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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Find the mean deviation about the mean for data : $x_i :5,7,9,10,12,15 \qquad f_i : 8,6,2,2,2,6$

$\begin{array}{1 1}(A)\;3.23\\(B)\;3\\(C)\;2\\(D)\;7\end{array} $

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1 Answer

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Step 1:
$x_i =5,\qquad f_i=8, \qquad CF=8,\qquad |x_i -M|=|5-7|=2 \qquad f_i |x_i-M|=16$
$x_i =7,\qquad f_i=6, \qquad CF=14,\qquad |x_i -M|=|7-7|=0 \qquad f_i |x_i-M|=00$
$x_i =9,\qquad f_i=2, \qquad CF=16,\qquad |x_i -M|=|9-7|=2 \qquad f_i |x_i-M|=04$
$x_i =10,\qquad f_i=2, \qquad CF=18,\qquad |x_i -M|=|10-7|=3 \qquad f_i |x_i-M|=06$
$x_i =12,\qquad f_i=2, \qquad CF=20,\qquad |x_i -M|=|12-7|=5 \qquad f_i |x_i-M|=10$
$x_i =15,\qquad f_i=6, \qquad CF=26,\qquad |x_i -M|=|5-7|=8 \qquad f_i |x_i-M|=48$
Total $\sum f_i=26$
$\sum f_i |x_i -M|=84$
Here, $N= \sum f_i =26(even)$
Median $M= \large\frac{N/2 th \;observation+ (n/2+1) th \;observation}{2}$
$\qquad= \large\frac{26/2 th \;observation+ (26/2+1) th \;observation}{2}$
$\qquad= \large\frac{13th \;observation+14th\;observation}{2}$
$\qquad = \large\frac{7+7}{2}$
$\qquad= \frac{14}{2}$
$\qquad= 7$
Step 2:
Mean deviation about median $= \large\frac{\sum f_i |x_i-M|}{\sum f_i}=\frac{84}{26}$$=3.23$
Hence A is the correct answer.
answered Jun 25, 2014 by meena.p
 
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