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Variance is

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

But why square the difference? Why not cube it, or any other exponent?

Related question.

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    $\begingroup$ If you cube it then some errors will cancel others due to negative signs and that is not good. $\endgroup$ Commented Sep 27, 2019 at 20:44
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    $\begingroup$ First of all, your formula is not the variance, but rather the sample variance. The variance of a random variable $X$ is defined as $$\mathbf{Var}(X)=\mathbb{E}[(X-\mathbb{E}[X])^2].$$ Now you may still ask what is the advantage of squaring the deviation $X-\mathbb{E}[X]$. My personal impression on this question is two-fold: (1) Variance is one of the two parameters that characterize the normal distribution. (2) Variance comes from covariance, which gives rise to the (degenerate) inner-product structure on the space of square-integrable random variables. $\endgroup$ Commented Sep 27, 2019 at 20:53

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You can also use other exponents, but then it is not called variance anymore. See this wikipedia page, for example.

Variance comes up quite naturally in various contexts, and carries a lot of information about the distribution.

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Still, I am not convinced by this answer. The naming convention can be separate topics to discuss. Here what is the issue if we take cubes or quad etc?

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