**Intro by Reid Barton**

I think the answer should involve the additivity of variance for independent variables and the central limit theorem.  Maybe someone can flesh this out.

**Expanded answer by ilya and Reid**

Indeed, the variance is defined to have the **additivity property**: if `r_1` is a random variable with mean `m_1` and variance `d_1` and `r_2` is a random variable with mean `m_2` and variance `d_2` and *these two variables are independent* then the new random variable `r = r_1+r_2` has the mean `m_1+m_2` and variance `d_1+d_2`.

This will obviously fail for any other function of variance, be it square, cube or something else. Answers that stress convenience are, unfortunately, missing the crucial point.

Moreover, suppose we sum a large number N of independent copies of our random variable `r` with mean `m` and variance `d`.  Under mild assumptions, the central limit says the distribution will approach a normal distribution, which by the above has mean `Nm` and variance `Nd`.  Observe that a normal distribution is completely determined by its mean and variance.  We conclude that the **only** parameters of a distribution that we can observe from the sum of many independent copies of the distribution are the mean and variance.

To get back to something in the same units as the original variable, we take the square root of the variance and call it the standard deviation.