Question

Vail Resorts pays part-time seasonal employees at ski resorts on an hourly basis. At a certain mountain, the hourly rates have a normal distribution with σ = $3.00. If 20 percent of all part-time seasonal employees make more than $13.16 an hour, what is the average hourly pay rate at this mountain? (Round your answer to 2 decimal places.) Average hourly pay rate $

Answer 1

`z=0.842<(13.16-\mu)/(3)`

And if we solve for a we got

`\mu=13.16 +0.842*3=15.69`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the hourly rates of a population, and for this case we know the distribution for X is given by:

`X \sim N(\mu,3)`

For this part we want to find a value a, such that we satisfy this condition:

`P(X>13.16)=0.20` (a)

`P(X<13.16)=0.80` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.20

If we use condition (b) from previous we have this:

`P(X<13.16)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.80`

`P(z<(13.16-\mu)/(\sigma))=0.80`

But we know which value of z satisfy the previous equation so then we can do this:

`z=0.842<(13.16-\mu)/(3)`

And if we solve for a we got

`\mu=13.16 +0.842*3=15.69`