Question

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000 with a standard deviation of $40,000. Assume the population distribution is normal. A random sample of 100 households is taken.

1. What is the probability that sample mean will be less than $120,000?
2. What is the probability that sample mean will be between $105,000 and $120,000?

Answer 1

μ = 110000

σ = 40000

n = 100

Standard error of mean (SE) = σ / √n

= 40000 / √100

= 4000

A)

P( X < 120000)

Z = X - μ / SE

= 120000 - 110000 / 4000

= 2.5

So, P(z < 2.5)

= 0.5 + 0.4938 [standard normal distribution table]

= 0.9938

B)

P(105000 < x < 120000)

= P( 105000 - 110000/4000 < z < 120000 - 110000 /4000 )

= P( -1.25 < z < 2.5 )

= 0.3944 + 0.4938 [standard normal distribution table]

= 0.8882