
Find the likelihood of selecting a sample with a mean of more than $120,000 but less than $130,000. (Round your z value to 2 decimal places and final answer to 4 decimal places.)
µ = 128000
sd = 33000
a) n = 66
Standard error of the mean = sd / sqrt(n) = 33000 / sqrt(66) = 4062
b) Expected shape of sampling distribution of the sample mean is approximately bell shaped i.e. normal distribution
c)
= P(Z > 0.49)
= 1 - P(Z < 0.49)
= 1 - 0.6879
= 0.3121
d)
= P(Z > -1.97)
= 1 - P(Z < -1.97)
= 1 - 0.0244
= 0.9756
e)
= P(-1.97 < Z < 0.49)
= P(Z < 0.49) - P(Z < -1.97)
= 0.6879 - 0.0244
= 0.6635