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I recently came across the following formula referenced in this particular answer here: https://math.stackexchange.com/a/493434/106050.

This is pretty much perfect to what I would like to use in a statistical library I'm working on, but I was wondering if there is a way to simplify this such that f(x) would yield the decimal percentage of the population covered by x standard deviations within a normal distribution; i.e.

f(1) = .6827
f(2) = .9545
f(3) = .9973, etc...

I'd like this to work for decimal standard deviations as well... what's the simplest equation that could satisfy this? I'm hoping it won't require integral calculation of any kind...

Eliseo D'Annunzio
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1 Answers1

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Unfortunately, it does require integration.

A normal distribution can be modeled as

$$\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}.$$

So, the probability that a point lies in the range $(-t\sigma+\mu, t\sigma+\mu)$ is

$$\int_{-t\sigma+\mu}^{t\sigma+\mu} \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx = \int_{-t}^t \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} dy.$$

This can be expressed in terms of the error function, but cannot be written in terms of elementary functions.

Carl Schildkraut
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  • From what I recall in integration the 1/sqrt(2pi) can be moved out, leaving the exponent term... but from memory, doesn't the rest of the integral yield, in turn, 2 * erf(t/sqrt(2))? I'm guessing that can't be simplified, given what I've just read in the Wikipedia entry you've supplied... – Eliseo D'Annunzio Apr 10 '18 at 02:21
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    It's possible to prove formally that $\operatorname{erf}$ doesn't have a closed-form representation. – Connor Harris Apr 10 '18 at 02:24