**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 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.

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 or 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.

(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.)

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.

Standard deviation, defined this way, is very useful. For example, a large class of random distributions are completely defined by their mean and their standard deviation.