Question

The annual salaries of all employees at a financial company are normally distributed with a mean Mu = $34,000 and a standard deviation Sigma = $4,000. What is the z-score of a company employee who makes an annual salary of $28,000?


1.5

-3.5

-1.5

-5

Answer 1

The z-score of a company employee who makes an annual salary of $28,000 is -1.5.

Step-by-step explanation:

We are given that the annual salaries of all employees at a financial company are normally distributed with a mean Mu = $34,000 and a standard deviation Sigma = $4,000.

Let X = annual salaries of all employees at a financial company

SO, X ~ Normal($\mu=\ 34,000 , \sigma^{2} = \ 4,000^{2}$)

The z score probability distribution for the normal distribution is given by;

                            Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean annual salary of emloyees = $34,000

            $\sigma$ = standard deviation = $4,000

Now, the z-score of a company employee who makes an annual salary of $28,000 is given by ;

                       Z  =  $\frac{28,000-34,000}{4,000}$

                           =  $\frac{-6}{4}$  =  -1.5

So, the required z-score is -1.5.

Answer 2

Answer:

The answer is C on edge 2021

Step-by-step explanation: