Suppose ACT Science scores are normally distributed with a mean of 20.8 and a standard deviation of 5.2. A university plans to send letters of recognition to students whose scores are in the top 12%. What is the minimum score required for a letter of recognition?

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 scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(20.8,5.2)[/tex]

Where [tex]\mu=20.8[/tex] and [tex]\sigma=5.2[/tex]

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

[tex]P(X>a)=0.12[/tex] (a)

[tex]P(X<a)=0.88[/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 0.8 of the area on the left and 0.12 of the area on the right it's z=1.175. On this case P(Z<1.175)=0.88 and P(z>1.175)=0.12

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

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.8[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.88[/tex]

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

[tex]z=1.175<\frac{a-20.8}{5.2}[/tex]

And if we solve for a we got

[tex]a=20.8 +1.175*5.2=26.91[/tex]

So the value of height that separates the bottom 88% of data from the top 12% is 26.91.