Sample Variance and Standard Deviation

Estimates the spread of a population from a sample. Dividing by count − 1 instead of the count gives the unbiased sample variance. At least two values are needed.

Explanation

When your data is a sample drawn from a larger population, the unbiased sample variance is what estimates that population's spread.

s^2 = \dfrac{1}{n - 1}\sum_{i=1}^{n} (x_i - \bar{x})^2 \qquad s = \sqrt{s^2}

The key is dividing by $n - 1$ rather than $n$ . That quantity is called the number of degrees of freedom.

Why n − 1

The deviations are measured from the sample mean $\bar{x}$ , and the sample mean sits as close to its own data as possible. The sum of squared deviations therefore comes out systematically too small compared with measuring from the true population mean.

Dividing by $n$ would understate the population variance. Dividing by the slightly smaller $n - 1$ cancels that bias exactly, so that on average the estimate is right. That is what "unbiased" means.

Example

For 10, 20, 30, 40, 50 the mean is 30 and the sum of squared deviations is $400 + 100 + 0 + 100 + 400 = 1000$ .

s^2 = \dfrac{1000}{5 - 1} = 250 \qquad s = \sqrt{250} = 15.8114

Dividing by $n$ instead would have given a variance of 200 and a standard deviation of 14.1421. The unbiased estimate is a little larger.

Which one to use

Population variance (divide by $n$ ) — your data is the whole group, such as every score in one class
Sample variance (divide by $n - 1$ ) — your data is a sample and you want to say something about the population

When in doubt, use the sample variance. Nearly every statistics package defaults to $n - 1$ .