Question

Home prices in a certain community have a distribution that is skewed right. The mean of the home prices is $498,000 with a standard deviation of $25,200.

a. Suppose we take a random sample of 30 homes in this community. What is the probability that the mean of this sample is between $500,000 and $510,000?

b. Suppose we take a random sample of 10 homes in this community. Can we find the approximate probability that the mean of the sample is more than $510,000? If so, find it. If not, explain why not.

Answer 1

Solution:

Given that mean  µ = 498,000

Standard deviation s = 25,200.

a) probability that the mean of this sample is between $500,000 and $510,000

here sample n = 30

P( 500,000 < X < 510,000 ) = P( 500,000 - 498000 / 25200 / √30  < Z < 510,000 - 498000 / 25200 / √30 )

= P( 2000 / 4,600.86 < Z < 12,000‬ / 4,600.86)

= P ( 0.4347 < Z < 2.6082)

Using standard normal table

P ( 0.4347 < Z < 2.6082) = 0.3291

b) probability that the mean of the sample is more than $510,000

here sample n = 30

P ( X > 510,000 ) = P( Z > 510,000 - 498000 / 25200 / √10 )

= P ( Z >0.1505 )

Using standard normal table

P ( Z > 0.1505 ) = 1−P ( Z< 0.1505 )=1−0.5596=0.4404