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Z-score/percentile rank question

  1. Jul 6, 2011 #1
    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. jcsd
  3. Jul 7, 2011 #2

    chiro

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    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).
     
  4. Jul 7, 2011 #3

    uart

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    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

    [tex]Z_3 = \frac{0.54 Z_1 + 0.46 Z_2 }{0.7094}[/tex]

    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.
     
  5. Jul 7, 2011 #4

    chiro

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    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.

    I'd be interested in what the OP has to say about this issue.
     
  6. Jul 9, 2011 #5
    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!
     
  7. Aug 25, 2011 #6
    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)????
     
  8. Aug 25, 2011 #7

    uart

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    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.
     
  9. Aug 31, 2011 #8
    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!
     
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