How do I calculate the variance of {3,6,7,8,9}?

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

Explanation:

Where:

${s}^{2}$ = variance

$\sum$ = sum of all values in the sample

$n$ = sample size

$\overline{x}$ = mean

${x}_{i}$ = Sample observation for each term

Step 1 - Find the mean of your terms.

$\frac{3 + 6 + 7 + 8 + 9}{5} = 6.6$

Step 2 - Subtract the sample mean from each term ($\overline{x} - {x}_{i}$).

$\left(3 - 6.6\right) = - 3.6$

${\left(6 - 6.6\right)}^{2}$$= - 0.6$

${\left(7 - 6.6\right)}^{2}$$= 0.4$

${\left(8 - 6.6\right)}^{2}$$= 1.4$

${\left(9 - 6.6\right)}^{2}$$= 2.4$

Note: The sum of these answers should be $0$

Step 3 - Square each of the results. (Squaring makes negative numbers positive.)

-${3.6}^{2} = 12.96$

-${0.6}^{2} = 0.36$

${0.4}^{2} = 0.16$

${1.4}^{2} = 1.96$

${2.4}^{2} = 5.76$

Step 4 - Find the sum of the squared terms.

$\left(12.96 + 0.36 + 0.16 + 1.96 + 5.76\right) = 21.2$

Step 5 - Finally, we'll find the variance. (Make sure to -1 from the sample size.)

${s}^{2} = \frac{21.2}{5 - 1}$

${s}^{2} = 5.3$

An extra, if you'd care to expand - from this point, if you take the square root of the variance, you'll get the standard deviation (a measure of how spread out your terms are from the mean).

I hope this helps. I'm sure that I didn't need to write out every step, but I wanted to make sure you knew exactly where each number was coming from.