Find the mean deviation about the median for the following data:
$\begin{array}{ccccccc}{x}_i& 5& 7& 9& 10& 12& 15\\ f_i& 8& 6& 2& 2& 2& 6\end{array}$

Finding cumulative frequencies

$\begin{array}{ccc}{x}_i& f_i& c.f\\ 5& 8& 8\\ 7& 6& 8+6=14\\ 9& 2& 14+2=16\\ 10& 2& 16+2=18\\ 12& 2& 18+2=20\\ 15& 6& 20+6=26\end{array}$

$N=\sum f_i=26$

$\begin{array}{ccc}{x}_i& f_i& c.f\\ 5& 8& 8\\ 7& 6& 8+6=14\end{array}$

N = 26 (even)

$\therefore$ Median = Mean of ${\left(\frac{N}{2}\right)}^{th}$ term and ${(\frac{N}{2}+1)}^{th}$ term

Median $=\frac{{\left(\frac{26}{2}\right)}^{th}\text{term}+{(\frac{26}{2}+1)}^{th}\text{term}}{2}$

$=\frac{(13{)}^{th}\text{term}+(14{)}^{th}\text{term}}{2}$

As both observations lie in the c.f. of 14.
$\therefore {13}^{th}$ observation $={14}^{th}$ observation = 7

$\therefore$Median $=\frac{7+7}{2}=7$

Median $M=7$
$\begin{array}{cccc}{x}_i& f_i& |x_i-M|& f_i|x_i-M|\\ 5& 8& 2& 16\\ 7& 6& =0& 0\\ 9& 2& 2& 4\\ 10& 2& 3& 6\\ 12& 2& 5& 10\\ 15& 6& 8& 48\\ & \sum f_i=26& & \sum f_i|x_i-M|=84\end{array}$

Mean deviation about median $=\frac{\sum f_i|x_i-M|}{\sum f_i}$

Hence, $M.D.\left(M\right)=\frac{84}{26}=3.23$