# Find the mean deviation about the mean for the data $13,17,16,14,11,13,10,16,11,18,12,17$

$\begin{array}{1 1}(A)\;2.33\\(B)\;4.55\\(C)\;9.56\\(D)\;7.56 \end{array}$

Toolbox:
• First arrange the given observations in ascending order and then calculate the median (n) after that mean deviation about the mean can be calculated by the reaction $=\large\frac{\sum |x -M|}{n}$ where $n$=number of terms
Step 1:
$\therefore$ Arranging the data in ascending order.
$10,11,11,12,13,13,14,16,16,17,17,18$
Number of observation $=12 (even)$
Median $M=\Large\frac {N/2 observation + (N/2+1)th observation}{2}$
$\qquad= \large\frac{12/2 observation +7th \;observation }{2}$
$\qquad=\large\frac{13+14}{2} =\frac{27}{2}$$=13.5$
$M= 13.5$
Step 2:
$X_i =10 , \qquad |X_i -M |=|10-13.5|=3.5$
$X_i =11, \qquad |X_i -M |=|11-13.5|=2.5$
$X_i =11 , \qquad |X_i -M |=|11-13.5|=2.5$
$X_i =12 , \qquad |X_i -M |=|12-13.5|=1.5$
$X_i =13 , \qquad |X_i -M |=|13-13.5|=0.5$
$X_i =13, \qquad |X_i -M |=|13-13.5|=0.5$
$X_i =14, \qquad |X_i -M |=|14-13.5|=0.5$
$X_i =16 , \qquad |X_i -M |=|16-13.5|=2.5$
$X_i =16 , \qquad |X_i -M |=|16-13.5|=2.5$
$X_i =17 , \qquad |X_i -M |=|17-13.5|=3.5$
$X_i =17 , \qquad |X_i -M |=|17-13.5|=3.5$
$X_i =18 , \qquad |X_i -M |=|18-13.5|=4.5$
$\qquad \sum |x_i-M|=28$
Step 3:
Mean deviation about median $=\large\frac{ \sum |x_i-M|}{n}$
$\qquad= \large\frac{28}{12}$
$\qquad=2.33$
Hence A is the correct answer.