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Q. Statement I Mean deviation about the mean for the data $6,7,10,12,13,4,8,12$ is $2.75$.
Statement II Mean deviation about the mean for the data $12,3,18,17,4,9,17,19,20,15,8,17,2,3$, $16,11,3,1,0,5$ is $3.1$.

Statistics

Solution:

I. We proceed stepwise and get the following
Step I Mean of the given data is
$\bar{x}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9$
Step II The deviations of the respective observations from the mean $\bar{x}$, i.e., $x_i-\bar{x}$ are
$6-9,7-9,10-9,12-9,13-9,4-9,8-9,12-9 \text {, }$
or $-3,-2,1,3,4,-5,-1,3$.
Step III The absolute values of the deviations, i.e., $\left|x_i-\bar{x}\right|$ are $3,2,1,3,4,5,1,3$.
Step IV The required mean deviation about the mean is
$\operatorname{MD}(\bar{x}) =\frac{ \displaystyle\sum_{i=1}^8\left|x_i-\bar{x}\right|}{8} $
$ =\frac{3+2+1+3+4+5+1+3}{8}=\frac{22}{8}=2.75$
Note Instead of carrying out the steps every time, we can carry on calculation, stepwise without referring to steps.
II. We have to first find the mean of the given data
$(\bar{x})=\frac{1}{20} \sum_{i=1}^{20} x_i=\frac{200}{20}=10$
The respective absolute values of the deviations from mean, i.e., $\left|x_i-\bar{x}\right|$ are $2,7,8,7,6,1,7,9,10,5,2,7,8,7$, $6,1,7,9,10,5$
Therefore, $ \displaystyle\sum_{i=1}^{20}\left|x_i-\bar{x}\right|=124$
and
$\operatorname{MD}(\bar{x}) =\frac{124}{20} $
$=6.2$