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 Statistical Tables

a. Suppose 35 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 37 manufacturing workers are selected randomly from across Japan. What is the probability that the sample average will exceed $21.00?
c. Suppose 49 manufacturing workers are selected randomly from across the United States. What is the probability that the sample average will be less than $22.95?

(Round the values of z to 2 decimal places. Round your answers to 4 decimal places.)

Answer 1

a):  The hourly wage for Switzerland is $30.67 .

Values are given as,
Mean,= 30.67,
sd= 4.00

n= 35

= 0.6761

We need to find P(30 < x < 31)
Using below formula

Put the values in the formula,
P[(30 - 30.67)/0.6761 < z < (31 - 30.67)/0.6761]
= (-0.99 < z < 0.49)
z scores are (-0.99, 0.49)

Using below formula,
P(a < z < b) = P(z < b) - P(z < a)

Put the values in above formula,
P(-0.99 < z < 0.49) = P(z < 0.49) - P(z < -0.99)

Using z table, we get,
So, P(-0.99 < z < 0.49) = 0.6879 - 0.1611= 0.5268

Required answer :P(-0.99 < z < 0.49) = 0.5268

b): Values are given as,
Mean,= 20.2,
sd= 4.00

n= 37
= 0.6576

We need to find P(x > 21)
Using below formula


Put the values in the formula,
P[(z < (21 - 20.2)/0.6576]
= (z < 1.22)
z score is 1.22

Using z table, we get,
P(z < 1.22) = 0.88877

Now P(z > 1.22) = 1 - P(z < 1.22)
Now P(z > 1.22) = 0.1112

Required answer : P(z > 1.22) = 0.1112

c): Values are given as,
Mean, = 23.82,
n= 49

sd= 4.00

= 0.5714

We need to find P(x < 22.95)
Using below formula


Put the values in the formula,
P[(z < (22.95 - 23.82)/0.5714]
= (z < -1.52)
z score is -1.52

Using z table, we get,
P(z < -1.52) = 0.0643

Required answer : P(z < -1.52) = 0.0643