Q.
Calculate the mean deviation about median for the following data.
Class
0-
10
10-
20
20-
30
30-
40
40-
50
50-
60
Frequency
6
7
15
16
4
2
| Class | 0- 10 | 10- 20 | 20- 30 | 30- 40 | 40- 50 | 50- 60 |
| Frequency | 6 | 7 | 15 | 16 | 4 | 2 |
Statistics
Solution:
Form the following table from the given data.
Class
Frequency $f_i$
Cumulative
frequency $(c.f.)$
Mid
points $x_i$
$|x_i -M|$
$f_i|x_i- M|$
0-10
6
6
5
23
138
10-20
7
13
15
13
91
20-30
15
28
25
3
45
30-40
16
44
35
7
112
40-50
4
48
45
17
68
50-60
2
50
55
27
54
50
508
Here $N = 50$
$\therefore \frac{N}{2}=25$
Therefore, $20-30$ is the median class.
Median $= l+\frac{\frac{N}{2}-C}{f}\times h$
Here $l = 20$, $C = 13$, $f= 15$, $h = 10$ and $N = 50$
Therefore, Median $= 20 +\frac{25-13}{15}\times10$
$= 20+8=28$
Thus, Mean deviation about median is given by
$M.D. \left(M\right)= \frac{1}{N} \sum\limits^{6}_{i = 1}f_{i}\left|x_{i}-M\right|$
$= \frac{1}{50}\times508 = 10.16$
| Class | Frequency $f_i$ | Cumulative frequency $(c.f.)$ | Mid points $x_i$ | $|x_i -M|$ | $f_i|x_i- M|$ |
|---|---|---|---|---|---|
| 0-10 | 6 | 6 | 5 | 23 | 138 |
| 10-20 | 7 | 13 | 15 | 13 | 91 |
| 20-30 | 15 | 28 | 25 | 3 | 45 |
| 30-40 | 16 | 44 | 35 | 7 | 112 |
| 40-50 | 4 | 48 | 45 | 17 | 68 |
| 50-60 | 2 | 50 | 55 | 27 | 54 |
| 50 | 508 |