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

Respuesta :

Solution:

Given:

[tex]\begin{gathered} \mu=50000 \\ \sigma=5000 \end{gathered}[/tex]

Using the Z-score formula;

[tex]Z=\frac{x-\mu}{\sigma}[/tex]

when x = 40000

[tex]\begin{gathered} Z=\frac{40000-50000}{5000} \\ Z=\frac{-10000}{5000} \\ Z=-2 \end{gathered}[/tex]

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

when x = 55000

[tex]\begin{gathered} Z=\frac{55000-50000}{5000} \\ Z=\frac{5000}{5000} \\ Z=1 \end{gathered}[/tex]

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