Question

Suppose that Jason recently landed job offers at two companies. Company A reports an average salary of $51,500 with astandard deviation of $2,050. Company B reports an average salary of $46,820 with a standard deviation of $5,755. Assumethat salaries at each company are normally distributed.Jason's goal is to secure a position that pays $55,000 per year. What are the z-scores for Jason's desired salary at Company Aand Company B? Please round your answers to two decimal places.

Answer 1

We are given the following information;

Company A;

$\begin{gathered} \text{Mean}=51500 \\ SD=2050 \end{gathered}$

Comapny B;

$\begin{gathered} \text{Mean}=46820 \\ SD=5755 \end{gathered}$

We have a normal distribution here. We can derive a standard normal distribution and the results would give us the z scores

That is;

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

Therefore, if

$X=x$

The z score would be;

$z=\frac{(x-\mu)}{\sigma}$

However, in this case, the random variable is the target salary which is 55,000.

With X as the random variable we would have;

$X\rightarrow N(51500,2050)$

This is the tracking salary amount of employees in company A

Similarly;

With Y as the random variable we would have;

$Y\rightarrow N(46820,5755)$

This on the other hand is the tracking salary amount of employees in company B.

If Jason's target is to get an annual salary of $55,000, the z score for company A would be;

$\begin{gathered} Z=\frac{X-\mu}{\sigma} \\ Z=\frac{55000-51500}{2050} \\ Z=\frac{3500}{2050} \\ Z=1.7073 \\ Z\approx1.71 \end{gathered}$

The z score for company B would be;

$\begin{gathered} Z=\frac{X-\mu}{\sigma} \\ Z=\frac{55000-46820}{5755} \\ Z=\frac{8180}{5755} \\ Z=1.4213 \\ Z\approx1.42 \end{gathered}$

ANSWER:

The z scores for both companies are;

$\begin{gathered} \text{Company A} \\ Z=1.71 \\ \text{Company B} \\ Z=1.42 \end{gathered}$

Both answers are rounded to 2 decimal places