# What are the variance and standard deviation of {1, 1, 1, 1, 1, 80, 1, 1, 1, 1, 1, 1}?

Jul 14, 2016

The population variance is:

${\sigma}^{2} \cong 476.7$

and the populations standard deviation is the square root of this value:

$\sigma \cong 21.83$

#### 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

$\mu = \frac{1 + 1 + 1 + 1 + 1 + 80 + 1 + 1 + 1 + 1 + 1 + 1}{12} = \frac{91}{12} = 7.58 \overline{3}$

Now we can proceed with the variance calculation:

${\sigma}^{2} = \frac{11 \cdot {\left(1 - 7.58 \overline{3}\right)}^{2} + {\left(80 - 7.58 \overline{3}\right)}^{2}}{12}$

${\sigma}^{2} \cong 476.7$

and the standard deviation is the square root of this value:

$\sigma \cong 21.83$