Information from the American Institute of Insurance indicat

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $122,000. This distribution follows the normal distribution with a standard deviation of $31,000. a. If we select a random sample of 60 households, what is the standard error of the mean? (Round your answer to the nearest whole number.) Standard error of the mean b. What is the expected shape of the distribution of the sample mean? Sample mean c. What is the likelihood of selecting a sample with a mean of at least $130,000? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability d. What is the likelihood of selecting a sample with a mean of more than $114,000? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability e. Find the likelihood of selecting a sample with a mean of more than $114,000 but less than $130,000. (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability

Solution

a)
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )=31000
Sample Size(n)=60
Standard Error = ( 31000/ Sqrt ( 60) )
= 4002.083
b)
Approximately Normal

c)
P(X < 130000) = (130000-122000)/31000/ Sqrt ( 60 )
= 8000/4002.0828= 1.999
= P ( Z <1.999) From Standard NOrmal Table
= 0.9772                  
              
P(X > = 130000) = 1 - P(X < 130000)
= 1 - 0.9772 = 0.0228                  
                  
d)
P(X > 114000) = (114000-122000)/31000/ Sqrt ( 60 )
= -8000/4002.083= -1.999
= P ( Z >-1.999) From Standard Normal Table
= 0.9772