Question

Calculate the mean deviation for the following data about median.

Class interval	$0-4$	$5-9$	$10-14$	$15-19$	$20-24$	$25-29$
Frequency	$11$	$12$	$17$	$12$	$20$	$28$

• A. $10.22$
• B. $7.57$
• C. $5.55$
• D. $8.45$

Answer 1

The correct option is B $7.57$

$\text{Class Interval}$	$f_i$	Cummulative Frequency$\left(cf\right)$	$\text{Mid-Value}\left(x_i\right)$	$|x_i-\text{Median}|$	$f_i|x_i-\text{Median}|$
$0-4$	$11$	$11$	$2$	$16.67$	$183.33$
$5-9$	$12$	$23$	$7$	$11.67$	$140$
$10-14$	$17$	$40$	$12$	$6.67$	$113.33$
$15-19$	$12$	$52$	$17$	$1.67$	$20$
$20-24$	$20$	$72$	$22$	$3.33$	$66.67$
$25-29$	$28$	$100$	$27$	$8.33$	$233.33$
	$n=100$				$\sum f_i|x_i-\text{Median}|=756.67$

Median class $=$ value of $(\frac{n}{2}{)}^{th}$ observation $=$ value of ${50}^{th}$ observation $=$ class $(15-19)$

Median class will be $14.5-19.5$.

$L$ $=$ Lower boundary point of median class $=$ $14.5$

$n$ $=$ Total frequency $=$ $100$

$F$ $=$ Cummulative frequency of class before median class $=$ $40$

$f$ $=$ Frequency of the median class $=$ $12$

$h$ $=$ class length of median class $=$ $5$

Median $=$ $L+\frac{\frac{n}{2}-F}{f}\times h=14.5+\frac{50-40}{12}\times 5=14.5+4.17=18.67$

Mean deviation $\frac{\sum f_i|x_i-\text{Median}|}{n}=\frac{756.67}{100}\approx 7.57$.