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I'm aware that this is a very elementary problem but I can't seem to find an answer to this exact question and I'm pretty inexperienced with statistics in general.

I have a value which represents my "score" on a test and I want to figure out which percentile I'm in (IE: top 1.4%). I have the mean score for the test and the standard deviation from that score, assuming standard deviation how can I find my percentile?

Example:

I scored 181 on a test where the average score is 40 and the standard deviation is 16.7. Is it possible to find from these numbers that, for example, my score is in the top .005%? Or do I need more data points than just the mean?

Thanks!

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I scored 181 on a test where the average score is 40 and the standard deviation is 16.7. Is it possible to find from these numbers that, for example, my score is in the top .005%?

Without more information, no. If you assume a distribution then you can find the quantiles. Without such an assumption you can obtain a bound:

For example the one-sided equivalent of the Chebyshev bound may be written

$P(X-\mu\geq a) \leq \frac{\sigma^2}{\sigma^2+a^2}$

or equivalently

$P((X-\mu)/\sigma \geq k) \leq \frac{1}{1+k^2}$

So for your values, you must have no more than $ 1/(1+(141/16.7)^2)$ of the distribution greater than or equal to your score ... but that's 0.0138339, a fair bit larger than 0.005.

It's also possible that the distribution is such that none of the distribution is up that far (i.e. it could be as low as 0).

If you add additional information the bounds may come in somewhat, though often the additional information has less impact than you might expect. e.g. adding higher moments doesn't do as much as people tend to imagine they would, and even assuming symmetry doesn't halve the one-sided bound. Stronger assumptions (like symmetric, continuous unimodal; pretty strong assumptions that I wouldn't expect to be true) would bring it down below the value you mentioned I think.

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  • $\begingroup$ Apologies, typo in my original post, I mean to clarity that standard distribution should work fine in this case. Also, the .005% example was just a guess, not meant to be accurate sample data. I appreciate your answer, I'll read more into the bounds problem and try to come up with a script-friendly formula if one doesn't already exist! $\endgroup$ – Egrodo Jul 15 '18 at 18:02
  • $\begingroup$ What standard distribution are you talking about? Do you mean to assume normality? While I regard that as highly dubious - especially in the extreme tails - if we do assume it then that case the percentile of the value you quoted is extremely small; the z-score is $141/16.7\approx 8.44$, so the upper tail area is roughly $\frac{1}{8.44}\times\frac{1}{\sqrt{2\pi}}e^{-8.44^2/2}\approx 1.6\times 10^{-17}$ (this approximation works okay for large z-values). More generally you use tables or computer functions to do that calculation -- but extreme tail values depend heavily on the distribution $\endgroup$ – Glen_b Jul 16 '18 at 1:11
  • $\begingroup$ For example in most spreadsheets (including Excel) there are normal distribution functions you can use for that. But the bounds are "safer". $\endgroup$ – Glen_b Jul 16 '18 at 1:14

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