Question

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $115,000. This distribution follows the normal distribution with a standard deviation of $37,000.

3. What is the likelihood of selecting a sample with a mean of at least $120,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

4. What is the likelihood of selecting a sample with a mean of more than $106,000? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

5. Find the likelihood of selecting a sample with a mean of more than $106,000 but less than $120,000. (Round your z value to 2 decimal places and final answer to 4 decimal places.)

Answer 1

Solution :

Given that ,

mean = = $115000

standard deviation = = $37000

c.

P(x $120000) = 1 - P(x   120000)

= 1 - P[(x - ) / (120000 - 115000) / 37000]

= 1 -  P(z 0.14)   

= 1 - 0.5557

= 0.4443

The likelihood of selecting a sample with a mean of at least $120,000 is 0.4443

d.

P(x > $106000) = 1 - P(x < 106000)

= 1 - P[(x - ) / < (106000 - 115000) / 37000)

= 1 - P(z < -0.24)

= 1 - 0.4052

= 0.5948

The likelihood of selecting a sample with a mean of more than $106,000 is 0.5948

e.

P($106000 < x < $120000) = P[(106000 - 115000)/ 37000) < (x - ) /  < (120000 - 115000) / 37000) ]

= P(-0.24 < z < 0.14)

= P(z < 0.14) - P(z < -0.24)

= 0.5557 - 0.4052

= 0.1505

The likelihood of selecting a sample with a mean of more than $106,000 but less than $120,000 is 0.1505