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 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 36 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 $22.95?

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

Answer 1

a)

X ~ N ( µ = 30.67 , σ = 3 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 30 - 30.67 ) / ( 3 / √(37))
Z = -1.36
Z = ( 31 - 30.67 ) / ( 3 / √(37))
Z = 0.67
P ( 30 < X̅ < 31 ) = P ( Z < 0.67 ) - P ( Z < -1.36 )
P ( 30 < X̅ < 31 ) = 0.7486 - 0.0869 (From Z table)
P ( 30 < X̅ < 31 ) = 0.6617

b)

X ~ N ( µ = 20.2 , σ = 3 )
P ( X > 21 ) = 1 - P ( X < 21 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 21 - 20.2 ) / ( 3 / √ ( 36 ) )
Z = 1.6
P ( ( X - µ ) / ( σ / √ (n)) > ( 21 - 20.2 ) / ( 3 / √(36) )
P ( Z > 1.6 )
P ( X̅ > 21 ) = 1 - P ( Z < 1.6 )
P ( X̅ > 21 ) = 1 - 0.9452 (From Z table)
P ( X̅ > 21 ) = 0.0548

c)

X ~ N ( µ = 23.82 , σ = 3 )
P ( X < 22.95 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 22.95 - 23.82 ) / ( 3 / √47 )
Z = -1.99
P ( ( X - µ ) / ( σ/√(n)) = ( 22.95 - 23.82 ) / ( 3 / √(47) )
P ( X < 22.95 ) = P ( Z < -1.99 )
P ( X̅ < 22.95 ) = 0.0233 (From Z table)