Question

A company pays its employees an average wage of $14.90 an hour with a standard deviation of $1.50. If the wages are approximately normally distributed and paid to the nearest cent, the highest 7% of the employees hourly wages is greater than what amount? Report your code, as well as your final answer (2 decimals).

Answer 1

`z=1.478<(a-14.9)/(1.5)`

And if we solve for a we got

`a=14.9 +1.478*1.5=17.12`

So the value of height that separates the bottom 93% of data from the top 3% is 17.12.

Step-by-step 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 wages per hour of a population, and for this case we know the distribution for X is given by:

`X \sim N(14.90,1.50)`

Where
`\mu=14.90` and
`\sigma=1.50`

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

`P(X>a)=0.07` (a)

`P(X<a)=0.93` (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 0.93 of the area on the left and 0.07 of the area on the right it's z=1.478. On this case P(Z<1.478)=0.93 and P(z>1.478)=0.07

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.93`

`P(z<(a-\mu)/(\sigma))=0.93`

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

`z=1.478<(a-14.9)/(1.5)`

And if we solve for a we got

`a=14.9 +1.478*1.5=17.12`

So the value of height that separates the bottom 93% of data from the top 3% is 17.12.