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$$\text{Mean deviation from mean}=\frac1N\sum_{i=1}^nf_i|x_i-\bar x|$$and $$\text{Standard Deviation ($\sigma$)}=\sqrt{\frac1N\sum_{i=1}^nf_i(x_i-\bar x)^2}$$ The step of squaring the deviations in SD overcomes the drawback of ignoring the signs of mean deviation.

How is the problem overcome? In the SD too, we ignore the signs by squaring, aren't we?

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Not sure if this is relevant here, but in least squares regression, the reason that you square instead of take the absolute value is so that the function remains differentiable everywhere. –  Jeremy Dec 14 '13 at 17:23
@Jeremy : I don't think that's the reason. –  Michael Hardy Dec 14 '13 at 17:28

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up vote 2 down vote accepted

The principal reason for the widespread use of mean square deviation instead of mean absolute deviation is that if random variables $X_1,\ldots,X_n$ are independent, then $$ \operatorname{var}(X_1+\cdots+X_n)=\operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n). $$ That makes it possible to related the dispersion of the sum, and hence the average, of the members of a sample to the dispersion in the population from which the sample was taken, and in particular, we'd have no central limit theorem otherwise. Without the central limit theorem, the usual method of finding a confidence interval for a population mean would not be justified.

Whoever told you that the reason is to overcome an alleged drawback of ignoring signs was probably confused.

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Thank you so much, Sir! I will have to read books very carefully. –  Silent Dec 15 '13 at 3:05
Sir, how does $\operatorname{var}(X_1+\cdots+X_n)=\operatorname{var}(X_1)+\cdots+\operatorname‌​{var}(X_n)$ hold for independent $X_1,\ldots,X_n$? –  Silent Dec 15 '13 at 3:13
@Sush : By mathematical induction, you can reduce the problem to that of proving that it works for only two random variables. Then $\operatorname{var}(X+Y)=\sum\limits_{x,y}((x+y)-(\mu+\nu))^2\Pr(X=x\ \&\ Y=y)$. By independence, you can change that probability to $\Pr(X=x)\cdot\Pr(Y=y)$ and then do some algebra. If you post that as a separate question, I'll give a more detailed answer. –  Michael Hardy Dec 15 '13 at 20:27
@Sush : OK, shorter answer: Show that $\operatorname{var}(X+Y)=\operatorname{var}(X)+2\operatorname{cov}(X+Y)+ \operatorname{var}(Y)$. Then show that if $X,Y$ are independent then the covariance is $0$. –  Michael Hardy Dec 15 '13 at 21:53
Thank you so much,Sir. –  Silent Dec 16 '13 at 3:39

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