# How does central tendency relate to normal distribution?

Oct 5, 2017

Any normal distribution has a graph that is perfectly symmetric about a vertical line through its peak. Therefore, all measures of central tendency (most commonly, the mean, median, and mode) give the same answer: the $x$-value of the peak.

#### Explanation:

A normal distribution with mean $\mu$ and standard deviation $\sigma$ has formula $f \left(x\right) = \frac{1}{\sigma \sqrt{2 \pi}} {e}^{- {\left(x - \mu\right)}^{2} / 2}$.

The graph of this function is the classical "bell-shaped curve". The standard normal distribution with $\mu = 0$ and $\sigma = 1$ is shown below.

graph{(1/sqrt(2pi))*e^(-x^2/2) [-2.5, 2.5, -1.25, 1.25]}

The mean of a general normal distribution is equal to the improper integral

${\int}_{- \infty}^{\infty} x f \left(x\right) \mathrm{dx} = \frac{1}{\sigma \sqrt{2 \pi}} {\int}_{- \infty}^{\infty} x {e}^{- {\left(x - \mu\right)}^{2} / 2} \mathrm{dx}$.

This integral is not too hard to do when $\mu = 0$ (use a substitution $u = - {x}^{2} / 2$ in that case). It can't be done with elementary functions when $\mu \ne 0$, but other techniques (moment generating functions ) can still be used to show it equals $\mu$ in that case (in other words, the mean is the mean!! lol!!...see above for the name of $\mu$).

The median is the value of $M$ such that ${\int}_{- \infty}^{M} f \left(x\right) \mathrm{dx} = \frac{1}{2}$. The convergence of the integral and the symmetry of the graph about the vertical line at $x = \mu$ can be used to say $M = \mu$ as well.

The mode is the $x$-value of the global maximum of this graph. Setting $f ' \left(x\right) = 0$ and solving for $x$ implies this global maximum occurs at $x = \mu$ (you should check this). The mode is therefore $x = \mu$ as well.

All the measures of central tendency therefore give the same answer: the $x$-value of the peak of the graph.