Question

Find the mean, variance and the standard deviation for the following data:

5, 9, 8, 12, 6, 10, 6, 8

Answer 1

Here n = 8.

Mean $\overline{x}=\frac{1}{8}(5+9+8+12+6+10+6+8)=\frac{64}{8}=8$

The values of $(x_i-\overline{x})$ are

-3, 1, 0, 4, -2, 2, -2, 0.

$\therefore (x_i-\overline{x})=(9+1+0+16+4+4+4+0)=38$

$\therefore$ variance,${\sigma }^2=\frac{\sum (x_i-\overline{x}{)}^2}{n}=\frac{38}{8}=\frac{19}{4}=4.75$

and, standard deviation, $\sigma =\sqrt{4.75}=2.17$

$\therefore$ mean = 8, variance = 4.75 and standard deviation = 2.17.

Short Cut Method

For simple ungrouped data, we have

${\sigma }^2=\frac{1}{n}.\left\{{\sum }_{i=1}^n\right(x_i-\overline{x}{)}^2\}$

$=\frac{1}{n}.\left\{{\sum }_{i=1}^n\right(x_i^2-2x_i\overline{x}+{\overline{x}}^2\left)\right\}$

$=\frac{\sum x_i^2}{n}-2\overline{x}\left(\frac{\sum x_i}{n}\right)+\frac{n{\overline{x}}^2}{n}$

$=\frac{\sum x_i^2}{n}-2\overline{x}.\overline{x}+{\overline{x}}^2=\frac{\sum x_i^2}{n}-{\overline{x}}^2$