What is the standard deviation for {1,3,4,6,8}?

Nov 17, 2016

$s = \setminus \sqrt{5.84} \setminus \approx 2.42$

Explanation:

First, we evaluate the mean:
$\setminus \overline{x} = \setminus \frac{{x}_{1} + {x}_{2} + \setminus \ldots + {x}_{N}}{N} = \setminus \frac{1 + 3 + 4 + 6 + 8}{5} = 4.4$
Now, it's often easier to find the value of the variance, then the standard deviation is it's square root:
${s}^{2} = \setminus \frac{{\left({x}_{1} - \setminus \overline{x}\right)}^{2} + {\left({x}_{2} - \setminus \overline{x}\right)}^{2} + \setminus \ldots + {\left({x}_{N} - \setminus \overline{x}\right)}^{2}}{N} =$
$= \setminus \frac{{\left(1 - 4.4\right)}^{2} + {\left(3 - 4.4\right)}^{2} + {\left(4 - 4.4\right)}^{2} + {\left(6 - 4.4\right)}^{2} + {\left(8 - 4.4\right)}^{2}}{5} = 5.84$
$s = \setminus \sqrt{{s}^{2}} = \setminus \sqrt{5.84} \setminus \approx 2.42$

Note:
There'a also a simpler formula for the variance:
${s}^{2} = \setminus \frac{{x}_{1}^{2} + {x}_{2}^{2} + \setminus \ldots + {x}_{N}^{2}}{N} - \setminus {\overline{x}}^{2} = \setminus \frac{{1}^{2} + {3}^{2} + {4}^{2} + {6}^{2} + {8}^{2}}{5} - {4.4}^{2} = 5.84$