How do you find the standard deviation for the population 9, 4, 6, 5, 4, 5, 5, 10?

Mar 5, 2017

$T h e S . D . \approx 2.12 \left(2 \mathrm{dp}\right) ,$

Explanation:

We have, the following $n = 8$ observations ${x}_{i} , 1 \le i \le 8 :$

${x}_{i} : 9 , 4 , 6 , 5 , 4 , 5 , 5 , 10.$

$\therefore {\sum}_{i = 1}^{i = 8} {x}_{i} = 9 + 4 + 6 + 5 + 4 + 5 + 5 + 10 = 48.$

$\Rightarrow \text{the Mean } \overline{x} = \frac{\sum {x}_{i}}{n} = \frac{48}{8} = 6.$

Recall that, the Standard Deviation $\sigma$ is given by,

$\sigma = \sqrt{{\sum}_{i = 1}^{i = 8} {\left({x}_{i} - \overline{x}\right)}^{2} / n} ,$ we proceed as below :-

$\left({x}_{i} - \overline{x}\right) = \left({x}_{i} - 6\right) : 3 , - 2 , 0 , - 1 , - 2 , - 1 , - 1 , 4.$

${\left({x}_{i} - \overline{x}\right)}^{2} : 9 , 4 , 0 , 1 , 4 , 1 , 1 , 16.$

$\therefore \sum {\left({x}_{i} - \overline{x}\right)}^{2} = 36.$

$\Rightarrow \sigma = \sqrt{\frac{36}{8}} = \sqrt{\frac{9}{2}} = \sqrt{4.5} \approx 2.12 \left(2 \mathrm{dp}\right) .$