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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} $

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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.

  1. Why would we use the average squared deviation? What makes variance interesting? Among other things, because of a basic fact about variances - that the variance of a sum of uncorrelated variables is the sum of the individual variances. (This is covered in a number of questions e.g. here on CrossValidated. This handy feature is not shared, for example, by the mean absolute deviation.
  2. Why take the square root of that? Because then it's in the same units as the original observations. It measures a particular kind of 'typical distance' from the mean (as mentioned, the RMS distance) - but because of the above property of variance - one that has some nice features.
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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.

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  • $\begingroup$ maybe it should be mentioned that this variance formula is only true for discrete uniforms. otherwise it might confuse the distinction between sample and population variance $\endgroup$ – Taylor Sep 8 '16 at 3:32
  • $\begingroup$ @Taylor, I don't know what you mean. The formula for the variance is unrelated to the distribution. $\endgroup$ – gung - Reinstate Monica Sep 8 '16 at 12:08
  • $\begingroup$ the formula for (sample) variance is unrelated to the distribution (en.wikipedia.org/wiki/Expected_value#Definition) $\endgroup$ – Taylor Sep 8 '16 at 14:57
  • $\begingroup$ @Taylor, I still don't know what you mean. The formula for the variance is unrelated to the distribution. To quote from the Wikipedia page, "The variance of a random variable, X, is the expected value of the squared deviation from the mean of X ... $\operatorname{Var}( X ) = E⁡[(X − μ)^2]$. This definition encompasses random variables that are generated by processes that are discrete, continuous, neither, or mixed." The formula does not only hold for the discrete uniform. $\endgroup$ – gung - Reinstate Monica Sep 8 '16 at 15:54
  • $\begingroup$ Yes, that's right, if you take $\mu = EX$, but $E[(X-\mu)^2]$ does not necessarily equal, for any random variable $X$, $\frac{1}{N}\sum_i(x_i - \mu)^2$. For one, the first is a constant and the second is random. Actually it's not clear whether the sum runs over the support of $X$ or the number of samples. If the latter, it's weird that you know $\mu$, which is rare in practice. If the former, then yes, it's only true for discrete (because it's a sum) uniforms (because the weights are all uniform). $\endgroup$ – Taylor Sep 8 '16 at 18:30
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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.

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@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 , $\sqrt{12.5\diagup{2}}=2.5$

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