Demonstration - normal distribution I came across a figure saying that the distance from mean-sd to mean+sd in a normal distribution has 68.2% of the elements.
How can we demonstrate that mathematically?


If we integrate the probability density function of the normal distribution from $\mu-\sigma$ to $\mu+\sigma$, we end up with about $0.683$:
$$ \int_{\mu-\sigma}^{\mu+\sigma} \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}}\,\mathrm dx \approx 0.683. $$
There is no closed form for this integral, but you can integrate it numerically with error bound guarantees, e.g., using rectangular approximations.
If you want to, you can use your favorite statistical software to evaluate this numerically for fixed $\mu$ and $\sigma$. For instance, here is R for $\mu=1$ and $\sigma=1$:
> pnorm(1,0,1)-pnorm(-1,0,1)
[1] 0.6826895

