We often get reference tables to match z-scores with equivalent percentiles.
Can anyone help me calculate percentiles from z-scores? What's the formula?
Thank you.
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$\begingroup$ Did you see this page? $\endgroup$– LE RogersonNov 17, 2016 at 12:06
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$\begingroup$ The reason why you get given tables instead of a formula is that there isn't a closed-form formula. The usual way to calculate it using computers is via calling functions that use one of several kinds of approximations, but these tend to be optimized for fast, accurate calculation on a computer, and are not especially suitable for working on a calculator or by hand. $\endgroup$– Glen_bNov 17, 2016 at 23:59
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$\begingroup$ Thank you that is very helpful, and the links too. I forgot to mention that I can do this in stats software, but a computer-programmer (using Java) had asked me how to implement it using a formula. I didn't know if there were equivalent functions in Java and I didn't know how our stats software did it. The best I could offer him was using an approximation of the curve (using exponentiated polynomial), which I assume can be used as a very simple approximation? $\endgroup$– bobmcpopNov 18, 2016 at 14:04
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$\begingroup$ I think "exact" duplicate is off, but the desired answer is the same. The primary answer on that page has a lot of useful information for this question. I was a bit surprised that the state of the art was based on a paper from 1969 because I have seen unexpected changes to these values in the last ten years. $\endgroup$– TavrockNov 30, 2016 at 21:32
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1 Answer
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The following formula can be found in the Handbook of Mathematical Functions
$$\hat{Z}\left ( x \right )=\left ( b_1t+b_2t^2+b_3t^3+b_4t^4+b_5t^5 \right )+\varepsilon \left ( x \right )$$
Where $t={1}/{\left (1+px \right)}$, $\left | \varepsilon (x) \right |< 7.5\times 10^{-8}$, $p=.2316419$, $b_1=.319381530$, $b_2=-.356563782$, $b_3=1.781477937$, $b_4=-1.821255978$, and $b_5=1.330274429$.
Most modern software that allows the user to calculate the percentile from the $Z_{score}$ provide more accurate results than this old approximation.
