Question

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $134,000. This distribution follows the normal distribution with a standard deviation of $30,000.

3. What is the likelihood of selecting a sample with a mean of at least $137,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

d. What is the likelihood of selecting a sample with a mean of more than $126,000? (Round your z value to 2 decimal places and final answer

5. Find the likelihood of selecting a sample with a mean of more than $126,000 but less than $137,000. (Round your z value to 2 decimal places and final answer to 4 decimal places.)

to 4 decimal places.)

Answer 1

Solution :

Given that,

mean = = 134000

standard deviation = = 30000

a ) P (x 137000 )

= 1 - P (x 137000 )

= 1 - P ( x -  / ) ( 137000 - 134000 / 30000)

= 1 - P ( z 3000 / 30000 )

= 1 - P ( z 0.1)

Using z table

= 1 - 0.5398

= 0.4602

Probability =0.4602

b ) P (x > 126000 )

= 1 - P (x < 126000 )

= 1 - P ( x -  / ) < ( 126000 - 134000 / 30000)

= 1 - P ( z <- 8000 / 30000 )

= 1 - P ( z < -0.27)

Using z table

= 1 - 0.3936

= 0.6064

Probability = 0.6064

c ) P (126000< x < 137000 )

P ( 126000 -134000 / 30000 ) < ( x -  / ) < ( 137000 - 134000 / 30000 )

P ( - 8000 / 30000 < z < 3000 / 30000 )

P (-0.27 < z < 0.1 )

P ( z < 0.1 ) - P ( z < -0.27)

Using z table

= 0.5398 - 0.3936

= 0.1462

Probability = 0.1462