Help on finding the math behind the 68/95/99 rule for Normal distribution I also see on the section on the Central Limit Theorem that the normal distribution is used and gives the graph of the 68/95/99 rule.  However, I haven't seen the actual math used to prove this or how it gets computed.  Can anyone give me a link or explain/show how the math behind the statement is done?
 A: Recall that the pdf for the normal distribution with standard deviation $1$ and mean $0$ is given by
$$f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$$
The way to find the probability that our random variable lies between $a$ and $b$ is to integrate from $a$ to $b$; in particular, the probability that $X$ lies within one standard deviation of the mean is
$$\int_{-1}^1 f(x) dx$$
This can be done numerically, as here, for an approximate value of $0.683$, or about $68\%$. The same process will give the $95$ and $99$ for the intervals $[-2, 2]$ and $[-3, 3]$.
A: If you don't want to just use numerical integration: for large $t$ the standard normal CDF has the asymptotic series (which can be established by integration by parts)
$$
\Phi(t) \approx 1 + \frac{\exp(-t^2/2)}{\sqrt{2\pi}} \left(- \frac{1}{t} +  \frac{1}{t^3} - \frac{3}{t^5} + \frac{3 \cdot 5}{t^7} - \frac{3 \cdot 5 \cdot 7}{t^9} + \ldots\right) $$
and thus
  $$P(-t \le X \le t) = 2 \Phi(t) - 1 = 1 + 2 \frac{\exp(-t^2/2)}{\sqrt{2\pi}} \left(- \frac{1}{t} +  \frac{1}{t^3} - \frac{3}{t^5} + \frac{3 \cdot 5}{t^7} - \frac{3 \cdot 5 \cdot 7}{t^9} + \ldots\right)$$
This is only an asymptotic series, which does not converge for any $t$.  But
if $t$ is large you can get good approximations using appropriate partial sums: the value lies between any two adjacent partial sums.
Thus $P(-2 \le X \le 2)$ lies between $1 + 2 \dfrac{\exp(-2)}{\sqrt{2\pi}} \left(-\dfrac{1}{2} + \dfrac{1}{8}\right) \approx 0.9595$ and 
$1 + 2 \dfrac{\exp(-2)}{\sqrt{2\pi}} \left(-\dfrac{1}{2} + \dfrac{1}{8} - \dfrac{3}{32}\right) \approx 0.9494$ (the actual value is $.9544997360$ to $10$ decimal places).  Similarly (taking $5$ or $6$ terms of the sum) $P(-3 \le X \le 3)$ lies between approximately $.99728$ and $.99733$ (the actual value is $.9973002039$).
A: When you have a standard normal distribution which has a mean of zero and a standard deviation of 1, which is denoted $N(0,1)$, the probability of an event happening between one standard deviation from the mean (which means both positive and negative) is $68$ percent, the probability of an event happening between two standard deviations is $95$ percent, and the probability of an event happening between three standard deviations from the mean is $99.7$ percent.
