The annual salaries of all employees at a financial company are normally distributed with a mean of [tex]$34,000 and a standard deviation of $[/tex]4,000.

What is the [tex]$z$[/tex]-score of a company employee who makes an annual salary of $28,000?

A. -5
B. -3.5
C. -1.5
D. 1.5



Answer :

To determine the [tex]$z$[/tex]-score of a given employee's salary, we use the following formula for the [tex]$z$[/tex]-score calculation in a normal distribution:

[tex]$ z = \frac{X - \mu}{\sigma} $[/tex]

Where:
- [tex]\( X \)[/tex] is the value for which we are calculating the [tex]$z$[/tex]-score (the individual's salary).
- [tex]\( \mu \)[/tex] is the mean of the distribution (the average salary).
- [tex]\( \sigma \)[/tex] is the standard deviation of the distribution (the standard deviation of salaries).

In this situation:
- The mean salary, [tex]\(\mu\)[/tex], is \[tex]$34,000. - The standard deviation, \(\sigma\), is \$[/tex]4,000.
- The employee's salary, [tex]\(X\)[/tex], is \[tex]$28,000. Plugging these values into the formula, we get: $[/tex][tex]$ z = \frac{28000 - 34000}{4000} $[/tex][tex]$ Simplify the numerator: $[/tex][tex]$ z = \frac{-6000}{4000} $[/tex][tex]$ Perform the division: $[/tex][tex]$ z = -1.5 $[/tex][tex]$ Thus, the $[/tex]z[tex]$-score for an employee with a salary of \$[/tex]28,000 is [tex]\(-1.5\)[/tex].

Therefore, the correct answer is:
[tex]\[ -1.5 \][/tex]