Question

The U.S. Bureau of Labor Statistics released hourly wage figures for various countries for workers in the manufacturing sector. The hourly wage was $30.67 for Switzerland, $20.20 for Japan, and $23.82 for the U.S. Assume that in all three countries, the standard deviation of hourly labor rates is $4.00. Appendix

a. Suppose 39 manufacturing workers are selected randomly from across Switzerland and asked what their hourly wage is. What is the probability that the sample average will be between $30.00 and $31.00?

b. Suppose 32 manufacturing workers are selected randomly from across Japan. What is the probability that the sample average will exceed $21.00?

c. Suppose 50 manufacturing workers are selected randomly from across the United States. What is the probability that the sample average will be less than $23.00?

Answer 1

a)

Here, μ = 30.67, σ = 0.6405, x1 = 30 and x2 = 31. We need to compute P(30<= X <= 31). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (30 - 30.67)/0.6405 = -1.05
z2 = (31 - 30.67)/0.6405 = 0.52

Therefore, we get
P(30 <= X <= 31) = P((31 - 30.67)/0.6405) <= z <= (31 - 30.67)/0.6405)
= P(-1.05 <= z <= 0.52) = P(z <= 0.52) - P(z <= -1.05)
= 0.6985 - 0.1469
= 0.5516

b)

Here, μ = 20.2, σ = 0.7071 and x = 21. We need to compute P(X >= 21). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (21 - 20.2)/0.7071 = 1.13

Therefore,
P(X >= 21) = P(z <= (21 - 20.2)/0.7071)
= P(z >= 1.13)
= 1 - 0.8708 = 0.1292

c)

Here, μ = 23.82, σ = 0.5657 and x = 23. We need to compute P(X <= 23). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (23 - 23.82)/0.5657 = -1.45

Therefore,
P(X <= 23) = P(z <= (23 - 23.82)/0.5657)
= P(z <= -1.45)
= 0.0735