Question

Suppose ACT Composite scores are normally distributed with a mean of 21.2 and a standard deviation of 5.2. A university plans to offer tutoring jobs to students whose scores are in the top 1%.What is the minimum score required for the job offer? Round your answer to 1 decimal place.

Answer 1

So the value of score that separates the bottom 99% of data from the top 1% is 33.316.

And rounded to 1 decimal place would be 33.3

Explanation:

1) 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".

2) Solution to the problem

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

`X \sim N(21.5,5.2)`

Where
`\mu=21.5` and
`\sigma=5.2`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

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

`P(X>a)=0.01` (a)

`P(X<a)=0.99` (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 0.99 of the area on the left and 0.01 of the area on the right it's z=2.33. On this case P(Z<2.33)=0.99 and P(Z>2.33)=0.01

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.99`

`P(Z<(a-\mu)/(\sigma))=0.99`

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

`Z=2.33<(a-21.2)/(5.2)`

And if we solve for a we got

`a=21.2 +2.33*5.2=33.316`

So the value of score that separates the bottom 99% of data from the top 1% is 33.316.

And rounded to 1 decimal place would be 33.3