4.110 Manufacturing hourly pay rate. Government data indicate that the mean hourly wage for manufacturing workers in the United States is \(20.10 (Bureau of Labor Statistics, January 2016). Suppose the distribution of manufacturing wage rates nationwide can be approximated by a normal distribution with a standard deviation \)1.25 per hour. The first manufacturing firm contacted by a particular worker seeking a new job pays \(21.40 per hour.

a. If the worker were to undertake a nationwide job search, approximately what proportion of the wage rates would be greater than \)21.40 per hour?

b. If the worker were to randomly select a U.S. manufacturing firm, what is the probability the firm would pay more than $21.40 per hour?

c. The population median, call it $\mathit{\eta }$, of a continuous random

variable xis the value such that $\mathit{P}\mathbf{(}\mathit{x}\mathbf{\ge }\mathit{\eta }\mathbf{)}\mathbf{=}\mathit{P}\mathbf{(}\mathit{x}\mathbf{\le }\mathit{\eta }\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$that is, the median is the value such that half the area under the probability distribution lies above and half lies below it. Find the median of the random variable corresponding to the wage rate and compare it with the mean wage rate.

The mean hourly wage for manufacturing workers in the United States is $20.10 (Bureau of Labor Statistics, January 2016). It is supposed that the distribution of manufacturing wage rates nationwide can be approximated by a normal distribution with a standard deviation of $1.25 per hour.

a.

Let x be the random variable that represents the hourly wage for a nationwide job search.

X follows a normal distribution with a mean of $20.10 and a standard deviation of $1.25

$$\begin{gathered} P(x>21.40)=P\left(\frac{x-20.10}{1.25}>\frac{21.40-20.10}{1.25}\right) \\ =P(z>1.04) \\ =1-P(z\le 1.04) \\ =1-0.85083 \\ =0.14917 \end{gathered}$$

So, the proportion of wage rate greater than $21.40 is 0.14917

b.

Let x be the random variable that represents the hourly wage for a nationwide job search.

X follows a normal distribution with a mean $20.10 and a standard deviation $1.25

$$\begin{gathered} P(x>21.40)=P\left(\frac{x-20.10}{1.25}>\frac{21.40-20.10}{1.25}\right) \\ =P(z>1.04) \\ =1-P(z\le 1.04) \\ =1-0.85083 \\ =0.14917 \end{gathered}$$

So,, the probability of firm would pay more than $21.40 is 0.14917

c.

Let $\eta$be the median of the random variable x.

The median is the value of the variable which divides the distribution into two equal parts.

$$\begin{gathered} P(x\le \eta )=0.50 \\ P\left(\frac{x-20.10}{1.25}\le \frac{\eta -20.10}{1.25}\right)=0.50 \\ P\left(z\le \frac{\eta -20.10}{1.25}\right)=0.50 \\ \Phi \left(\frac{\eta -20.10}{1.25}\right)=0.50 \\ \frac{\eta -20.10}{1.25}={\Phi }^{-1}(0.50) \\ \frac{\eta -20.10}{1.25}=0 \\ \eta -2010=0 \\ \eta =20.10 \end{gathered}$$

So, the median value of wage is $20.10

Here the mean and median value is the same. Since the distribution of wages is normal.