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 $3.00.

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

a. Suppose 43 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 34 manufacturing workers are selected randomly from across Japan. What is the probability that the sample average will exceed $21.00?

c. Suppose 47 manufacturing workers are selected randomly from across the United States. What is the probability that the sample average will be less than $22.75?

Answer 1

Question (a)

Given n = 43

Hourly wages = 30.67

Standard deviation of the sampling distribution will be / n = 3 / 43

Z = (x - ) / ( / n)

Z-value (at x=30) = (30 - 30.67) / (3 / 43 )

= -0.67 / 0.45749

= -1.464495

= -1.46 rounded to 2 decimals

The area to the left of z-value can be found from the negative z-table attached below. the area is 0.07215

Z-value (at x=31) = (31 - 30.67) / (3 / 43 )

= 0.33 / 0.45749

= 0.721318

= 0.72 rounded to 2 decimals

The area to the left of z-value can be found from the positive z-table.attached below the area is 0.76424

Since both the values are the areas left to the respective z-values. to get the probability between those 2 values we need to subtract 0.07215 from 0,76424

So probability that the sample average will be between $30.00 and $31.00 = 0.76424 - 0.07215

= 0.69209

Question (b)

Given n = 34

Hourly wages = 20.20

Standard deviation of the sampling distribution will be / n = 3 / 34

Z = (x - ) / ( / n)

Z-value (at x=21) = (21 - 20) / (3 / 34 )

= 1 / 0.514496

= 1.943651

= 1.94 rounded to 2 decimals

The area to the left of z-value can be found from the positive z-table.attached below the area is 0.97381

but we want the area to the right of z-score since we want probability that the sample average will exceed $21.00

So probability that the sample average will exceed $21.00 = 1 - 0.97381

= 0.02619

Question (c)

Given n = 47

Hourly wages = 23.82

Standard deviation of the sampling distribution will be / n = 3 / 47

Z = (x - ) / ( / n)

Z-value (at x=22.75) = (22.75 - 23.82) / (3 / 47 )

= -1.07 / 0.437595

= -2.445183

= -2.45 rounded to 2 decimals

The area to the left of z-value can be found from the negative z-table.attached below the area is 0.00714

So the probability that the sample average will be less than $22.75 = 0.00714