Q.
Consider the following data
$36,72,46,42,60,45,53,46,51,49$
Then, the mean deviation about the median for the data is
Statistics
Solution:
The given data is $36,72,46,42,60,45,53,46,51,49$
Arranging the data in ascending order,
$36,42,45,46,46,49,51,53,60,72$
Number of observations $=10$ (even)
Median $M =\frac{\left(\frac{N}{2}\right)^{\text {th }} \text { observation }+\left(\frac{N}{2}+1\right)^{\text {th }} \text { observation }}{2} $
$ =\frac{\left(\frac{10}{2}\right)^{\text {th }} \text { observation }+\left(\frac{10}{2}+1\right)^{\text {th }} \text { observation }}{2} $
$ =\frac{5^{t h} \text { observation }+6^{\text {th }} \text { observation }}{2} $
$ =\frac{46+49}{2}=47.5$
$x_i$
$|x_i - M|$
36
|36-47.5|=11.5
42
|42-47.5| = 5.5
45
|45-47.5 | = 2.5
46
| 46- 47.5| = 1.5
46
|46 - 47.5 | = 1.5
49
|49 - 47.5| = 1.5
51
|51 - 47.5| = 3.5
53
|53 - 47.5| = 5.5
60
|60 - 47.5| = 12.5
72
|72 - 47.5| = 24.5
$\Sigma|x_i - M| = 70$
$\therefore$ Mean deviation about median $=\frac{\Sigma\left|x_i-M\right|}{n}=\frac{70}{10}=7$
| $x_i$ | $|x_i - M|$ |
|---|---|
| 36 | |36-47.5|=11.5 |
| 42 | |42-47.5| = 5.5 |
| 45 | |45-47.5 | = 2.5 |
| 46 | | 46- 47.5| = 1.5 |
| 46 | |46 - 47.5 | = 1.5 |
| 49 | |49 - 47.5| = 1.5 |
| 51 | |51 - 47.5| = 3.5 |
| 53 | |53 - 47.5| = 5.5 |
| 60 | |60 - 47.5| = 12.5 |
| 72 | |72 - 47.5| = 24.5 |
| $\Sigma|x_i - M| = 70$ |