Find the mean deviation about the median for the following data:

$x_i$	$3$	$6$	$9$	$12$	$13$	$15$	$21$	$22$
$f_i$	$3$	$4$	$5$	$2$	$4$	$5$	$4$	$3$

The given observations are already in ascending order. Adding a row corresponding to cumulative frequencies $(c.f.)$ to the given data:

$x_i$	$3$	$6$	$9$	$12$	$13$	$15$	$21$	$22$
$f_i$	$3$	$4$	$5$	$2$	$4$	$5$	$4$	$3$
$c.f.$	$3$	$7$	$12$	$14$	$18$	$23$	$27$	$30$

Now,$N = 30$,which is even.
Median is the mean of the $15^{\text{th}}$ and $16^{\text{th}}$ observations. Both of these observations lie in the cumulative frequency $18$,for which the corresponding observation is $13$.
Therefore,Median $M = \frac{15^{\text{th}} \text{ observation} + 16^{\text{th}} \text{ observation}}{2} = \frac{13+13}{2} = 13$.
Now,absolute values of the deviations from median,$|x_i - M|$,are calculated:

$|x_i - M|$	$|3-13|=10$	$|6-13|=7$	$|9-13|=4$	$|12-13|=1$	$|13-13|=0$	$|15-13|=2$	$|21-13|=8$	$|22-13|=9$
$f_i$	$3$	$4$	$5$	$2$	$4$	$5$	$4$	$3$
$f_i|x_i - M|$	$30$	$28$	$20$	$2$	$0$	$10$	$32$	$27$

We have $\sum f_i = 30$ and $\sum f_i|x_i - M| = 30+28+20+2+0+10+32+27 = 149$.
Therefore,$M.D.(M) = \frac{1}{N} \sum f_i|x_i - M| = \frac{149}{30} = 4.966... \approx 4.97$.