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Mean deviation for n observation $x_1,x_2........x_n$ from their mean $\bar{x}$ is given by

$\begin{array}{1 1}(A)\; \sum \limits_{i=1}^n (x_i -\bar {x})\\(B)\;\large\frac{1}{n} \sum \limits_{i=1}^{n} |x_i -\bar {x}|\\(C)\; \sum \limits_{i=1} (x_i -\bar{x})^2\\(D)\;\frac{1}{n} \sum \limits_{i=1}^n (x_i -\bar {x})^2\end{array} $

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The n observations are $x_1,x_2,x_3........x_n$
Let mean of these n observation be $\large\frac{x_1+x_2+x_n}{n} $$=\bar{x}$
Deviation of each $x_i$ from $\bar {x}$ are $x_1- \bar{x},x_2 -\bar{x},x_3-\bar{x},........x_n-\bar{x}$ and their absolute value will be $|x_1- \bar{x} |,|x_2- \bar{x}|,|x_3 -\bar{x} |,...........|x_n -\bar {x} |$
Mean of the absolute of the deviations = $ \large\frac{|x_1-\bar{x}|+|x_2- \bar {x} |+ |x_3 -\bar{x}|+.........|x_n -\bar{x} |}{n}$
Mean deviation about mean $= \large\frac{ \sum \limits_{i=1} ^n |x_i -\bar{x}|}{n}$
$\qquad= \large\frac{1}{n} \sum \limits_{i=1}^{n} |x_i -\bar{x}|$
Hence B is the correct answer.
answered Jul 8, 2014 by meena.p

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