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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Find the mean deviation about the mean for the data $38,70,48,40,42,55,63,46,54,44$

$\begin{array}{1 1}(A)\;8.4\\(B)\;3\\(C)\;3.8\\(D)\;4.2\end{array} $

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1 Answer

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Toolbox:
  • Mean deviation about can be calculated by the formula $= \large\frac{\sum |X_i- \bar{X}|}{n}$ where $\bar {X}=$ mean of the given observations.
Step 1:
Mean of the given series $\bar {X}=\large\frac{Sum\;of\;terms}{Number\;of\;terms}$
$\qquad= \large\frac{\sum X_i}{n}$
$\qquad= \large\frac{38+70+48+40+42+55+63+46+54+44}{10}$$=50$
$X_i =38 , \qquad |X_i -\bar{X} |=|38-50|=12$
$X_i =70 , \qquad |X_i -\bar{X} |=|70-50|=20$
$X_i =48 , \qquad |X_i -\bar{X} |=|48-50|=02$
$X_i =40 , \qquad |X_i -\bar{X} |=|40-50|=10$
$X_i =42 , \qquad |X_i -\bar{X} |=|42-50|=8$
$X_i =55 , \qquad |X_i -\bar{X} |=|55-50|=5$
$X_i =63 , \qquad |X_i -\bar{X} |=|63-50|=13$
$X_i =46 , \qquad |X_i -\bar{X} |=|46-50|=4$
$X_i =54 , \qquad |X_i -\bar{X} |=|54-50|=4$
$X_i =44 , \qquad |X_i -\bar{X} |=|44-50|=6$
$\sum x_i =500 \qquad \sum |x_i -\bar{x}|=84$
Step 2:
$\therefore$ meandeviation about mean $=\large\frac{\sum |x_i - \bar {x}|}{n}$
$\qquad= \large\frac{84}{10}$$=8.4$
Hence A is the correct answer.
answered Jun 25, 2014 by meena.p
 
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