Question

How does central tendency relate to normal distribution?

Answer 1

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(x)=1/(sigma sqrt{2pi})e^(-(x-mu)^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}xf(x)dx=1/(sigma sqrt{2pi}) int_{-infty}^{infty}xe^(-(x-mu)^2/2)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!=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(x)dx=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'(x)=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.