# Z-score/percentile rank question

1. Jul 6, 2011

### ACHQ

SO the entrance to a college is based off of two factors: 54% grade point average, and 46% supplemental application. Both factors are standardized by the college using a z-score.

If a student is ranked in the only ~55 percentile for his grade point average, what percentile rank (approximately) would be needed on their supplemental application in order to be among the ~85 percentile of applicants?

(This is not homework and relates to an actual application)

Thanks

2. Jul 7, 2011

### chiro

I'm going to base my calculations off the assumption that your percentile is calculated using the formula:

Final Percentile = 100 x [ (Percentile Probability of GPA From Normal Distribution) x 0.54
+ (Percentile of Supplemental Application) x 0.46 ]
= 100 x [0.54 x P1 + 0.46 x P2] %

We are given in your example P1 = 0.55, and you want Final Percentile (FP) = 85%.

Using some algebra we get:

85 = 100 x [0.54 x 0.55 + 0.46 x P2] (P2 is supp application percentile probabilty)

Arranging terms and solving for P2 we get

P2 = 1.202173913

This means with 55 percentile for GPA you can not possibly get 85% percentile.

If you want to know the maximum percentile let P2 = 1 (100% percentile) and you find that your maximum overall percentile is 0.46 + 0.55 x 0.54 = 0.757 or roughly 76% percentile (rounded up)

In the above I'm assuming 99 percentile is top 1% and that the percentiles are weighted like you said above (54% GPA, the rest application).

3. Jul 7, 2011

### uart

I'm reading the question a little different chiro. I'm thinking that if they normalize both the GPA and the Supp to Z scores and then form the weighted average of those two Z scores, then the result will no longer be a Z score and will have to get normalized again - which changes the outcome somewhat.

Here's my thinking on this.

Start with Z score, N(0,1), and times by 0.54. The resulting variance is 0.54^2 x 1^2. And similar for the other one.

Let X3 = 0.54 Z1 + 0.46 Z2. This the sum of normal RV's variance 0.54^2 and 0.46^2 therefore variance of X3 is 0.54^2 + 0.46^2 = 0.5032. (stdev = 0.7094).

So X3 is not as yet a Z score since the variance in not unity. So we need to use Z3 = X3/sd3 which is normal.

Putting it all together, for the percentiles the OP states when can look up the normal cdf and find we have Z1=0.15, Z2=? and Z3=1.04

$$Z_3 = \frac{0.54 Z_1 + 0.46 Z_2 }{0.7094}$$

Putting in the given data and solving for Z2 I get Z2=1.43, corresponding to about the 92nd to 93rd percentile.

I think this is correct but to be honest I'm not 100% sure.

4. Jul 7, 2011

### chiro

Yeah I think you're right uart. The assumption that the weights are purely linear is probably not the best way to describe the problem.

5. Jul 9, 2011

### ACHQ

The linear assumption isn't really correct. The reason being is that they normalize both the GPA and the Supp to Z scores and then take the top ~15% of the "weighted average score". uart seems to be on the right track, I myself couldn't figure out the math behind it, but that number 92-93rd percentile seems to make sense.

I know for a fact that individuals that score in the 55-65 percentile for the GPA component do get into that college, based off of their supplemental application. I just wanted to put a number estimate on how stellar of a supplemental application one would need if they fell into that GPA range.

Thanks!

6. Aug 25, 2011

### ACHQ

So if the college's admission criteria were: 27% GPA, 27% Standardized test, 46% Supplementary application (again all have been Z-scored individually), would the same formula still apply, except instead of the variance being 0.54^2 + 0.46^2 = 0.5032 (stdev = 0.7094). It would be 0.27^2 + 0.27^2 +0.46^2 = 0.3974 (stdev = 0.5978)????

7. Aug 25, 2011

### uart

Yes, that would be correct. The weighted sum of the Z-scores would not itself be a z-score but could be considered a random variable with zero mean and stdev of approx 0.5978 as you calculated. So dividing it by this stdev will turn it back into a z-score.

8. Aug 31, 2011

### ACHQ

Ok I get everything except for one thing...

Why is the variance of each sum equal to : 0.27^2, 0.27^2, 0.46^2 respectively (i.e. why is the variance equal to the %weight squared?)

Thanks!