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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Find the mean deviation about the mean for data : $x_i :5,10,15,20,25 \qquad f_i : 7,4,6,3,5$

$\begin{array}{1 1}(A)\;6.32\\(B)\;7.5\\(C)\;1.76\\(D)\;7.1\end{array} $

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1 Answer

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Step 1:
$x_i =5,\qquad f_i=7, \qquad f_ix_i =35, \qquad |x_i-\bar{x}|=|5-14|=9 \qquad f_i |x_i -\bar{x}|=63$
$x_i =10,\qquad f_i=4, \qquad f_ix_i =40, \qquad |x_i-\bar{x}|=|10-14|=4 \qquad f_i |x_i -\bar{x}|=16$
$x_i =15,\qquad f_i=6, \qquad f_ix_i =90, \qquad |x_i-\bar{x}|=|15-14|=1 \qquad f_i |x_i -\bar{x}|=06$
$x_i =20, \qquad f_i=3, \qquad f_ix_i =60, \qquad |x_i-\bar{x}|=|20-14|=1 \qquad f_i |x_i -\bar{x}|=18$
$x_i =25,\qquad f_i=5, \qquad f_ix_i =125, \qquad |x_i-\bar{x}|=|25-14|=11 \qquad f_i |x_i -\bar{x}|=55$
Total $\sum t_i=25 \qquad=350$
Mean $\bar{X}=\large\frac{\sum f_ix_i}{\sum f_i}$
$\qquad=\large\frac{158}{25}$$=6.32$
Hence A is the correct answer.
answered Jun 25, 2014 by meena.p
 
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