Find the mean deviation about the median for the following data:
$\begin{array}{ccccccccc}{x}_i& 3& 6& 9& 12& 13& 15& 21& 22\\ f_i& 3& 4& 5& 2& 4& 5& 4& 3\end{array}$

Step 1: Finding cumulative frequencies and median
The given observations are already in ascending order. So, finding cumulative frequencies to given data,
$\begin{array}{ccc}{x}_i& f_i& c.f.\\ 3& 3& 3\\ 6& 4& 3+4=7\\ 9& 5& 7+5=12\\ 12& 2& 12+2=14\\ 13& 4& 14+4=18\\ 15& 5& 18+5=23\\ 21& 4& 23+4=27\\ 22& 3& 27+3=30\end{array}$

$N=\sum f_i=30$

$\begin{array}{ccc}{x}_i& f_i& c.f\\ 3& 3& 3\\ 6& 4& 7\\ 9& 5& 12\\ 12& 2& 14\\ 13& 4& 18\end{array}$

$\because N=30$ (even)

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

$=\frac{(15{)}^{th}\text{term}+(16{)}^{th}\text{term}}{2}$

As both observations lie in the c.f. of 18.

$\therefore {15}^{th}$ observation $={16}^{th}$ observation = 13

Median $=\frac{13+13}{2}=13$

Median, $M=13$
$\begin{array}{cccc}{x}_i& f_i& |x_i-M|& f_i\times |x_i-M|\\ 3& 3& |3-13|=10& 3\times 10=30\\ 6& 4& |6-13|=7& 4\times 7=28\\ 9& 5& |9-13|=4& 5\times 4=20\\ 12& 2& |12-13|=1& 2\times 1=2\\ 13& 4& |13-13|=0& 4\times 0=0\\ 15& 5& |15-13|=2& 5\times 2=10\\ 21& 4& |21-13|=8& 4\times 8=32\\ 22& 3& |22-13|=9& 3\times 9=27\\ & \sum f_i=30& & \sum f_i|x_i-M|=149\end{array}$

$\sum f_i=30\sum f_i|x_i-M|=149$

Mean deviation about median $=\frac{\sum f_i|x_i-M|}{\sum f_i}$
Hence, $M.D.\left(M\right)=\frac{149}{30}=4.97$