Question

Please include bell shape curve for B & C.

According to the Organization for Economic Co-Operation and Development (OECD), adults in the United States worked an average of 1,805 hours in 2007. Assume the population standard deviation is 395 hours and that a random sample of 70 U.S. adults was selected. a. Calculate the standard error of the mean. b. What is the probability that the sample mean will be more than 1,775 hours? c. What is the probability that the sample mean will be between 1,765 and 1,820 hours? d. Would a sample mean of 1,815 hours support the claim made by the OECD? Explain?

Answer 1

answer)

As the population s.d is known, so we can use standard normal z table to estimate the answers

Given mean = 1805

S.d = 395

N = 70

Z = (x-mean)/(s.d/√n)

A)

Standard error = s.d/√n = 395/√70 = 47.2115300687

B)

We need to find

P(x>1775)

Z = (1775-1805)/(395/√70)

Z = -0.64

From z table, P(Z>-0.64) = 0.7389

C)

P(1765<X<1820) = P(X<1820) - P(X<1765)

P(X<1820) = (1820-1805)/(395/√70)

Z = 0.32

From z table, P(Z<0.32) = 0.6255

P(X<1765)

Z = -0.85

From z table, P(Z<-0.85) = 0.1977

Required probability is = 0.6255-0.1977 = 0.4278

D)

P(X>1815)

Z = (1815-1805)/(395/√70)

Z = 0.21

From z table, P(Z>0.21) = 0.4168

As the probability is quite large

Yes a sample mean of 1815 supports the claim.