Mean and Variance

Basics

Given $n$ values of $x$

$$\begin{array}{rl}& X=\{x_1,x_2,...x_n\}\end{array}$$

Mean of $x$:

$$\begin{array}{rl}& {\mu }_x=\frac{1}{n}\sum _{i=1}^n{x}_i\end{array}$$

Variance of $x$:

$$\begin{array}{rl}& {\sigma }_x{}^2=\frac{1}{n}[\sum _{i=1}^n(x_i-{\mu }_x{)}^2]\end{array}$$

Standard Deviation of $x$:

$$\begin{array}{rl}& {\sigma }_x=\sqrt{\frac{1}{n}[\sum _{i=1}^n(x_i-{\mu }_x{)}^2]}\end{array}$$

Mean centring

Example

Given this set of values:

$$\begin{array}{rl}Z& =\{1,5,12\}\\ \\ {\mu }_z& =6\\ \\ {\sigma }_z{}^2& =\frac{1}{3}\cdot (25+1+36)\\ \\ & =20.667\end{array}$$

Now subtract the mean from each value:

$$\begin{array}{r}\\ Z^{\prime }& =\{1-6,5-6,12-6\}\\ & =\{-5,1,6\}\\ \\ {\mu }_z^{\prime }& =0\\ \\ {\sigma }_z{}^2& =\frac{1}{3}\cdot (25+1+36)\\ \\ & =20.667\end{array}$$

So mean-centring shifts the data horizontally along $\text{R}$ but doesn't change the variance

Based on course material by Paul Bendich and Daniel Egger from Data Science Maths Skills