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).

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:

[tex]X \sim N(14.90,1.50)[/tex]

Where [tex]\mu=14.90[/tex] and [tex]\sigma=1.50[/tex]

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

[tex]P(X>a)=0.07[/tex] (a)

[tex]P(X<a)=0.93[/tex] (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:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.93[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.93[/tex]

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

[tex]z=1.478<\frac{a-14.9}{1.5}[/tex]

And if we solve for a we got

[tex]a=14.9 +1.478*1.5=17.12[/tex]

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