Question

7.5 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 $3.00.

a. Suppose 37 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 35 manufacturing workers are selected randomly from across Japan. What is the probability that the sample average will exceed $21.00?

c. Suppose 48 manufacturing workers are selected randomly from across the United States. What is the probability that the sample average will be less than $22.85?

Answer 1

a)

Here, μ = 30.67, σ = 0.4932, 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)/0.4932 = -1.36
z2 = (31 - 30.67)/0.4932 = 0.67

Therefore, we get
P(30 <= X <= 31) = P((31 - 30.67)/0.4932) <= z <= (31 - 30.67)/0.4932)
= P(-1.36 <= z <= 0.67) = P(z <= 0.67) - P(z <= -1.36)
= 0.7486 - 0.0869
= 0.6617

b)

Here, μ = 20.2, σ = 0.5071 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)/0.5071 = 1.58

Therefore,
P(X >= 21) = P(z <= (21 - 20.2)/0.5071)
= P(z >= 1.58)
= 1 - 0.9429 = 0.0571

c)
Here, μ = 23.82, σ = 0.433 and x = 22.85. We need to compute P(X <= 22.85). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (22.85 - 23.82)/0.433 = -2.24

Therefore,
P(X <= 22.85) = P(z <= (22.85 - 23.82)/0.433)
= P(z <= -2.24)
= 0.0125