Information from the American Institute of Insurance indicates the mean amount o
ID: 3219950 • Letter: I
Question
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $120,000. This distribution follows the normal distribution with a standard deviation of $38,000.
If we select a random sample of 58 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)
What is the likelihood of selecting a sample with a mean of at least $124,000? (Round 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 $112,000? (Round 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 $112,000 but less than $124,000. (Round z value to 2 decimal places and final answer to 4 decimal places.)
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $120,000. This distribution follows the normal distribution with a standard deviation of $38,000.
If we select a random sample of 58 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)
What is the likelihood of selecting a sample with a mean of at least $124,000? (Round 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 $112,000? (Round 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 $112,000 but less than $124,000. (Round z value to 2 decimal places and final answer to 4 decimal places.)
Explanation / Answer
a)as we know std error =std deviation/(n)1/2 =4989.64~4990
b)here P(X>124000)=1-P(X<124000)=1-P(Z<(124000-120000)/4989.64)=1-P(Z<0.8017)=1-0.7886 =0.2114
c) P(X>112000) =1-P(Z<-1.6033)=1-0.0544=0.9456
d) P(112000<X<124000)=P(-1.6033<Z<0.8017)=0.7886-0.0544=0.7342