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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Calculate the mean deviation from the median of the following frequency data:

$\begin{array}{1 1}(A)\;4.15\\(B)\;7.08\\(C)\;790\\(D)\;9.2\end{array} $

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1 Answer

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Toolbox:
  • The formula to solve this problem:
  • Median $= l+\large\frac{\Large\frac {N}{2}-C}{f} $$\times h$
  • Mean deviation about the mean $= \large\frac{\sum f_i |x_i - \bar {x} |}{\sum f_i}$
Step 1:
As $ N=20 $ is even
$\therefore \large\frac{(N/2)\; th \;observation + (N/2+)th\; observation }{2}$
$\qquad= \large\frac{(20/2)\; th \;observation + (20/2+)th\; observation }{2}$
$\qquad= \large\frac{10 th \;observation + 11th\; observation }{2}$
$\qquad= \large\frac{5+5}{2}$$=5$
The class is $6-12$
$l=6,c=4,f=5,h=6$
Step 2:
Median $= l+\large\frac{\Large\frac {N}{2}-C}{f} $$\times h$
$\qquad= 6+ \large\frac{\Large\frac{20}{2} -4}{5}$$ \times 6$
$\qquad= 6+\large\frac{36}{5}$
$\qquad= 6+7.2$
$\qquad= 13.2$
Step 3:
Mean deviation about the mean $= \large\frac{\sum f_i |x_i - \bar {x} |}{\sum f_i}$
$\qquad= \large\frac{141.6}{20} $$=7.08$
Hence B is the correct answer.
answered Jul 4, 2014 by meena.p
 

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