Question

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $130,000. This distribution follows the normal distribution with a standard deviation of $39,000.

a) If we select a random sample of 68 households, what is the standard error of the mean?

Standard Error of the Mean:

b) What is the expected shape of the distribution of the sample mean?

The distribution will be:

c) What is the likelihood of selecting a sample with a mean of at least $135,000?

Probability:

d) What is the likelihood of selecting a sample with a mean of more than $121,000?

Probability:

e) Find the likelihood of selecting a sample with a mean of more than $121,000 but less than $135,000.

Probability:

Answer 1

Solution:

Given that,

mean = = $130,000.

standard deviation =   = 1$39,000.

a ) n = 68

= 130,000.

=  ( /n) = (39,000./ 68 ) = 4729.4447

b )The distribution of the sample mean is approximatly normal

c ) P ( 135000)

= 1 - P (   135000 )

= 1 - P ( -  / ) ( 135000 - 130,000./ 4729.4447)

= 1 - P ( z 5000 / 4729.4447)

= 1 - P ( z   1.06)

Using z table

= 1 - 0.8554

= 0.1446

Probability = 0.1446

d ) P ( > 121000)

= 1 - P ( < 121000 )

= 1 - P ( -  / ) < ( 121000 - 130,000./ 4729.44 47)

= 1 - P ( z < - 9000 / 4729.4447)

= 1 - P ( z < -1.90)

Using z table

= 1 - 0.0287

= 0.9713

Probability = 0.9713

e )P (  121000 > < 135000)

= P ( -  / ) < ( 135000 - 130,000./ 4729.44 47)

= P ( z  < 5000 / 4729.4447)

= P ( z < 1.06)

Using z table

= 0.8554

0.0287 + 0.8554 =0.8841

Probability = 0.8841