I have a distribution that looks like this: U

In other words small and big values are more frequent than middle values. A better graphical example of the distribution is this: Here

So I have a couple of questions:

  1. Is there a function that can transform this distribution to a normal distribution?

I want to normalise my data because I want to calculate z-scores: $$z = (x - mean)/std$$ which assumes a normal distribution.

My other question is:

  1. Is there a variation of the z-scores calculation that do not assume a normal distribution of the data, or better that is aimed to the arcsine distribution?

1 Answer 1


If $F$ is the arcsine cdf and $\Phi$ is the Normal $\mathcal{N}(0,1)$ cdf, when $X\sim F(x)$, then $$\Phi^{-1}(F(X))\sim \mathcal{N}(0,1)$$

  • 1
    $\begingroup$ Would you mind explaining me in more detail how could I compute this transformation? $\endgroup$
    – alejandro
    Feb 18, 2016 at 18:22
  • 2
    $\begingroup$ @AlejandroJimenez-Sanchez All the details are in the answer already - you take your random variable, and transform it by its own cdf (in this case, the cdf for a Beta$(\frac12,\frac12)$, yielding a uniform result. You then transform again, applying the quantile function (inverse cdf) of the desired distribution (in this case by the standard normal quantile function /inverse of the normal cdf, producing a variable with a standard normal distribution). ctd... $\endgroup$
    – Glen_b
    Feb 19, 2016 at 1:36
  • 1
    $\begingroup$ ctd... e.g. The same approach is used in the opening paragraph of this answer to convert a chi-squared(1) variate to a normal, and in the opening paragraph of this answer to convert a Rayleigh-distributed random variate with known parameter to normal. $\endgroup$
    – Glen_b
    Feb 19, 2016 at 1:40

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