Question

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

• A. 6
• B. 8
• C. 7
• D. None of these

Answer 1

Answer: (C)

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 $M=\frac{{\left(\frac{N}{2}\right)}^{\text{th }}\text{ observation }+{\left(\frac{N}{2}+1\right)}^{\text{th }}\text{ observation }}{2}$
$=\frac{{\left(\frac{10}{2}\right)}^{\text{th }}\text{ observation }+{\left(\frac{10}{2}+1\right)}^{\text{th }}\text{ observation }}{2}$
$=\frac{5^{th}\text{ observation }+6^{\text{th }}\text{ observation }}{2}$
$=\frac{46+49}{2}=47.5$

$x_i$	$|x_i-M|$
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 $=\frac{\Sigma |x_i-M|}{n}=\frac{70}{10}=7$