# I'm new to studying z-scores and I've been told that for a gaussian statistic, around 95% of the val

Callum Dudley 2022-07-04 Answered
I'm new to studying z-scores and I've been told that for a gaussian statistic, around 95% of the values lie within the area two standard deviations above and below the mean, which (in accordance to my interpretation) would imply,
${\int }_{\mu -2\sigma }^{\mu +2\sigma }A{e}^{-\left(\left(x-\mu \right)/\sigma {\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=0.95\ast {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}A{e}^{-\left(\left(x-\mu \right)/\sigma {\right)}^{2}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$
Firstly, am I correct in my presumption? and secondly, is there any way to calculate the integral on the left to prove this point mathematically?
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Caiden Barrett
Using a change of variables $x=\mu +\sigma t$, and inserting the missing 1/2 in the exponentials, the left side can be evaluated using the error function
$A{\int }_{-2}^{2}{e}^{-{t}^{2}/2}dt=\sqrt{2\pi }A\text{erf}\left(\sqrt{2}\right)$
while the right side is $\sqrt{2\pi }A$. In fact $\text{erf}\left(\sqrt{2}\right)\approx 0.9544997360$.