Question

Mean deviation about the median for the data $13,17,16,14,11,13,10,16,11,18,12,17$ is

• A. $2.44$
• B. $2.33$
• C. $1.44$
• D. $1.33$

Answer 1

Answer: (B)

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 $M=\frac{{\left(\frac{N}{2}\right)}^{\text{th }}\text{ observation }+{\left(\frac{N}{2}+1\right)}^{th}\text{ observation }}{2}$
$=\frac{{\left(\frac{12}{2}\right)}^{\text{th }}\text{ observation }+{\left(\frac{12}{2}+1\right)}^{\text{th }}\text{ observation }}{2}$
$=\frac{6^{\text{th }}\text{ observation }+7^{\text{th }}\text{ observation }}{2}$
$=\frac{13+14}{2}=\frac{27}{2}$
$\Rightarrow M=13.5$

$x_i$	$|x_i-M|$
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 $=\frac{\Sigma |x_i-M|}{n}$
$=\frac{28}{12}=2.33$