# Is there a rule that can be used to easily approximate the pdf(x) for normal distribution?

Given the Normal Distribution with mean Mu and variance Sigma. With the respect to the rule of 3 Sigma, can one use similar estimations for the value of probability density function within 1, 2, ... Sigmas away from the mean?

• also, a word of warning, mean is often denoted by mu ($\mu$) and standard deviation is often denoted by sigma ($\sigma$). It is more often that variance is denoted by sigma squared ($\sigma^2$). Careful not to confuse standard deviation and variance as it can mess up your calculations if you use the wrong one. Commented Dec 14, 2014 at 20:24

The probability distribution function (pdf) of a normal distribution with mean $\mu$ and standard deviation $\sigma$ is defined to be:

$$f(x) = \dfrac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$

For $x= \mu\pm\sigma$ this simplifies to $f(\mu\pm\sigma) = \dfrac{1}{\sigma\sqrt{2\pi e}}$

For $x= \mu\pm 2\sigma$ this simplifies to $f(\mu\pm 2\sigma) = \dfrac{1}{\sigma e^2\sqrt{2\pi}}$

For $x= \mu\pm 3\sigma$ this simplifies to $f(\mu\pm 3\sigma) = \dfrac{1}{\sigma e^4\sqrt{2\pi e}}$

As you can see, it doesn't simplify completely and will depend on what $\sigma$ is (unlike the cdf such as in the "rule of 3").

For the standard normal distribution where $\mu = 0$ and $\sigma = 1$ it can simplify a bit further.

$N(0) = \dfrac{1}{\sqrt{2\pi}}\approx 0.399$

$N(\pm 1) = \dfrac{1}{\sqrt{2\pi e}} \approx 0.24$

$N(\pm 2) = \dfrac{1}{e^2\sqrt{2\pi}} \approx 0.054$

$N(\pm 3) = \dfrac{1}{e^4\sqrt{2\pi e}} \approx 0.0044$

Personally, I don't find these numbers helpful or useful to memorize, so I'm not sure what you would want to do with them except use them to draw a more accurate normal curve. Any math using normal distributions generally uses calculus and integrals to find area under the curve or tables which already computed those areas for us. Specific values of the function are less relevant.