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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)\;21.65,1.25\end{array} $

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1 Answer

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Toolbox:
  • The formula used to solve this problem are
  • Mean $\bar{x}=\large\frac{\sum f_ix_i}{\sum f_i}$
  • Mean deviation about mean $=\large\frac{\sum f_i | x_i - \bar{x}|}{\sum f_i }$
Step 2:
Mean $\bar {x} =\large\frac{\sum f_i x_i}{\sum f_i} $
$\qquad= \large\frac{433}{20} $$=21.65$
Step 2:
Mean deviation about mean $=\large\frac{\sum f_i | x_i - \bar{x}|}{\sum f_i }$
$\qquad= \large\frac{25}{20} $$=1.25$
Hence D is the correct answer.
answered Jul 2, 2014 by meena.p
 

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