Why is variance squared?

The mean absolute deviation is:

$$\dfrac{\sum_{i=1}^{n}|x_i-\bar x|}{n}$$

The variance is: $$\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}$$

• So the mean deviation and the variance are measuring the same thing, yet variance requires squaring the difference. Why? Squaring always gives a non-negative value, but the absolute value is also a non-negative value.
• Why isn't it $|x_i-\bar x|^2$, then? Squaring just enlarges, why do we need to do this?

A similar question is here, but mine is a little different.

Thanks.

A late answer, just for completeness with a different view on the thing.

You might look at your data as measured in a multidimensional space, where each subject is a dimension and each item is a vector in that space from the origin towards the items' measurement over the full subject's space.
Additional remark: this view of things has an additional nice flavour because it uncovers the condition, that the subjects are assumend independent of each other. This is to have the data-space euclidean; changes in that independence-condition require then changes in the mathematics of the space: it has correlated (or "oblique") axes.

Now the distance of one vector-arrowhead to another is just the formula for distances in the Euclidean space, the squarerroot of squares of distances-of-coordinates (from the Pythagorean theorem) : $$d = \sqrt { (x_1-y_1)^2+(x_2-y_2)^2+ \cdots+(x_n-y_n)^2}$$ And the standard-deviation is that value, normed by the number of subjects, if the mean-vector is taken as the $y$-vector. $$\text{sdev} = \sqrt { {(x_1- \bar x)^2 +(x_2-\bar x)^2+ \cdots +(x_n-\bar x)^2 \over n} }$$

They don't measure the same thing. To see this, think about physical units.

Suppose the value of $x$ is measured in seconds. For example, $n$ people do a 100-meter race and the values $x_i$ are how many seconds it took each one to finish.

The formula $|x_i - \bar x|$ measures the difference of two times, so it's also measured in seconds.

The mean absolute deviation is therefore an average of second-values, so it's also measured in seconds.

However, the formula $(x_i - \bar x)^2$ squares the difference of two times, so it's measured in seconds squared. The variance is therefore also in seconds squared. They don't belong to the same physical space of variables, so they measure different things.

The standard deviation, however (the square root of the variance) is again measured in seconds, so it measures something similar (at least, physically similar).

As for why we like the square-root-of-average-of-squares better than the average-of-absolute-values - the square has better mathematical properties, as shown in other answers and in the link you referred to (particularly Rich's answer).