Why is Standard Deviation Calculated With the Square Root of the Sum of Squares Instead of Sum of Square Roots of Squares

The standard deviation definition I saw in my textbook and heard in the lessons is that it is a measure of how spread-out a data set is and as such it calculated as the mean of the distances between a data point and the arithmetic mean. Yet the formula for standard deviation does not reflect this. If it is really the average of the distances it should be something more like this shouldn't it?

$$\frac{\sum{|{x_i}-{\overline{x}}|}}{n}$$

I've seen and heard some saying that absolute values make it harder to do algebraic calculations and thats why the square root is used but even then, why aren't the individual roots are added up? If the whole purpose of the square inside the square root is to get a distance to the mean without using an absolute value, shouldn't the formula look something more like this?

$$\frac{\sum{(\sqrt{(x-{\overline{x}})^2}})}{n}$$

$$\sqrt{\frac{\sum{(x-{\overline{x}})^2}}{n}}$$

Wouldn't this lead to errors? Wouldn't taking the number of terms also in the root give a wrong result?

If anyone could explain this to a high school student, me, without any too advanced math I would be really grateful.

• At a very high level because $L^p$ is a Hilbert space only for $p=2$. Commented Feb 26, 2022 at 15:46
• Commented Feb 26, 2022 at 15:54
• It's not wrong or right to use one metric or another to measure spread. The metric you are referring to is the average absolute deviation. Commented Feb 26, 2022 at 15:56
• Related viz. @MathematicsStudent1122's point.
– J.G.
Commented Feb 26, 2022 at 16:09
• Commented Feb 27, 2022 at 14:30

To me the most convincing reason to use $$\sigma^2=\mathbb E\Big[\Big(X-\mathbb E[X]\Big)^2\Big]$$ as a definition of variance of a random variable $$X$$ is that we can estimate it from a sample $$\{Y_1,...,Y_n\}$$ of size $$n>1$$ of realizations of $$X$$ by $$\sum_{i=1}^n\frac{(Y_i-\overline{Y})^2}{n-1}$$ where $$\mathbb E\Bigg[\sum_{i=1}^n\frac{(Y_i-\overline{Y})^2}{n-1}\Bigg]=\sigma^2\,.$$ See Sample Variance. If someone can show that other measures of dispersion have such a simple property I would be very interested.