If you apply Bessel's correction --- dividing by $5-1$ rather than by $5$ when you have $5$ numbers --- then some of the otherwise right things stated in some of the answers are wrong. Bessel's correction is intended to be used *only* when the variance one is computing is based on a sample to be used to *estimate* the variance of the whole population.

I won't be surprised if nobody used the variance and standard deviation before Abraham de Moivre did so in the 18th century. De Moivre considered this question: If you toss a fair coin $1800$ times, what is the probability that the number of heads is in some specified range? You have a binomial distribution, and computing its exact values was not feasible. De Moivre approximated the distribution of the number of heads with a normal distribution with the same mean and the same standard deviation. In so doing, he was the first to introduce the normal distribution, and the first to prove a special case of the central limit theorem. The normal distribution with mean $0$ and variance $1$ is
$$
\varphi(x)\,dx=\frac 1 {\sqrt{2\pi}} e^{-x^2/2}\,dx
$$
and with mean $\mu$ and variance $\sigma^2$ it is
$$
\varphi\left(\frac{x-\mu}\sigma\right)\, \frac{dx}\sigma.
$$
It's easy to find the mean and standard deviation for the number of heads when one fair coin is tossed: they're both $1/2$. **How do you do it for the sum of $1800$ independent copies of that random variable?** De Moivre found that the mean-square deviation is **additive**: for independent random variables $X_1,\ldots,X_{1800}$ one has $\operatorname{var}(X_1+\cdots+X_{1800})=\operatorname{var}(X_1)+\cdots+\operatorname{var}(X_{1800})$. You cannot do that with mean absolute deviation. If I'm recalling some details correctly, he published these findings in a paper in Latin while he lived in France, and at that time he gave the normal distribution as
$$
C e^{-x^2/2}\,dx
$$
where he could find $C$ only numerically. Later he went to England to escape the persecution of Protestants and met James Stirling, who showed that $C=1/\sqrt{2\pi}$. De Moivre wrote a book in English called *The Doctrine of Chances*, which I think was 18th-century English for the theory of probability. Some have speculated that the Reverend Thomas Bayes may have studied under him, but I don't know that that's gone beyond speculation.

(If you want to know the probability that the number of heads is $\ge894$, note that that's the same as $\text{“}{>893}\text{''}$, and find the probability that the normally distributed random variable with the same mean and variance is $>893.5$. That is a "continuity correction" and works surprisingly well even for fairly small samples.)

On to Bessel's correction: When does one use
$$
\frac{(x_1-\bar x)^2+\cdots+(x_n-\bar x)^2}{n-1},
$$
where $\bar x=(x_1+\cdots+x_n)/n$, with $n-1$ rather than $n$ in the denominator? As you can see from simple examples, that will **not** serve de Moivre's purpose described above: it's not additive.

If $X_1,\ldots,X_n$ are an independent sample from a population with mean $\mu$ and variance $\sigma^2$, then the expected value of
$$
\frac{(X_1-\mu)^2+\cdots+(X_n-\mu)^2} n \tag 1
$$
is $\sigma^2$. But if one has only the sample and not the whole population, one doesn't know $\mu$ and one can use the sample mean $\bar X$ as an estimate of $\mu$. But the expected value of
$$
\frac{(X_1-\bar X)^2+\cdots+(X_n-\bar X)^2} n
$$
is smaller than the expected value of $(1)$. Specifically, a bit of algebra shows that
$$
\sum_{i=1}^n (X_i-\mu)^2 = \left( \sum_{i=1}^n (X_i-\bar X)^2 \right) + n(\bar X-\mu)^2, \tag 2
$$
and since the expectation of the last term is $\sigma^2$, that of the first term on the right in $(2)$ must be $(n-1)\sigma^2$. Thus Bessel's correction gives an unbiased estimate of the population variance $\sigma^2$. (But its square root does not give an unbiased estimate of the population standard deviation. And unbiasedness is at best somewhat overrated anyway, and in some cases is a very very bad thing (I had a paper in the *American Mathematical Monthly* a few years ago demonstrating how bad it can sometimes be).

wronganswer though — there are many better reasons for the deviation formula than convenience. $\endgroup$2more comments