If $m$ and M denote the mean deviations about mean and about median respectively of the data $20,5.15,2,7,3…
If $m$ and M denote the mean deviations about mean and about median respectively of the data $20,5.15,2,7,3,11$ then the mean deviation about the mean of $m$ and $M$ is
$\frac{1}{7}$
$\frac{38}{7}$
$\frac{36}{7}$
$\frac{37}{7}$
Solution
Since, mean of the given data
$\bar{x}=\frac{20+5+15+2+7+3+11}{7}=9$
Now, mean deviation about mean
$\begin{aligned}
= & \frac{|9-20|+|9-5|+|9-15|+|9-2|+|9-7|+|9-3|+|9-11|}{7}=\frac{38}{7}=m
\end{aligned}$
Since, Median $=\left(\frac{7+1}{2}\right)^{\text {th }}$ term $=4^{\text {th }}$ term $=7$ (after ascending order)
Now, mean deviation about median
$=\frac{|20-7|+|5-7|+|15-7|+|2-7|+|7-7|+|3-7|+|11-7|}{7}=\frac{36}{7}=M$
Let $\bar{x}^{\prime}=\frac{1}{2}\left(\frac{36}{7}+\frac{38}{7}\right)=\frac{74}{2 \times 7}=\frac{37}{7}$
So, mean deviation about mean
$=\frac{1}{2}\left(\left|\frac{37}{7}-\frac{38}{7}\right|+\left|\frac{37}{7}-\frac{36}{7}\right|\right)=\frac{1}{2}\left(\frac{2}{7}\right)=\frac{1}{7} .$