Why is sample variance divided by $n-1$ and not $n$ Sample Variance, customarily denoted, $s^2$, as in the formula below, is the _average of the squared deviations_ , except that we divide by $n-1$ instead of $n$.

Why is sample variance divided by $n-1$ and not $n$ Sample Variance, customarily denoted, $s^2$, as in the formula below, is the _average of the squared deviations_ , except that we divide by $n-1$ instead of $n$. > $$s^2 = \frac{1}{n-1} \sum\limits_{i=1}^{n} (X_i - \overline{X})^2$$ _My question is_ : Why do we divide by $n-1$ instead of $n$? Does it have anything to with having a "binary" operation, meaning two ( _bi_ -nary) operands? This may be a duplicate question.

It is natural to wonder why the sum of the squared deviations is divided by $n − 1$ rather than $n$. The purpose in computing the sample standard deviation is to estimate the amount of spread in the population from which the sample was drawn.

Ideally, therefore, we would compute deviations from the mean of all the items in the population, rather than the deviations from the sample mean.

However, the population mean is in general unknown, so the sample mean is used in its place.

**It is a mathematical fact that the deviations around the sample mean tend to be a bit smaller than the deviations around the population mean and that _dividing by $n − 1$ rather than $n$ provides exactly the right correction._**