Finding Gaussian integrals with non-infinite limits to justify z-scores

Question

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, $$\int_{\mu-2\sigma}^{\mu+2\sigma}Ae^{-((x-\mu)/\sigma)^2}\,\mathrm{d}x=0.95*\int_{-\infty}^{+\infty}Ae^{-((x-\mu)/\sigma)^2}\,\mathrm{d}x$$ 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?

Answer 1

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 $$ A \int_{-2}^2 e^{-t^2/2} dt = \sqrt{2\pi} A \text{erf}(\sqrt{2})$$ while the right side is $\sqrt{2\pi} A$. In fact $\text{erf}(\sqrt{2}) \approx 0.9544997360$.