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.
Indeed, the variance is defined to have has the additivity property: if
r_1 is a random variable with mean
m_1 and variance
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
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.
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.