# What is the variance of {-4, 5, 8 ,-1, 0 ,4 ,-12, 4}?

Jun 24, 2016

The population variance of the data set is

${\sigma}^{2} = 35$

#### Explanation:

First, let's assume that this is the entire population of values. Therefore we are looking for the population variance . If these numbers were a set of samples from a larger population, we would be looking for the sample variance which differs from the population variance by a factor of $n / \left(n - 1\right)$

The formula for the population variance is

${\sigma}^{2} = \frac{1}{N} {\sum}_{i = 1}^{N} {\left({x}_{i} - \mu\right)}^{2}$

where $\mu$ is the population mean, which can be calculated from

$\mu = \frac{1}{N} {\sum}_{i = 1}^{N} {x}_{i}$

In our population the mean is

$\mu = \frac{- 4 + 5 + 8 - 1 + 0 + 4 - 12 + 4}{8} = \frac{4}{8} = \frac{1}{2}$

Now we can proceed with the variance calculation:

${\sigma}^{2} = \frac{{\left(- 4 - \frac{1}{2}\right)}^{2} + {\left(5 - \frac{1}{2}\right)}^{2} + {\left(8 - \frac{1}{2}\right)}^{2} + {\left(- 1 - \frac{1}{2}\right)}^{2} + {\left(0 - \frac{1}{2}\right)}^{2} + {\left(4 - \frac{1}{2}\right)}^{2} + {\left(- 12 - \frac{1}{2}\right)}^{2} + {\left(4 - \frac{1}{2}\right)}^{2}}{8}$

${\sigma}^{2} = 35$