Q.
Consider the following data
$x_i$
15
21
27
30
35
$f_i$
3
5
6
7
8
Then, the mean deviation about the median for the data is
| $x_i$ | 15 | 21 | 27 | 30 | 35 |
| $f_i$ | 3 | 5 | 6 | 7 | 8 |
Statistics
Solution:
$x_i$
$f_i$
$cf$
$|x_i - M|$
$f_i |x_i - M|$
15
3
3
|15 - 30| = 15
45
21
5
8
|21 -30| = 9
45
27
6
14
|27- 30| = 3
18
30
7
21
|30-30| = 0
40
35
8
29
|35-30|=5
40
Total
$\Sigma f_i = 29$
$\Sigma f_i |x_i - M| = 148$
Here, $ N=\Sigma f=29$ (odd)
$\therefore$ Median $M=\left(\frac{N+1}{2}\right)^{\text {th }}$ observation
$=\left(\frac{29+1}{2}\right)^{\text {th }}$ observation $=15^{\text {th }}$ observation
$\Rightarrow M=30$
$\therefore$ Mean deviation about median
$=\frac{\Sigma f_i\left|x_i-M\right|}{\Sigma f_i}=\frac{148}{29}=5.1$
| $x_i$ | $f_i$ | $cf$ | $|x_i - M|$ | $f_i |x_i - M|$ |
|---|---|---|---|---|
| 15 | 3 | 3 | |15 - 30| = 15 | 45 |
| 21 | 5 | 8 | |21 -30| = 9 | 45 |
| 27 | 6 | 14 | |27- 30| = 3 | 18 |
| 30 | 7 | 21 | |30-30| = 0 | 40 |
| 35 | 8 | 29 | |35-30|=5 | 40 |
| Total | $\Sigma f_i = 29$ | $\Sigma f_i |x_i - M| = 148$ |