Trying to understand this is in a high school level.
I understand that the how $\frac {\Sigma|x-\bar x|}{n}$ calculates the mean of the distances of each score to the mean.
I use this idea to map out how standard deviation works, but how does the formula $\sqrt\frac {\Sigma(x-\bar x)^2}{n}$ actually map out the percentages in the normal distribution? ie. 68%, 95%, 99.7%??
Also is it possible to prove this?
On
The formula doesn't actually have anything specifically to do with the normal distribution (as opposed to any other distribution). It simply yields the standard deviation, a measure of how spread out the values are from their average value (their mean). If the values are distributed according to the normal distribution, then about $68$ percent of the values will be within $1$ standard deviation of the mean, about $95$ percent will be within $2$ standard deviations, and so on.
However, those percentages come not from the formula for standard deviation, but for the formula for the normal distribution:
$$ f_X(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$
If you were to calculate the area under this curve between $-1$ and $1$, you would get approximately $0.68$; between $-2$ and $2$, approximately $0.95$; and so on. If the distribution is not normal, then those proportions will be different. In many cases (as with a bimodal or two-peaked distribution), the standard deviation may even be a misleading measure of the distribution. The standard deviation is merely a tool for characterizing a distribution, and should be applied intelligently.
Suppose you have a Normal distribution $X \sim N(\mu ,\sigma^2)$ we can transform the random variable into a standard Normal distribution using $Z=\frac{X-\mu}{\sigma}$. This is done so that we have $Z \sim N(0,1)$ which is easier to work with.
The cumulative distribution function of the standard Normal distribution is given by: $$\Phi (z)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^ze^{-(1/2)t^2}dt$$
Given $z>0$ and using symmetry ($p(Z\le 0)=0.5$) this is the same as: $$\Phi (z)=\frac{1}{2}+\frac{1}{\sqrt{2 \pi}}\int_{0}^ze^{-(1/2)t^2}dt$$
As mentioned in the comments the integral isn't nice but one method would be to use integration by parts. This will result in an approximation but that may be all that's required. I will perform a few calculations to illustrate: $$\int_{0}^ze^{-(1/2)t^2}dt=\left[te^{-(1/2)t^2}\right]_0^z-\int_0^z -t^2e^{-(1/2)t^2}dt\\\int_{0}^ze^{-(1/2)t^2}dt=\left[te^{-(1/2)t^2}\right]_0^z+\int_0^z t^2e^{-(1/2)t^2}dt\\ \int_{0}^ze^{-(1/2)t^2}dt=ze^{-(1/2)z^2}+\int_0^z t^2e^{-(1/2)t^2}dt$$
Now keep repeating the integration by parts process as far as necessary. The next result is $$\int_{0}^ze^{-(1/2)t^2}dt=ze^{-(1/2)z^2}+\frac{z^3}{3}e^{-(1/2)t^2}+\int_0^z \frac{t^4}{3}e^{-(1/2)t^2}dt$$
Having done the integration a few times a pattern emerges which we may utilise $$\int_{0}^ze^{-(1/2)t^2}dt=ze^{-(1/2)z^2}+\frac{z^3}{3}e^{-(1/2)t^2}+\frac{z^5}{15}e^{-(1/2)t^2}+...$$
We may tidy this up by factoring$$\int_{0}^ze^{-(1/2)t^2}dt=e^{-(1/2)z^2}\left(z+\frac{z^3}{3}+\frac{z^5}{15}+...\right)$$
Putting this altogether we have a way to compute the cdf for $z>0$ namely$$\Phi (z)=\frac{1}{2}+\frac{1}{\sqrt{2 \pi}}e^{-(1/2)z^2}\left(z+\frac{z^3}{3}+\frac{z^5}{15}+...\right)$$
Finally if we want the probability that $z$ is within a standard deviation $1$ of the mean $(0)$ we compute $2(\Phi(1)-0.5)$ This leads to $$p(-1<Z<1)\approx 2\times 0.338759=0.677581$$
We have established the fact that around $\frac{2}{3}$ of the distribution lies within a standard deviation of the mean. The other results you mention could likewise be formulated.
Note I don't claim integrating by parts is the most efficient numerical approximation there will be better more sophisticated methods.
Hope this helps.