Question

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

• A. 5
• B. $5.3$
• C. $5.1$
• D. $5.2$

Answer 1

Answer: (C)

$x_i$	$f_i$	$cf$	$|x_i-M|$	$f_i|x_i-M|$
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
$=\frac{\Sigma f_i|x_i-M|}{\Sigma f_i}=\frac{148}{29}=5.1$