logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Statistics
0 votes

Following are the marks obtained by $9$ students in a mathematics test: $50,69,20,33,53,39,40,65,59$ The mean deviation from the median is :

$\begin{array}{1 1}(A)\; 9\\(B)\;10.5\\(C)\;12.67\\(D)\;14.76\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Median M $ = \bigg( \large\frac{N+1}{2} \bigg) $ th observation
  • Mean deviation about mean = $ \large\frac{\sum|x_i -M|}{n}$
Median M $ = \bigg( \large\frac{N+1}{2} \bigg) $ th observation
N=9
$\therefore \bigg(\large\frac{9+1}{2}\bigg)$th observation.
$\qquad= \large\frac{10}{2} $th observation
$\qquad = 5$ th observation
Median (M)=53
$x_i =50; \qquad |x_i-M|=|50-53|=3$
$x_i =69; \qquad |x_i-M|=|69-53|=16$
$x_i =20; \qquad |x_i-M|=|20-53|=33$
$x_i =33; \qquad |x_i-M|=|33-53|=20$
$x_i =53; \qquad |x_i-M|=|53-53|=0$
$x_i =39; \qquad |x_i-M|=|39-53|=14$
$x_i =40; \qquad |x_i-M|=|40-53|=13$
$x_i =65; \qquad |x_i-M|=|65-53|=12$
$x_i =59; \qquad |x_i-M|=|59-53|=6$
$\qquad =117$
Mean deviation about mean = $ \large\frac{\sum|x_i -M|}{n}$
$\qquad= \large\frac{117}{9} $
$\qquad= 13$
The nearest option is $12.67$
Hence C is the correct answer.
answered Jul 8, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...