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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Find the mean deviation about the mean for data : $x_i :10,30,50,70,90 \qquad f_i : 4,24,28,16,8$

$\begin{array}{1 1}(A)\;26\\(B)\;16\\(C)\;76\\(D)\;7\end{array} $

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1 Answer

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Step 1:
$x_i =10,\qquad f_i=4, \qquad f_ix_i =40, \qquad |x_i-\bar{x}|=|10-50|=40 \qquad f_i |x_i -\bar{x}|=160$
$x_i =30,\qquad f_i=24, \qquad f_ix_i =720, \qquad |x_i-\bar{x}|=|30-50|=20 \qquad f_i |x_i -\bar{x}|=480$
$x_i =50,\qquad f_i=28, \qquad f_ix_i =1400, \qquad |x_i-\bar{x}|=|50-50|=0 \qquad f_i |x_i -\bar{x}|=0$
$x_i =70,\qquad f_i=16, \qquad f_ix_i =1120, \qquad |x_i-\bar{x}|=|70-50|=20 \qquad f_i |x_i -\bar{x}|=320$
$x_i =90,\qquad f_i=8, \qquad f_ix_i =720, \qquad |x_i-\bar{x}|=|90-50|=40 \qquad f_i |x_i -\bar{x}|=320$
Total $\sum f_i =80 \qquad \sum f_ix_i=4000$
Mean $= \large\frac{\sum f_ix_1}{\sum f_i} =\frac{4000}{80}$$=50$
Mean deviation about mean $=\large\frac{\sum f_i |x_i -\bar{x}|}{\sum f_i }=\frac{1280}{80} $$=16$
Hence B is the correct answer.
answered Jun 25, 2014 by meena.p
 
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