Question

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $132,000. This distribution follows the normal distribution with a standard deviation of $40,000.

If we select a random sample of 70 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 $139,000? (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 $127,000? (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 $127,000 but less than $139,000. (Round your z value to 2 decimal places and final answer to 4 decimal places.)

Answer 1

Solution:

Given that,

mean =  = 132,000

standard deviation =  = 40,000

n = 70

The expected shape of the distribution of the sample mean

= 132,000

The standard error of the mean IS  

=  ( /n) = (40000 / 70 ) = 4781

the standard error of the mean = 4781

A ) p (   139,000 )

= 1 - P (     139,000 )

1- P ( - /) (139,000 - 132,000 / 4781)

1 - P ( z 7000 / 4781 )

1 - P ( z 1.46 )

Using z table

= 1 - 0.9279

= 0.0721

Probability = 0.0721

B ) P (   > 127,000 )

= 1 - P (   < 127,000 )

1- P ( - /) < (127,000  - 132,000 / 4781)

1 - P ( z < - 5000 / 4781 )

1 - P ( z < - 1. 05 )

Using z table

= 1 - 0.1469

= 0.8531

Probability = 0.8531

C ) P ( 12700 > <  139,000 )

P (   > 127,000 )

= 1 - P (   < 127,000 )

1- P ( - /) < (127,000  - 132,000 / 4781)

1 - P ( z < - 5000 / 4781 )

1 - P ( z < - 1. 05 )

Using z table

= 1 - 0.1469

= 0.8531

P (   < 139,000 )

P ( - /) <  (139,000  - 132,000 / 4781)

P ( z < 7000 / 4781 )

P ( z < 1.46 )

Using z table

= 0.9279

= 0.8531 +  0.9279

= 1.781

Probability = 1.781