# Find the mean deviation about the mean for the data $4,7,8,9,10,12,13,17$

$\begin{array}{1 1}(A)\;1\\(B)\;2\\(C)\;3\\(D)\;4\end{array}$

Toolbox:
• Mean deviation about can be calculated by the formula $= \large\frac{\sum |X_i- \bar{X}|}{n}$ where $\bar {X}=$ mean of the given observations.
Step 1:
Mean of the given series $\bar {X}=\large\frac{Sum\;of\;terms}{Number\;of\;terms}$
$\qquad= \large\frac{\sum X_i}{n}$
$\qquad= \large\frac{4+7+8+9+10+12+13+17}{8}$$=10$
$X_i =4 , \qquad |X_i -\bar{X} |=|4-10|=6$
$X_i =7 , \qquad |X_i -\bar{X} |=|7-10|=3$
$X_i =8 , \qquad |X_i -\bar{X} |=|8-10|=2$
$X_i =9 , \qquad |X_i -\bar{X} |=|9-10|=1$
$X_i =10 , \qquad |X_i -\bar{X} |=|10-10|=0$
$X_i =12 , \qquad |X_i -\bar{X} |=|12-10|=2$
$X_i =13 , \qquad |X_i -\bar{X} |=|13-10|=3$
$X_i =17 , \qquad |X_i -\bar{X} |=|17-10|=7$
$\sum x_i =80 \qquad \sum |x_i -\bar{x}|=24$
Step 2:
$\therefore$ meandeviation about mean $=\large\frac{\sum |x_i - \bar {x}|}{n}$
$\qquad= \large\frac{24}{8}$
$\qquad=3$
Hence C is the correct answer.