Solution:-

Draw a table of the given data and append other columns after calculations.

Now, N = 29, which is odd.

The cumulative frequency greater than \[14.5\] is \[21\], for which the corresponding observation is \[30\].

Median = (\[15\] observation + \[16\] observation)/ \[2\]

= \[(30+30)/2\]

= \[60/2\]

= \[30\]

Therefore, \[\sum\limits_{i=1}^{5}{{{f}_{i}}}=29\]and \[\sum\limits_{i=1}^{5}{{{f}_{i}}}\left| {{x}_{i}}-M \right|=148\]

Mean deviation(M) =\[\frac{1}{N}\sum\limits_{i=1}^{5}{{{f}_{i}}}\left| {{x}_{i}}-M \right|\]

=\[(1/29)\times 148\]

=\[5.1\]

Therefore, the mean deviation about the median is \[5.1\]