Question

The distribution of grade point averages​ (GPAs) for medical school applicants of a certain year were approximately​ Normal, with a mean of 3.48 and a standard deviation of 0.36. Suppose a medical school will only consider candidates with GPAs in the top 20​% of the applicant pool. An applicant has a GPA of 3.75. Does this GPA fall in the top 20​% of the applicant​ pool?

Answer 1

`a=3.48 +0.842*0.36=3.783`

So the score that separates the bottom 80% of data from the top 20% is 3.783. Since the value obtained by the applicant is 3.75 <3.783 not falls on the top 20%.

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 grade point averages​ (GPAs) for medical school applicants of a certain year of a population, and for this case we know the distribution for X is given by:

`X \sim N(3.48,0.36)`

Where
`\mu=3.48` and
`\sigma=0.36`

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

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

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

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

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

Both conditions are equivalent on this case. We can use the z score 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. We can found this with the following excel code:"=NORM.INV(0.8,0,1)". On this case P(Z<0.842)=0.80 and P(Z>0.842)=0.2

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

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

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

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

`z=0.842<(a-3.48)/(0.36)`

And if we solve for a we got

`a=3.48 +0.842*0.36=3.783`

So the score that separates the bottom 80% of data from the top 20% is 3.783. Since the value obtained by the applicant is 3.75 <3.783 not falls on the top 20%.