Q. Mean deviation about the median for the data $13,17,16,14,11,13,10,16,11,18,12,17$ is
Statistics
Solution:
The given data is $13,17,16,14,11,13,10,16,11,18$, 12, 17
Arranging in ascending order,
$10,11,11,12,13,13,14,16,16,17,17,18$
Number of observations $=12$ (even)
Median $M=\frac{\left(\frac{N}{2}\right)^{\text {th }} \text { observation }+\left(\frac{N}{2}+1\right)^{t h} \text { observation }}{2}$
$=\frac{\left(\frac{12}{2}\right)^{\text {th }} \text { observation }+\left(\frac{12}{2}+1\right)^{\text {th }} \text { observation }}{2}$
$ =\frac{6^{\text {th }} \text { observation }+7^{\text {th }} \text { observation }}{2} $
$ =\frac{13+14}{2}=\frac{27}{2} $
$ \Rightarrow M =13.5$
$x_i$
$|x_i - M|$
10
$|10-13.5|=3.5$
11
|11-13.5| = 2.5
11
|11 -13.5 | = 2.5
12
| 12- 13.5| = 1.5
13
|13 - 13.5 | = 0.5
13
|13 - 13.5| = 0.5
16
|16 - 13.5| = 2.5
16
|16 - 13.5| = 2.5
17
|17 - 13.5| = 3.5
17
|17 - 13.5| = 3.5
18
|18 - 13.5| = 4.5
$\Sigma|x_i - M| = 28$
$ \therefore $ Mean deviation about median $ =\frac{\Sigma\left|x_i-M\right|}{n}$
$ =\frac{28}{12}=2.33$
| $x_i$ | $|x_i - M|$ |
|---|---|
| 10 | $|10-13.5|=3.5$ |
| 11 | |11-13.5| = 2.5 |
| 11 | |11 -13.5 | = 2.5 |
| 12 | | 12- 13.5| = 1.5 |
| 13 | |13 - 13.5 | = 0.5 |
| 13 | |13 - 13.5| = 0.5 |
| 16 | |16 - 13.5| = 2.5 |
| 16 | |16 - 13.5| = 2.5 |
| 17 | |17 - 13.5| = 3.5 |
| 17 | |17 - 13.5| = 3.5 |
| 18 | |18 - 13.5| = 4.5 |
| $\Sigma|x_i - M| = 28$ |