Question

The U.S. Census Bureau announced that the mean sales price of new houses sold in 2016 was $370,800. Suppose the sales price follows a normal distribution with a standard deviation of $90,000.

a. (5 points) If you select samples of n = 4, what is the standard error of the mean?

b. (5 points) If you select samples of n = 100, what is the standard error of the mean?

c. (5 points) If you select a random sample of n = 100, what is the probability that the sample mean will be less than $370,000?

d. (5 points) If you select a random sample of n = 100, what is the probability that that sample mean will be between $350,000 and $365,000?

e. (5 points) The U.S. Census Bureau announced that the median sales price of new houses sold in 2016 was $316,500. Can we still assume that the sales price follows a normal distribution? How will this change the shape of the sampling distribution of n = 4 and n = 100?

Answer 1

Solution:

We are given

Mean = µ = 370800

SD = σ = 90000

Part a

We are given n=4

Standard error = σ/sqrt(n) = 90000/sqrt(4) = 45000

Part b

We are given n=100

Standard error = σ/sqrt(n) = 90000/sqrt(100) = 9000

Part c

We have to find P(Xbar<370000)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (370000 - 370800)/( 90000/sqrt(100))

Z = -0.088888889

P(Z<-0.088888889) = 0.464585107 (by using z-table)

P(Xbar<370000) = 0.464585107

Required probability = 0.464585107

Part d

We have to find P(350000<Xbar<365000)

P(350000<Xbar<365000) = P(Xbar<365000) – P(Xbar<350000)

Find P(Xbar<365000)

Z = (365000 - 370800)/( 90000/sqrt(100))

Z = -0.64444

P(Z<-0.64444) = P(Xbar<365000) = 0.259644

(by using z-table or excel)

Find P(Xbar<350000)

Z = (350000 - 370800)/( 90000/sqrt(100))

Z = -2.31111

P(Z<-2.31111) = P(Xbar<350000) = 0.010413 (by using z-table or excel)

P(350000<Xbar<365000) = P(Xbar<365000) – P(Xbar<350000)

P(350000<Xbar<365000) = 0.259644 - 0.010413

P(350000<Xbar<365000) = 0.249231

Required probability = 0.249231

Part e

We still cannot assume that the sales price follows a normal distribution because the median value and mean value should be approximately equal or same. For the sampling distribution with n=4, distribution becomes skewed. For n=100 distribution will approach to normal distribution.