Question

information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $104,000. This distribution follows the normal distribution with a standard deviation of $31,000.  If we select a random sample of 42 households, what is the standard error of the mean?  What is the likelihood of selecting a sample with a mean of at least $110,000? What is the likelihood of selecting a sample with a mean of more than $94,000? Find the likelihood of selecting a sample with a mean of more than $94,000 but less than $110,000

Answer 1

Standard error =

                        = 31000/ = 4783.4038

a) P(> 110000)

= P(( - )/() > (110000 - )/())

= P(Z > (110000 - 104000)/4783.4038)

= P(Z > 1.25)

= 1 - P(Z < 1.25)

= 1 - 0.8944

= 0.1056

b) P( > 94000)

= P(( - )/() > (94000 - )/())

= P(Z > (94000 - 104000)/4783.4038)

= P(Z > -2.09)

= 1 - P(Z < -2.09)

= 1 - 0.0183

= 0.9817

c) P(94000 < X < 110000)

= P((94000 - )/() < (X - )/() < (110000 - )/())

= P((94000 - 104000)/4783.4038 < Z < (110000 - 104000)/4783.4038)

= P(-2.09 < Z < 1.25)

= P(Z < 1.25) - P(Z < -2.09)

= 0.8944 - 0.0183

= 0.8761