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 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 33 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?

Answer 1

The hourly wage is $30.67 for Switzerland. The hourly wage is $20.20 for Japan. The hourly wage is $23.82 for the US.

The standard deviation for hourly wages is $3.00 in all three countries.

Let, X denote the random variable denoting the hourly wage in Switzerland.

Let, Y denote the random variable denoting the hourly wage in Japan.

Let, Z denote the random variable denoting the hourly wage in US.

So, X follows Normal with mean 30.67 and sd 3.

Y follows Normal with mean 20.20 and sd 3.

Z follows Normal with mean 23.82 and sd 3.

(a) 42 workers are selected from Switzerland. Let, Xbar be the random variable denoting the hourly wage of this sample.

Then, Xbar follows Normal distribution with a mean of 30.67 and sd of (3/sqrt(42))=0.462.

To find P(30<Xbar<31)

=P(30-30.67<Xbar-30.67<31-30.67)

=P(-0.67<Xbar-30.67<0.33)

=P(-0.67/0.462<Z<0.33/0.462)

Where, Z is the standard normal variate.

=P(-1.45<Z<0.71)

=phi(0.71)-phi(-1.45)

Where, phi is the distribution function of the standard normal variate.

=phi(0.71)+phi(1.45)-1

=0.7611+0.9265-1

=0.6876.

So, the required probability is 0.6876.

(b) 33 workers are selected from Japan. Let, Ybar be the random variable denoting the hourly wage of the sample.

So, Ybar follows a normal distribution with mean 20.20 and sd=(3/sqrt(33))=0.522.

To find P(Ybar>21)

=P(Ybar-20.20>21-20.20)

=P(Ybar-20.20>0.80)

=P(Z>0.80/0.522)

=P(Z>1.53)

Where, Z is the standard normal variate.

=1-phi(1.53)

Where, phi is the distribution function of standard normal variate.

=1-0.9370

=0.063.

So, the required probability is 0.063.

(c) 49 workers are selected from across the US. Let, Zbar be the random variable denoting the hourly wage of this sample.

So, Zbar follows normal distribution with mean 23.82 and sd=(3/sqrt(49))=0.428.

To find, P(Zbar<22.80)

=P(Zbar-23.82<22.80-23.82)

=P(Zbar-23.82<-1.02)

=P(Z<-1.02/0.428)

=P(Z<-2.38)

Where, Z is the standard normal variate.

=phi(-2.38)

=1-phi(2.38)

Where, phi is the distribution function of the standard normal variate.

=1-0.9913

=0.0087.

So, the required probability is 0.0087.