Solution:
The frequency distribution is given
We now need to find the mean deviation from the median
Let’s construct a table of the given data and append other columns after calculations
$$\begin{tabular}{|l|l|l|l|}
\hline Class interval & Mid – Value $\left(\mathrm{x}_{\mathrm{i}}\right)$ & Frequency $\left(\mathrm{f}_{\mathrm{i}}\right)$ & $\mathrm{CF}$ \\
\hline $0-6$ & 3 & 4 & 4 \\
\hline $6-12$ & 9 & 5 & 9 \\
\hline $12-18$ & 15 & 3 & 12 \\
\hline $18-24$ & 21 & 6 & 18 \\
\hline $24-30$ & 27 & 2 & 20 \\
\hline & total & 20 & \\
\hline
\end{tabular}$$
Here $\mathrm{N}=20$, which is even.
Also here $\mathrm{N}=20$, which is even.
Here the median class $=\frac{\mathrm{N}}{2}=10^{\text {th }}$ term
This observation lies in the class interval 12-18, so we can write the median as,
$\mathrm{M}=1+\frac{\frac{\mathrm{N}}{2}-\mathrm{cf}}{\mathrm{f}} \times \mathrm{h}$
Here $l=12, c f=9, f=3, h=6$ and $N=20$, substituting these values, the above equation becomes,
$\mathrm{M}=12+\frac{\frac{20}{2}-9}{3} \times 6$
$\Rightarrow \mathrm{M}=12+\frac{10-9}{3} \times 6$
$\Rightarrow \mathrm{M}=12+\frac{1 \times 6}{3}$
$\Rightarrow \mathrm{M}=12+2=14$
$$\begin{tabular}{|l|l|l|l|l|l|}
\hline Class interval & Mid $-$ Value $\left(\mathrm{x}_{\mathrm{i}}\right)$ & Frequency $\left(\mathrm{f}_{\mathrm{i}}\right)$ & $\mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}$ & $\mathrm{d}_{\mathrm{i}}=\mid \mathrm{x}_{\mathrm{i}}-$ mean $\mid$ & $\mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}$ \\
\hline $0-6$ & 3 & 4 & 4 & 11 & \\
\hline $6-12$ & 9 & 5 & 9 & 5 & 44 \\
\hline $12-18$ & 15 & 3 & 12 & 1 & 25 \\
\hline $18-24$ & 21 & 6 & 18 & 7 & 3 \\
\hline $24-30$ & 27 & 2 & 20 & 13 & 42 \\
\hline & total & 20 & & & 26 \\
\hline
\end{tabular}$$
Therefore the Mean Deviation becomes,
$MD=\frac{\sum f_{i} d_{i}}{\sum f_{i}}=\frac{140}{20}=7$
As a result, 7 is the mean deviation about the median of the distribution.