Question

The average hourly wage of employees of a certain company is $14.67. Assume the variable is normally distributed. If the standard deviation is $4.42, find the probability that a randomly selected employee earns less than $8.26

Answer 1

\[z=\left|\frac{8.26-14.67}{4.42}\right|\]
This \(z\)-value gives the following area:

Look up the area (probability) \(A\) on a \(z\)-table.

The area under the curve to the left of the mean is \(\dfrac{1}{2}\), so the area (probability) corresponding to the desired interval (less that 8.26) is \(\dfrac{1}{2}-A\).