From what I understand, standard deviation is a way to quantify how individual results differ from their mean. Since $E[X - E[X]] = 0$, we square that difference and take the square root of the expectation to get a value that represents the "expected" deviation of a value from its mean.
But is there some specific reason we use $E[(X-E[X])^2]^{1/2}$, and not $E[(X-E[X])^3]^{1/3}$ or $E[(X-E[X])^n]^{1/n}$ for any other $n$ (e. g. a more analytical way to derive standard deviation). Does the case when $n=2$ represent something particularly useful, and what would other cases represent?
I also wonder what's the relationship between standard deviation and Euclidean distance / $L^2$-norm. $\left(\int_{\mathbb{R}} (x-E[X])^pf(x)\ dx\right)^{1/p}$ also looks similar to $\left(\int_{\mathbb{R}} |f|^p \ d\mu\right)^{1/p}$
