Q.
Calculate the mean deviation about the median for the following data.
Height
(in cm)
95--
105
105-
115
115-
125
125-
135
135-
145
145-
155
Number
of boys
9
13
25
30
13
10
| Height (in cm) | 95-- 105 | 105- 115 | 115- 125 | 125- 135 | 135- 145 | 145- 155 |
| Number of boys | 9 | 13 | 25 | 30 | 13 | 10 |
Statistics
Solution:
First we find the median
Class
Frequency $f_i$
Cumulative
Frequency $(C)$
Midvalue $(x_i)$
95 - 105
9
9
100
105-115
13
22
110
115-125
25
47
120
125-135
30
77
130
135 - 145
13
90
140
145 -155
10
100
150
$N = \Sigma f_{i} = 100$
Thus $N = 100$ and therefore, $\frac{N}{2} = 50$
$\Rightarrow $ median class is $125 - 135$.
$\Rightarrow l = 125$, $f= 30$, $h= 10$ and $C = 47$
$\therefore $ Median $= l + \frac{\left(\frac{N}{2}-C\right)}{f}\times h$
$= \left\{125+\frac{\left(50-47\right)}{30}\times10\right\}=126$
$\therefore M= 126$.
Now, we prepare the table given below.
$f_i$
$x_i$
$|x_i-M|$
$f_i|x_i-M|$
9
100
26
234
13
110
16
208
25
120
6
15
30
130
4
120
13
140
14
182
10
150
24
240
N=100
1134
$\therefore M.D. \left(M\right) = \frac{\Sigma f_{i}\left|x_{i}-M\right|}{N}$
$=\frac{1134}{100} = 11.34$.
| Class | Frequency $f_i$ | Cumulative Frequency $(C)$ | Midvalue $(x_i)$ |
|---|---|---|---|
| 95 - 105 | 9 | 9 | 100 |
| 105-115 | 13 | 22 | 110 |
| 115-125 | 25 | 47 | 120 |
| 125-135 | 30 | 77 | 130 |
| 135 - 145 | 13 | 90 | 140 |
| 145 -155 | 10 | 100 | 150 |
| $N = \Sigma f_{i} = 100$ |
| $f_i$ | $x_i$ | $|x_i-M|$ | $f_i|x_i-M|$ |
|---|---|---|---|
| 9 | 100 | 26 | 234 |
| 13 | 110 | 16 | 208 |
| 25 | 120 | 6 | 15 |
| 30 | 130 | 4 | 120 |
| 13 | 140 | 14 | 182 |
| 10 | 150 | 24 | 240 |
| N=100 | 1134 |