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 36 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?

Answer 1

Answer:

The probability that the sample average will be between $30.00 and $31.00 is 0.6553.

Step-by-step explanation:

Let X represent the hourly wage for Switzerland.

It is provided that the mean hourly wage for Switzerland was $30.67 and the standard deviation of hourly labor rates is $3.00.

A random sample of 36 manufacturing workers are selected randomly from across Switzerland and asked what their hourly wage.

According to the Central Limit Theorem if an unknown population is selected with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from this population with replacement, then the distribution of the sample means will be approximately normally.  

As n = 36 > 30, Central limit theorem is applicable.

Then, the mean of the sample means is given by,

$\mu_{\bar x}=\mu=30.67$

And the standard deviation of the sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{36}}=0.50$

Compute the probability that the sample average will be between $30.00 and $31.00 as follows:

$P(30<\bar X<31)=P(\frac{30-30.67}{0.50}<\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}<\frac{31-30.67}{0.50})$

                         $=P(-1.34<Z<0.66)\\\\=P(Z<0.66)-P(Z<-1.34)\\\\=0.74537-0.09012\\\\=0.65525\\\\\approx 0.6553$

Thus, the probability that the sample average will be between $30.00 and $31.00 is 0.6553.