Normal (Gaussian) distribution

The story

Add together lots of small, independent nudges, and the total tends to follow a bell curve whatever the nudges looked like individually. That is the Central Limit Theorem, and it is why the Normal turns up in measurement errors, heights and test scores, and underlies so much of classical statistics.

The curve is centred on $\mu$ , which is its mean, median and mode all at once, and $\sigma$ sets its width. A handy rule: about 68% of the probability lies within $1\sigma$ of the mean, about 95% within $2\sigma$ ( $1.96\sigma$ , to be precise), and about 99.7% within $3\sigma$ .

Formulas

Support

x \in \mathbb{R}

PDF

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

CDF

F(x) = \tfrac{1}{2}\left[1 + \operatorname{erf}\!\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right]

Expectation

\mathbb{E}[X] = \mu

Variance

\operatorname{Var}(X) = \sigma^2

Mode

\mu

Parameters

Parameter	What it controls
$\mu$	The mean, and the centre of symmetry. Changing it slides the bell left or right without changing its shape.
$\sigma$	The standard deviation. Larger $\sigma$ makes the bell wider and lower; the variance is $\sigma^2$ .

Where it shows up

Adult heights within a population.
Random measurement error in a laboratory instrument.
Standardised test scores, which are often scaled to be roughly Normal.
The noise term in linear regression, which is classically assumed to be Normal.

Try this

Shade the interval from $\mu - \sigma$ to $\mu + \sigma$ and read off about 0.683. Change $\mu$ and $\sigma$ and re-shade: the answer never changes, because every Normal is just a shifted and stretched copy of the standard one.
Set $\mu = 0$ , $\sigma = 1$ and shade from $-1.96$ to $1.96$ . The probability is 0.95, the source of the 1.96 in every 95% confidence interval.
Switch to Sample means and move $n$ from 1 to 25. The histogram stays Normal but narrows by a factor of $\sqrt{n}$ : at $n = 25$ its spread is $\sigma/5$ .

Related distributions

If $Z$ is standard Normal, then $Z^2$ is Chi-squared with 1 degree of freedom, and a sum of $k$ squared standard Normals is Chi-squared with $k$ .
The Student t tends to the Normal as $\nu \to \infty$ .
If $X$ is $\text{Normal}(\mu, \sigma^2)$ , then $e^X$ is Log-normal with parameters $\mu$ and $\sigma$ .
The Binomial and the Poisson are both well approximated by a Normal when $n$ or $\lambda$ is large.