Q.
Calculate the mean deviation about the median for the following data.
Class
16-
20
21-
25
26-
30
31-
35
36-
40
41-
45
46-
50
51-
55
Frequency
5
6
12
14
26
12
16
9
| Class | 16- 20 | 21- 25 | 26- 30 | 31- 35 | 36- 40 | 41- 45 | 46- 50 | 51- 55 |
| Frequency | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
Statistics
Solution:
Converting the given series into an exclusive series, we prepare the table given below.
Class
Frequency $f_i$
$C$
Midvalue $(x_i)$
15.5-20.5
5
5
18
20.5-25.5
6
11
23
25.5-30.5
12
23
28
30.5-35.5
14
37
33
35.5-40.5
26
63
38
40.5-45.5
12
75
43
45.5-50.5
16
91
48
50.5-55.5
9
100
53
N = 100
Thus $N = 100$ and therefore, $\frac{N}{2} = 50$
$\Rightarrow $ median class is $35.5 - 40.5$
$\Rightarrow l = 35.5$, $f= 26$, $h= 5$ and $C = 37$
$\therefore $ Median $= l + \frac{\left(\frac{N}{2}-C\right)}{f}\times h$
$=\left\{35.5+\frac{\left(50-37\right)}{26}\times5\right\}$
$\left(35.5+2.5\right)=38$.
Thus, $M= 38$.
Now, we prepare the table given below.
$f_i$
$x_i$
$|x_i-M|$
$f_i|x_i-M|$
5
18
20
100
6
23
15
90
12
28
10
120
14
33
5
70
26
38
0
0
12
43
5
60
16
48
10
160
9
53
15
135
N=100
735
Thus, $\sum f_{i} \left|x_{i}-M\right| = 735$ and $N = 100$
$\therefore M.D. \left(M\right) = \frac{\sum f_{i}\left|x_{i}-M\right|}{N}$
$=\frac{735}{100} = 7.35$.
Hence, the mean deviation about the median is $7.35$.
| Class | Frequency $f_i$ | $C$ | Midvalue $(x_i)$ |
|---|---|---|---|
| 15.5-20.5 | 5 | 5 | 18 |
| 20.5-25.5 | 6 | 11 | 23 |
| 25.5-30.5 | 12 | 23 | 28 |
| 30.5-35.5 | 14 | 37 | 33 |
| 35.5-40.5 | 26 | 63 | 38 |
| 40.5-45.5 | 12 | 75 | 43 |
| 45.5-50.5 | 16 | 91 | 48 |
| 50.5-55.5 | 9 | 100 | 53 |
| N = 100 |
| $f_i$ | $x_i$ | $|x_i-M|$ | $f_i|x_i-M|$ |
|---|---|---|---|
| 5 | 18 | 20 | 100 |
| 6 | 23 | 15 | 90 |
| 12 | 28 | 10 | 120 |
| 14 | 33 | 5 | 70 |
| 26 | 38 | 0 | 0 |
| 12 | 43 | 5 | 60 |
| 16 | 48 | 10 | 160 |
| 9 | 53 | 15 | 135 |
| N=100 | 735 |