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.

a. Suppose 44 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 36 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 $22.90?

Answer 1

a)

Here, μ = 30.67, σ = 0.603, 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)/0.603 = -1.11
z2 = (31 - 30.67)/0.603 = 0.55

Therefore, we get
P(30 <= X <= 31) = P((31 - 30.67)/0.603) <= z <= (31 - 30.67)/0.603)
= P(-1.11 <= z <= 0.55) = P(z <= 0.55) - P(z <= -1.11)
= 0.7088 - 0.1335
= 0.5753

b)

Here, μ = 20.2, σ = 0.6667 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)/0.6667 = 1.2

Therefore,
P(X >= 21) = P(z <= (21 - 20.2)/0.6667)
= P(z >= 1.2)
= 1 - 0.8849 = 0.1151

c)

Here, μ = 23.82, σ = 0.5657 and x = 22.9. We need to compute P(X <= 22.9). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (22.9 - 23.82)/0.5657 = -1.63

Therefore,
P(X <= 22.9) = P(z <= (22.9 - 23.82)/0.5657)
= P(z <= -1.63)
= 0.0516