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.

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 33 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 $22.95?

Answer 1

(a)

= 30.67

= 3

n = 35

SE = /

= 3/

= 0.5071

To find P(30 < xbar < 31):

For xbar =30:

Z = (30 - 30.67)/0.5071

= - 1.3213

By Technology, Cumulative Area Under Standard Normal Curve = 0.0932

For xbar =31:

Z = (31 - 30.67)/0.5071

= 0.6508

By Technology, Cumulative Area Under Standard Normal Curve = 0.7424

So,

P(30 < xbar < 31): = 0.7424 - 0.0932 = 0.6492

So,

Answer is:

0.6492

(b)

= 20.20

= 3

n = 33

SE = /

= 3/

= 0.5222

To find P(xbar>21):

Z = (21 - 20.20)/0.5222

= 1.5320

By Technology, Cumulative Area Under Standard Normal Curve = 0.9372

So,

P(xbar>21):= 1 - 0.9372 = 0.0628

So,

Answer is:

0.0628

(c)

= 23.82

= 3

n = 48

SE = /

= 3/

= 0.4330

To find P(xbar<22.95):

Z = (22.95 - 23.82)/0.4330

= - 2.0092

By Technology, Cumulative Area Under Standard Normal Curve = 0.0.0223

So,

P(xbar <22.95): = 0.0223

So,

Answer is:

0.0223