Question

Show that the two formulae for the standard deviation of ungrouped data

$\sigma =\sqrt{\frac{1}{n}\underset{}{\sum {\left(x_i-\overline{X}\right)}^2}}$ and $\sigma \text{'}=\sqrt{\frac{1}{n}\underset{}{\sum x_i^2-{\overline{X}}^2}}$ are equivalent, where $\overline{X}=\frac{1}{n}\sum _{}{x}_i$

Answer 1

$$\begin{gathered} \sigma =\sqrt{\frac{1}{n}\underset{}{\sum {\left(x_i-\overline{X}\right)}^2}} \\ =\sqrt{\frac{1}{n}\underset{}{\sum \left(x_i^2-2x_i\overline{X}+{\overline{X}}^2\right)}} \\ =\sqrt{\frac{1}{n}\underset{}{\sum x_i^2}-{\displaystyle \frac{1}{n}}\sum 2x_i\overline{X}+{\displaystyle \frac{1}{n}}\sum {\overline{X}}^2} \\ =\sqrt{\frac{1}{n}\underset{}{\sum x_i^2}-{\displaystyle \frac{1}{n}}\times 2\overline{X}\sum x_i+{\displaystyle \frac{1}{n}}\times {\overline{X}}^2\sum 1} \\ =\sqrt{\frac{1}{n}\underset{}{\sum x_i^2}-{\displaystyle \frac{1}{n}}\times 2\overline{X}\times n\overline{X}+{\displaystyle \frac{1}{n}}\times {\overline{X}}^2\times n}\left(\overline{X}=\frac{1}{n}\underset{}{\sum x_i}\right) \end{gathered}$$
$$\begin{gathered} =\sqrt{\frac{1}{n}\underset{}{\sum x_i^2-2{\overline{X}}^2+}{\overline{X}}^2} \\ =\sqrt{\frac{1}{n}\underset{}{\sum x_i^2-{\overline{X}}^2}} \\ =\sigma \text{'} \end{gathered}$$

Hence, the formulae $\sigma =\sqrt{\frac{1}{n}\underset{}{\sum {\left(x_i-\overline{X}\right)}^2}}$ and $\sigma \text{'}=\sqrt{\frac{1}{n}\underset{}{\sum x_i^2-{\overline{X}}^2}}$ are equivalent, where $\overline{X}=\frac{1}{n}\sum _{}{x}_i$.