Suppose you have some apples. You can weigh every apple, and write down all the masses. The apples in this circumstance constitute a population. If the apples have masses

\[x_1, \: x_2, \: x_3,..., \: x_n\]

.The population standard deviation is

\[\begin{equation} \begin{aligned} \sigma = \sigma_n &= \sqrt{\frac{\sum_i^n x_i^2 - (\sum_i^n x_i)^2/n}{n}} \\ &=\sqrt{\frac{\sum_i^n x_i^2 - n {\bar{x}}^2}{n}} \\ &= \sqrt{\frac{\sum_i^n x_i^2}{n} - (\frac{\sum_i^n x_i}{n})^2} \end{aligned} \end{equation}\]

.Suppose however that we were to take a sample of

\[m\]

apples. If we wanted to find the standard deviation of these apples, we would be finding the sample standard deviation \[s=\sigma_{n-1}\]

.The sample standard deviation is

\[\begin{equation} \begin{aligned} s &= \sigma_{m-1} \\ &= \sqrt{\frac{\sum_i^m x_i^2 - (\sum_i^m x_i)^2/m}{m-1}} \\ &=\sqrt{\frac{\sum_i^m x_i^2 - m {\bar{x}}^2}{m-1}} \\ &= \sqrt{\frac{\sum_i^m x_i^2}{m-1} - \frac{(\sum_i^m x_i)^2}{m(m-1)}} \end{aligned} \end{equation}\]

.The difference between the two definitions is the presence of

\[n-1\]

instead of \[n\]

in the denominator of the root. This is because by taking a sample we are introducing a random element. The standard deviation each time we choose a different sample, so the sample standard deviation is not a fixed number, and the change in the denominator takes account of this.