A worker is told that only $5\%$ of all workers make a higher wage. If the wages are normally distributed, what is the average hourly wage?

Question

So let's say a worker earns $\$16$ per hour at a plant and is told that only $5\%$ of all workers make a higher wage. If the wages are normally distributed with standard deviation of $\$5$ per hour then what is the average hourly wage of the plants workers. I've worked with the normal distribution before but I don't know how to do this question based on how it's worded.

Answer 1

If you use the standard normal distribution the random variable must be standardized.

$Z=\frac{X-\mu}{\sigma}$

Therefore

$P(X\leq x)=\Phi\left(\frac{x-\mu}{\sigma} \right)=\Phi\left(\frac{16-\mu}{5} \right)=0.95$

$\Phi\left(z\right)$ is the cdf of the standard normal distribution. $95\%(=100\%-5\%)$ of the other workers earn less than this specific worker.

Taking the inverse of the cdf.

$\frac{16-\mu}{5} =\Phi^{-1}(0.95)$

The table shows that $\Phi^{-1}(0.95)\approx 1.645$.

Now you can solve for $\mu$.

Answer 2

Hint.
Currently, the worker being considered makes $\$16$ and is told that only $5\%$ make more. Hence this person is in the $95$th percentile. In other words, his/her z-score is $1.644854$.