Solution:-
Draw the table of the given data and append other columns after calculations.
The class interval containing \[{{N}^{th}}/2\]or \[25\] item is \[20-30\]
So, \[20-30\]is the median class.
Then,
Median = l + (((N/ \[2\]) – c)/f) × h
Where, l = \[20\], c = \[14\], f = \[14\], h = \[10\] and n = \[50\]
Median = \[20+(((25-14))/14)\times 10\]
= \[20+7.85\]
= \[27.85\]
So \[\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-Med \right|=517.1}\]
And M.D.(M) = \[\frac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-Med \right|}\]
=\[(1/50)\times 517.1\]
=\[10.34\]
Therefore, the mean deviationabout the median is \[10.34\]