Solution:-

The given data is converted into continuous frequency distribution by subtracting \[0.5\] from the lower limit and adding the \[0.5\] to the upper limit of each class intervals and append other columns after calculations.

The class interval containing \[{{N}^{th}}/2\]or \[50\] item is \[35.5-40.5\]

So, \[35.5-40.5\]is the median class.

Then,

Median = l + (((N/ \[2\]) – c)/f) × h

Where, l = \[35.5\], c = \[37\], f = \[26\], h = \[5\] and n = \[100\]

Median = \[35.5+(((50-37))/26)\times 5\]

= \[35.5+2.5\]

= \[38\]

So \[\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-Med \right|=735}\]

And M.D.(M) = \[\frac{1}{N}\sum\limits_{i=1}^{6}{{{f}_{i}}\left| {{x}_{i}}-Med \right|}\]

=\[(1/100)\times 735\]

=\[7.35\]

Therefore, the mean deviationabout the median is \[7.35\]