# What is the standard deviation of the following numbers: 24, 36, 33, 21, 15, 11?

Dec 9, 2017

$\sigma = \frac{19 \sqrt{2}}{3}$

#### Explanation:

We know:

${\sigma}^{2} = \frac{\Sigma {x}^{2}}{n} - {\left(\frac{\Sigma x}{n}\right)}^{2}$

$\Sigma x = 24 + 36 + 33 + 21 + 15 + 11 = 140$

$\Sigma {x}^{2} = {24}^{2} + {36}^{2} + {33}^{2} + {21}^{2} + {15}^{2} + {11}^{2} = 3748$

$n$ is the number data in the sample, $\implies n = 6$

Hence plugging into the formula:

${\sigma}^{2} = \frac{3748}{6} - {\left(\frac{140}{6}\right)}^{2} = \frac{722}{9}$

$\implies \sigma = \frac{\sqrt{722}}{\sqrt{9}} = \frac{19 \sqrt{2}}{3}$

Dec 9, 2017

Standard Deviation $\sigma = \textcolor{p u r p \le}{8.9567}$

#### Explanation:

Mean for the given numbers
$= \frac{24 + 36 + 33 + 21 + 15 + 11}{6} = 23.33$

Variance ${\sigma}^{2} = \sum {\left(x - \overline{x}\right)}^{2} / \left(n\right)$

${\sigma}^{2} = \sum \frac{{\left(23.33 - 24\right)}^{2} + {\left(23.33 - 36\right)}^{2} + {\left(23.33 - 33\right)}^{2} + {\left(23.33 - 21\right)}^{2} + {\left(23.33 - 15\right)}^{2} + {\left(23.33 - 11\right)}^{2}}{6}$

${\sigma}^{2} = \frac{{\left(- 0.67\right)}^{2} + {\left(- 12.67\right)}^{2} + {\left(- 9.67\right)}^{2} + {\left(2.33\right)}^{2} + {\left(8.33\right)}^{2} + {\left(12.33\right)}^{2}}{6}$

${\sigma}^{2} = 80.2222$

Standard Deviation $\sigma = 8.9567$