Question

# The U.S. Bureau of Labor Statistics released hourly wage figures for various countries for workers in...

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 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 38 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 \$22.80?

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

a. enter the probability that the sample average will be between \$30.00 and \$31.00 supposing that 37 manufacturing workers are selected randomly from across Switzerland
b. enter the probability that the sample average will exceed \$21.00 supposing that 38 manufacturing workers are selected randomly from across Japan
c. enter the probability that the sample average will be less than \$22.80 supposing that 49 manufacturing workers are selected randomly from across the United States

Solution:-

a) The probability that the sample average will be between \$30.00 and \$31.00 is 0.5381.

n = 37, Mean = 30.67, S.D = 4

x1 = 30

x2 = 31

By applying normal distribution:- z1 = -1.02

z2 = 0.50

P( -1.019 < z < 0.502) = P(z > -1.019) - P(z > 0.502)

P( -1.019 < z < 0.502) = 0.8459 - 0.3078

P( -1.019 < z < 0.502) = 0.5381

b) The probability that the sample average will exceed \$21.00 is 0.1088.

n = 38, Mean = 20.20, S.D = 4

x = 21

By applying normal distribution:- z = 1.23

Use the z-score table or p-value calculator.

P(z > 1.233) = 0.1088

c) The probability that the sample average will be less than \$22.80 is 0.0371.

n = 49, Mean = 23.82, S.D = 4

x = 22.80

By applying normal distribution:- z = - 1.79

Use the z-score table or p-value calculator.

P(z < - 1.785) = 0.0371

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