Question

The weekly income of managers in a certain company is normally distributed with a mean of 50,000 and a standard deviation of 5,000. What is the z-score for weekly income ofa. 40,000?b. 55,000

Answer 1

Given:

`\begin{gathered} \mu=50000 \\ \sigma=5000 \end{gathered}`

Using the Z-score formula;

`Z=(x-\mu)/(\sigma)`

when x = 40000

`\begin{gathered} Z=(40000-50000)/(5000) \\ Z=(-10000)/(5000) \\ Z=-2 \end{gathered}`

Therefore, the Z-score for the weekly income of 40000 is -2

when x = 55000

`\begin{gathered} Z=(55000-50000)/(5000) \\ Z=(5000)/(5000) \\ Z=1 \end{gathered}`

Therefore, the Z-score for the weekly income of 55000 is 1