I was reading about variance from Head First Statistics :
And then -
Q. I find the reasoning a little absurd. Wouldn't just taking the absolute distance suffice if cancelling out of the terms was the reason ? Why do squares to make it positive and complicate the calculations further (In terms of computations on a computer, square would be costlier than subtraction, right?) ?
On
You can take the average absolute deviation from the mean as a measure of dispersion if you wish.
There are various reasons why the variance and its square root (the standard deviation) are often used, including:
the average square of the deviation is minimised when taken about the mean, while the average absolute deviation is minimised when taken about the median, so the variance is in this sense a natural partner of the mean
for a random variable with a normal (Gaussian) distribution, the mean and variance together are sufficient statistics, while the mean and average absolute distribution are not
the average square of the deviation is easily differentiable while the average absolute deviation, in the same way as $x^2$ is differentiable for all $x$ while $|x|$ is not
Actually the real "absolute distance" is the standard deviation (i.e $\sqrt{\text{Var}(X)}$) ! I'll make the exemple in the case where the expected value is $0$. The reason is that $$\sqrt{x_1^2+...+x_n^2}\leq |x_1|+...+|x_n|$$ therefore $\sqrt{x_1^2+...+x_n^2}$ gives a better approximation than $|x_1|+...+|x_n|$.