**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.

**Answer**

Indeed, the variance has the **additive property**: if $r_1$ and $r_2$ are random variables with means $\mu_1, \mu_2$  and variances $d_1, d_2$, and *these two variables are independent*, then the new random variable $r = r_1+r_2$ has the mean $\mu_1+\mu_2$ and variance $d_1+d_2$.

Moreover, suppose we sum a large number $N$ of independent copies of our random variable $r$ with mean $\mu$ and variance $d$.  Under mild assumptions, the central limit says the distribution will approach a normal distribution, which by the above has mean $N\mu$ 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.

Now that we established how good it is to **square** numbers, to get variance, the standard deviation has a very easy explanation: it's the *only way to get back from variance to something with the dimension of our original set*. That is, suppose you numbers are some lengths written in *meters*. Since the variance is *meters squared*, you *have* to take the **square root** to get something that can be compared with the original set.

Now, honestly, this not *the only way*, since you could also, e.g., multiply it by 2. That's why it's called *standard* deviation — to show that among different numerical constants we've chosen a specific one.