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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Find the mean deviation about the mean for data : $x_i :15,21,27,30,35 \qquad f_i : 3,5,6,7,8$

$\begin{array}{1 1}(A)\;5.1\\(B)\;3\\(C)\;2.5\\(D)\;7.6\end{array} $

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1 Answer

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Step 1:
$x_i =15,\qquad f_i=3, \qquad CF=3,\qquad |x_i -M|=|15-30|=15 \qquad f_i |x_i-M|=45$
$x_i =21,\qquad f_i=5, \qquad CF=8,\qquad |x_i -M|=|21-30|=3 \qquad f_i |x_i-M|=18$
$x_i =27,\qquad f_i=6, \qquad CF=14,\qquad |x_i -M|=|27-30|=3 \qquad f_i |x_i-M|=18$
$x_i =30,\qquad f_i=7, \qquad CF=21,\qquad |x_i -M|=|30-30|=0 \qquad f_i |x_i-M|=0$
$x_i =35,\qquad f_i=8, \qquad CF=29,\qquad |x_i -M|=|35-30|=5 \qquad f_i |x_i-M|=40$
Total $N= \sum f_i=29(odd)$
$Medium M= \bigg( \large\frac{N+1}{2} \bigg)$ th observation
$\qquad= \bigg( \large\frac{29+1}{2} \bigg)$ th value =15 th
Step 2:
Mean deviation about median $= \large\frac{\sum f_i |x_i-M|}{\sum f_i}$
$\qquad= \large\frac{148}{29}$
$\qquad= 5.1$
Hence A is the correct answer.
answered Jun 26, 2014 by meena.p
 
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