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 \[10.34\]