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Suppose that Jason recently landed job offers at two companies. Company A reports an average salary of $51,500 with a standard deviation of $2,175. Company B reports an average salary of $46,820 with a standard deviation of $5,920. Assume that salaries at each company are normally distributed. Jason's goal is to secure a position that pays $55,000 per year. What are the ???? ‑scores for Jason's desired salary at Company A and Company B? Please round your answers to two decimal places.

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Answer:

company B

Step-by-step explanation:

average salary of Company A(μ) = $51,500

standard deviation of Company A (σ)=  $2,175.

average salary of Company B(μ) = $46,820

standard deviation of Company B(σ) =$5,920

desired salary(x) = $55,000

z-score for company A = [tex]\dfrac{x-\mu}{\sigma}[/tex]

                                    = [tex]\dfrac{55000-51500}{ 2175} = 1.61[/tex]

z-score for company A = [tex]\dfrac{x-\mu}{\sigma}[/tex]

                                    =  [tex]\dfrac{55000-46820}{ 5920} = 1.38[/tex]

higher the value of z less chances of getting the desired salary hence company B has value of z is less so, the chances of getting desired salary is more in company B.

z scores of some value of some normal distribution is that value in standard normal distribution transformation of the specified normal distribution.

The z scores for Jason's decided salary at Company A and B are:

  • At Company A, z = 1.61 approximately.
  • At Company B, z = 1.38 approximately.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex]  and standard deviation [tex]\sigma[/tex])

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

So if X = x, the z score corresponding to this value is

[tex]z = \dfrac{x - \mu}{\sigma}[/tex]

For the given cases, let we have:

X = random variable tracking salary amount of employees in company A
Then, by the given information, we get

[tex]X \sim N(51,500, 2,175)[/tex]

and

Y = random variable tracking salary amount of employees in company B

[tex]Y \sim N(46,820, 5,920)[/tex]

Jason's goal is to get $55,000 / year salary.

So, its z score would be

  • For company A

[tex]Z = \dfrac{X - 51500}{2175}\\\\\\\rm At X = 55000\\\\ z = \dfrac{55000 - 51500}{2175} \approx 1.61[/tex]

  • For company B

[tex]Z = \dfrac{Y - 46820}{5920}\\\\\\\rm At X = 55000\\\\ z = \dfrac{55000 - 46820}{5920} \approx 1.38[/tex]

Thus,

The z scores for Jason's decided salary at Company A and B are:

  • At Company A, z = 1.61 approximately.
  • At Company B, z = 1.38 approximately.

Learn more about standard normal distribution here:

https://brainly.com/question/10984889

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