Question

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

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

• A. $4.97$
• B. $5.9$
• C. $2.99$
• D. $3.97$

Answer 1

Answer: (A)

The given observations are already in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get

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

Now, $N=30$, which is even.
$\therefore$ Median is the mean of the $15^{th}$ and $16^{th}$ observations. Both of the these observations lie in the cumulative frequency $18$, for which the corresponding observation is $13$.
$\therefore$ Median $M=\frac{\text{15th observation + 16th observation}}{2}$
$=\frac{13+13}{2}=13$
Now,

$|x_i-M|$	$10$	$7$	$4$	$1$	$0$	$2$	$8$	$9$
$f_i$	$3$	$4$	$5$	$2$	$4$	$5$	$4$	$3$
$f_i|x_i-M|$	$30$	$28$	$20$	$2$	$0$	$10$	$32$	$27$

We have, $\sum _{i=1}^8{f}_i=30$ and
$\sum _{i=1}^8{f}_i\left|x_i-M\right|=149$
Therefore M.D. (M) $=\frac{1}{N}\sum _{i=1}^8{f}_i\left|x_i-M\right|$ $=\frac{1}{30}\times 149=4.97$