Displacement law: $ \sum_{i=1}^{n} (x_i - \bar{x})^2 $ why is it squared? $ \sum_{i=1}^{n} (x_i - \bar{x})^2 $
where $ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i $ (mean value)
I would like to understand why the term inside the sum gets squared. Why isn't it enough to e.g. just use the absolute value here? 
Squaring this value means, that the bigger an element $ x_i $ differs from the mean $ \bar{x} $, the more weight it puts onto the sum.
Could anybody explain this to me? This question came up to me as I took a look at the variance: $ \text{Var(X) = E(X}^2 \text{) - E(X)}^2 $
 A: It is not so convenient to work with absolute values.
A: For your first question:
The term is squared because it is the estimate of the $2^{nd}$ central moment of the underlying distribution, or sample second central moment. Central moments are nice becasue they exhibit translation invariance, but they are also theoretically important, as moments are a key property of many distributions (e.g., moment generating functions).
So, from a pragmatic point of view, it is just one measure of the statistical disperson. However, from a theoretical point of view, it comes up in many areas of mathematical statistics. For example, the t-distribution uses the square root of your formula. In general, the moments of a distribution, when they exist, present many theoretical opportunities for developing tests and estimates. Probably one of the most imporant of which is the central limit theorem, which uses the mean, and the variance (i.e., second central moment) to approximate the underlying sampling distribution.
For your second question: Do you know how to calculate the variance? It is $E[(X-E[X])^2]$. From the linearity of expectation, this can be re-formulated as follows:
$E[(X-E[X])^2] \rightarrow E[X^2]-2E[X]^2+E[X]^2 \rightarrow E[X^2]-E[X]^2$
A: kmitov gives the "standard" answer: the average squared deviation has nicer analytic properties than the average absolute deviation. (It is differentiable, for example, whereas the average absolute deviation isn't.) But you're right that this isn't a very good answer, because it doesn't explain why we use squares rather than any even power. In fact, from an algebraic perspective, there are many different definitions of a measure of spread: average absolute deviation, average squared deviation, interquartile range, etc. Which definition you use depends on what you want to do with it. Unfortunately many introductory textbooks present the average squared deviation as the only possible measure of spread, without giving any sense of the bigger picture.
A better way to to explain what makes squaring seem "natural" is to think of the problem geometrically rather than algebraically. Suppose you have some data $\left\{x_i\right\}$ for $1\leq i\leq n$. You can encode this data in an $n$-dimensional vector $\vec{x}$. You can also encode the mean vector $\vec{\bar{x}}$, which has the mean $\bar{x}$ in each of the $n$ coordinates. Then the vector
$$\vec{d}:=\vec{x}-\vec{\bar{x}}$$
measures how far off the data are from the average vector $\vec{\bar{x}}$, and the magnitude of this vector is $$||\vec{d}||=\sqrt{\sum(x_i-\bar{x})^2}$$
which is exactly the quantity you asked about.
So the real answer to your question is twofold: (1) measures of spread depend on what you want to use them for, and there are plenty of situations in which one might not want to use squared deviations, but (2) squared deviations are natural from a geometric perspective. (In slightly more advanced lingo, you can do Euclidean geometry in a Hilbert space of random variables.)
POSTSCRIPT: The best justification of squaring will come when you study linear regression the right way, in terms of vectors and matrices. In that setting, regression amounts simply to "projecting" a vector of dependent variable data onto a certain subspace determined by the independent variable data. These projections, of course, are carried out in Euclidean space, which amounts to dealing with the Euclidean norm -- which is based on squares. See, for example, this quick overview.
A: I suspect that the popularity of the variance is due to the fact that $\bar x$ minimizes $\sum_{i=1}^n (x_i-t)^2$ with respect to $t$.  By the way, if you want to say something like $X$ is twice as variable as $Y$ you should use the standard deviation instead of the variance.
The variance is not the only measure of dispersion.  You could instead look at the mean absolute deviation or some such object (which would be minimized at the median).
