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Estimating the standard deviation of a distribution requires to choose a distance.
Any of the following distance can be used:

$d_{n}((X)_{i=1..I},\mu)=(\sum \left | X-\mu \right |^n)^{1/n}$

We usually use the natural euclidean distance (n=2), which is the one everybody uses in daily life. The distance that you propose is the one with $n=1$.
Both are good candidates but they are different.

One could decide to use $n=3$ as well.

I am not sure that you will like my answer, my point contrary to others is not to demonstrate that $n=2$ is better. I think that if you want to estimate the standard deviation of a distribution, you can absolutely use a different distance.

Estimating the standard deviation of a distribution requires to choose a distance.
Any of the following distance can be used:

$$d_n((X)_{i=1,\ldots,I},\mu)=\left(\sum | X-\mu|^n\right)^{1/n}$$

We usually use the natural euclidean distance ($n=2$), which is the one everybody uses in daily life. The distance that you propose is the one with $n=1$.
Both are good candidates but they are different.

One could decide to use $n=3$ as well.

I am not sure that you will like my answer, my point contrary to others is not to demonstrate that $n=2$ is better. I think that if you want to estimate the standard deviation of a distribution, you can absolutely use a different distance.

Estimating the standard deviation of a distribution requires to choose a distance.
Any of the following distance can be used:

$d_{n}((X)_{i=1..I},\mu)=(\sum \left | X-\mu \right |^n)^{1/n}$

We usually use the natural euclidean distance (n=2), which is the one everybody uses in daily life. The distance that you propose is the one with $n=0$.
Both are good candidates but they are different.

One could decide to use $n=3$ as well.

I am not sure that you will like my answer, my point contrary to others is not to demonstrate that $n=2$ is better. I think that if you want to estimate the standard deviation of a distribution, you can absolutely use a different distance.

Estimating the standard deviation of a distribution requires to choose a distance.
Any of the following distance can be used:

$d_{n}((X)_{i=1..I},\mu)=(\sum \left | X-\mu \right |^n)^{1/n}$

We usually use the natural euclidean distance (n=2), which is the one everybody uses in daily life. The distance that you propose is the one with $n=1$.
Both are good candidates but they are different.

One could decide to use $n=3$ as well.

I am not sure that you will like my answer, my point contrary to others is not to demonstrate that $n=2$ is better. I think that if you want to estimate the standard deviation of a distribution, you can absolutely use a different distance.

Estimating the standard deviation of a distribution requires to choose a distance.
Any of the following distance can be used:

$d_{n}(X,\mu)=(\sum \left | X-\mu \right |^n)^{1/n}$

We usually use the natural euclidean distance (n=2), which is the one everybody uses in daily life. The distance that you propose is the one with $n=0$.
Both are good candidates but they are different.

One could decide to use $n=3$ as well.

I am not sure that you will like my answer, my point contrary to others is not to demonstrate that $n=2$ is better. I think that if you want to estimate the standard deviation of a distribution, you can absolutely use a different distance.

Estimating the standard deviation of a distribution requires to choose a distance.
Any of the following distance can be used:

$d_{n}((X)_{i=1..I},\mu)=(\sum \left | X-\mu \right |^n)^{1/n}$

We usually use the natural euclidean distance (n=2), which is the one everybody uses in daily life. The distance that you propose is the one with $n=0$.
Both are good candidates but they are different.

One could decide to use $n=3$ as well.

I am not sure that you will like my answer, my point contrary to others is not to demonstrate that $n=2$ is better. I think that if you want to estimate the standard deviation of a distribution, you can absolutely use a different distance.