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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Find the mean deviation about the mean of the distribution

$\begin{array}{1 1}(A)\;10.10,1.99\\(B)\;8.4,5.6\\(C)\;3,8\\(D)\;12,1.25\end{array} $

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1 Answer

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Toolbox:
  • Formula used to solve this problem are :
  • Median $M =\large\frac{\bigg(\Large\frac{N}{2} \bigg)th\; observation +\bigg( \Large\frac{N}{2}+1 \bigg) th\; observation}{2}$
  • Mean deviation about mean $=\large\frac{\sum f_i | x_i - \bar{x}|}{\sum f_i }$
Step 1:
$N= \sum f_i =20(even)$
Median $M =\large\frac{\bigg(\Large\frac{N}{2} \bigg)th\; observation +\bigg( \Large\frac{N}{2}+1 \bigg) th\; observation}{2}$
$\qquad= \large\frac{\bigg(\Large\frac{20}{2} \bigg)th\; observation + \bigg( \Large\frac{20}{2} +1\bigg) th \;observation}{2}$
$\qquad= \large\frac{10th \;observation + 11th \;observation}{2}$
$\qquad= \large\frac{12+12}{2}$
$\qquad= \large\frac{24}{2}$
$\qquad= 12$
Step 2:
Mean deviation about mean $=\large\frac{\sum f_i | x_i - \bar{x}|}{\sum f_i }$
$\qquad= \large\frac{25}{20}$
$\qquad= 1.25$
Hence D is the correct answer.
answered Jul 2, 2014 by meena.p
 
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