Density of the normal distribution

Definition

Let $\mu \in \mathbb{R}$ and $\sigma^2 > 0$. The density of the normal distribution with mean $\mu$ and variance $\sigma^2$ is the function $\varphi_{\mu, \sigma^2} \colon \mathbb{R} \to (0, \infty)$ with $$\varphi_{\mu, \sigma^2}(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x - \mu)^2}{2\sigma^2}\right).$$ The standard normal density is $\varphi(x) = \varphi_{0, 1}(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$.

Remarks

The integral of $\varphi_{\mu, \sigma^2}$ over $\mathbb{R}$ is $1$. This follows from the Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$. So $\varphi_{\mu, \sigma^2}$ is a probability density function, the density of the normal distribution $\mathcal{N}(\mu, \sigma^2)$. The function is symmetric about $\mu$, has its maximum $1/\sqrt{2\pi\sigma^2}$ at $x = \mu$, and has inflection points at $\mu \pm \sigma$. It satisfies $\varphi_{\mu, \sigma^2}(x) = \frac{1}{\sigma} \varphi\left(\frac{x - \mu}{\sigma}\right)$ with $\sigma = \sqrt{\sigma^2}$.

Explanations

L1 human

This is the formula of the bell curve. The exponent $-(x - \mu)^2 / (2\sigma^2)$ makes the curve fall very fast away from $\mu$, and the constant in front makes the total area equal to $1$.