I'm trying to understand a very basic concept of standard deviation.
From the formula $\sigma= \sqrt{ \dfrac{ \sum\limits_{i=1}^n (x_i-\mu)^2} N } $
I can't understand why should we halve the population "N" i.e why do we want to take $\sqrt{N}$ when we didnt do ${N^2}$? Doesn't that skew the population that we are considering?
Shouldn't be the formula be $\sigma= \dfrac{ \sqrt{ \sum\limits_{i=1}^n (x_i-\mu)^2} } {N} $
You're trying to find a "typical" deviation from the mean.
The variance is "the average squared distance from the mean".
The standard deviation is the square root of that.
That makes it the root-mean-square deviation from the mean.
The standard deviation is the square root of the variance.
The variance is the average squared distance of the data from the mean. Since an average is the sum divided by the number of items summed, the formula for the variance is:
$$
\text{Var}(X)=\text{E}[(X-\mu)^2] = \frac{\sum_{i=1}^N(x_i-\mu)^2}{N}
$$
Since, again, the standard deviation is simply the square root of this, the formula for the standard deviation is:
$$
\text{S.D.}(X)=\sqrt{\text{Var}(X)} = \sqrt{\frac{\sum_{i=1}^N(x_i-\mu)^2}{N}}
$$
Nothing has been added or changed about the assumptions or the variance here, we simply took the square root of the variance, because that's what the standard deviation is.
First thing to understand is that standard deviation (std) is different than average absolute deviation. These two define different mathematical property about the data.
Unlike average absolute deviation, standard deviation (std) weighs more to the values that are far from mean, which is done by squaring the difference values.
E.g., For following four data points: \begin{array}{|c|c|c|} \hline Data (x)& |x - mean| & (x-mean)^2 \\ \hline 2 & 2 & 4\\ \hline -2 &2 &4\\ \hline -6 &6 &36\\ \hline 6 &6 &36\\ \hline \sum x =0 & \sum (|x-mean|) = 16 & \sum (x-mean)^2 = 80 \end{array}
average absolute deviation (aad) $= 16/4 = 4.0$, and
Standard deviation (std) = $\sqrt{80/4} = \sqrt 20 = 4.47 $
In the data, there are two points which are 6 distance away from mean, and two points which are 2 distance away from mean. So, deviation of 4.47 makes more sense than 4.
Since total observation are always $N$, for computing std we are not diving by $\sqrt N$, instead we divide the total variance by $N$, and take its square root, to bring it to the same unit as the original data.
I would agree that your definition of standard deviation $\sigma= \dfrac{ \sqrt{ \sum\limits_{i=1}^n (x_i-\mu)^2} } {N} $ could be used to measure the spread of a population. However it is a harder concept to come up with. (And I think it is also harder to use to, for example, prove other theories.)
I don't know the exact history behind the invention of variance and standard deviation, but it should be roughly similar to:
We need a positive number to measure the spread; let us define variance = $\dfrac{ \sum\limits_{i=1}^n (x_i-\mu)^2} N $
This variance looks much bigger than the original values due to squaring; let us take square root of it to 'undo' the effect of squaring: $\sigma = \sqrt{\frac{\sum\limits_{i=1}^n (x_i-\mu)^2}{N}}$
So now we have a very simple relation $std = \sqrt{variance}$, hence $N$ is under square root.
@Mahesh Subramaniya - This is just mathematical twist. When we have original value like $a/b = (-)d$. We can get same value using these two equations ${a}^2\diagup{b}=c$ and $\sqrt{c\diagup{b}}=d$.
E.g. just do it with ${-5}\diagup{2}$ = $-2.5$. But, we want only value not minus.
Now, ${-5}^2\diagup{2}=12.5$. And , $\sqrt{12.5\diagup{2}}=2.5$
