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.

a. Suppose 42 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 49 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

a)

µ = 30.67, σ = 4, n = 42

P(30 < X̅ < 31) =

= P( (30-30.67)/(4/√42) < (X-µ)/(σ/√n) < (31-30.67)/(4/√42) )

= P(-1.0855 < z < 0.5347)

= P(z < 0.5347) - P(z < -1.0855)

Using excel function:

= NORM.S.DIST(0.5347, 1) - NORM.S.DIST(-1.0855, 1)

= 0.5647

b)

µ = 20.2, σ = 4, n = 34

P(X̅ > 21) =

= P( (X̅-μ)/(σ/√n) > (21-20.2)/(4/√34) )

= P(z > 1.1662)

= 1 - P(z < 1.1662)

Using excel function:

= 1 - NORM.S.DIST(1.1662, 1)

= 0.1218

c)

µ = 23.82, σ = 4, n = 49

P(X̅ < 22.75) =

= P( (X̅-μ)/(σ/√n) < (22.75-23.82)/(4/√49) )

= P(z < -1.8725)

Using excel function:

= NORM.S.DIST(-1.8725, 1)

= 0.0306