Why standard deviation is calculated the way it is calculated?

Question

I do not have any problems with degree of freedoms or why squaring is better than taking absolute values but I have more fundamental problem with understanding of why formula looks this way.

$\frac{\sqrt{{\sum_i^n (x_{i}-\overline{x})^{2}}{}}}{\sqrt{n}}$

Why number of items in sample is also square rooted? I understand that this is related to Variance formula but what's the sense to take the square root of n? Unlike deviances it was not squared in the first place.

Thanks in advance for your responses!

Answer 1

The standard deviation is simply the square root of the variance. It is not defined by the formula above but rather by $\sigma = \sqrt{var}$. This means that any value that is present in the variance will (probably) be present in the standard deviation.

As $n$ is part of the definition for the variance (see e.g. Variance of a discrete random variable on Wikipedia) it is also in the standard deviation.

Answer 2

Suppose you have $n$ samples, half of which equal $-1$ and half of which equal $1$. Then $$\sqrt{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2} = 1$$ no matter what $n$ is. This makes sense: the "spread" of the data is the same, regardless of whether the sample size is $10$ or $10000$.

If you don't take the square root of $n$, then you get $$\frac{1}{n} \sqrt{\sum_{i=1}^n (x_i - \bar{x})^2} = \frac{1}{\sqrt{n}}$$ which shrinks to zero as $n \to \infty$, which would not really make sense for a notion of spread.