Find the mean deviation about the median of the following distribution:
$ \begin{array}{|l|l|l|l|l|l|} \hline \text{Marks obtained} & 10 & 11 & 12 & 14 & 15 \\ \hline \text{Number of students} & 2 & 3 & 8 & 3 & 4 \\ \hline \end{array} $

(C) First,we find the cumulative frequency $(cf)$ and the median $(M_e)$.
$ \begin{array}{|c|c|c|c|c|} \hline \text{Marks} (x_i) & f_i & cf & |x_i - M_e| & f_i |x_i - M_e| \\ \hline 10 & 2 & 2 & |10-12|=2 & 4 \\ \hline 11 & 3 & 5 & |11-12|=1 & 3 \\ \hline 12 & 8 & 13 & |12-12|=0 & 0 \\ \hline 14 & 3 & 16 & |14-12|=2 & 6 \\ \hline 15 & 4 & 20 & |15-12|=3 & 12 \\ \hline \text{Total} & \Sigma f_i = 20 & & & \Sigma f_i |x_i - M_e| = 25 \\ \hline \end{array} $
Total number of observations $N = \Sigma f_i = 20$.
Since $N$ is even,the median $M_e$ is the average of the $10^{th}$ and $11^{th}$ observations. Looking at the $cf$ column,both the $10^{th}$ and $11^{th}$ observations fall in the value $12$.
Therefore,$M_e = 12$.
The mean deviation about the median is given by:
$MD(M_e) = \frac{\Sigma f_i |x_i - M_e|}{\Sigma f_i} = \frac{25}{20} = 1.25$.