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?

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 =34,000andastandarddeviationSigma=4,000.

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

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

2

=$4,000

2

)

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

Z = \frac{X-\mu}{\sigma}

σ

X−μ

~ N(0,1)

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

\sigmaσ = 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}

4,000

28,000−34,000

= \frac{-6}{4}

4

−6

= -1.5

So, the required z-score is -1.5.