Q.
Find the mean deviation about the median for the following data.
$x_i$
$3$
$6$
$9$
$12$
$13$
$15$
$21$
$22$
$f_i$
$3$
$4$
$5$
$2$
$4$
$5$
$4$
$3$
| $x_i$ | $3$ | $6$ | $9$ | $12$ | $13$ | $15$ | $21$ | $22$ |
| $f_i$ | $3$ | $4$ | $5$ | $2$ | $4$ | $5$ | $4$ | $3$ |
Statistics
Solution:
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{$15^{th}$ observation + $16^{th}$ 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, $\sum\limits^{8}_{i = 1} f_{i} = 30$ and
$\sum\limits^{8}_{i = 1} f_{i} \left|x_{i}-M\right|=149$
Therefore M.D. (M) $=\frac{1}{N} \displaystyle \sum_{i=1}^{8}f_{i} \left|x_{i}-M\right|$
$= \frac{1}{30}\times149 = 4.97$
| $x_i$ | $3$ | $6$ | $9$ | $12$ | $13$ | $15$ | $21$ | $22$ |
| $f_i$ | $3$ | $4$ | $5$ | $2$ | $4$ | $5$ | $4$ | $3$ |
| $c.f.$ | $3$ | $7$ | $12$ | $14$ | $18$ | $23$ | $27$ | $30$ |
| $|x_i-M|$ | $10$ | $7$ | $4$ | $1$ | $0$ | $2$ | $8$ | $9$ |
| $f_i$ | $3$ | $4$ | $5$ | $2$ | $4$ | $5$ | $4$ | $3$ |
| $f_i|x_i-M|$ | $30$ | $28$ | $20$ | $2$ | $0$ | $10$ | $32$ | $27$ |