Question

Calculate mean deviation from median for the given data

Class	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$2$	$3$	$8$	$14$	$8$	$3$	$2$

• A. $50$
• B. $10$
• C. $15$
• D. $17$

Answer 1

Answer: (B)

$10$

$\text{Class-Interval}$	$f_i$	Cummulative Frequency$(c.f)$	$\text{Mid-Value}\left(x_i\right)$	$|x_i-\text{Median}|$	$f_i|x_i-\text{Median}|$
$10-20$	$2$	$2$	$15$	$30$	$60$
$20-30$	$3$	$5$	$25$	$20$	$60$
$30-40$	$8$	$13$	$35$	$10$	$80$
$40-50$	$14$	$27$	$45$	$0$	$0$
$50-60$	$8$	$35$	$55$	$10$	$80$
$60-70$	$3$	$38$	$65$	$20$	$60$
$70-80$	$2$	$40$	$75$	$30$	$60$

$N=\sum f_i=40$

Here, $N=40$ , so $\frac{40}{2}=20$ and the cummulative frquency is just greater than $\frac{N}{2}$ is $27$.

Thus, $40-50$ is median class.

$l$ $=$ lower limit of median class $=40$

$F$ $=$ Cummulative frequency before median class $=13$

Difference between the class $=$ $h=10$

Frequency of median class $=$ $f=14$

Median $=l+\frac{\frac{N}{2}-F}{f}\times h=40+\frac{20-13}{14}\times 10=45$

$\sum f_i\times |x_i-\text{Median}|=400$

Mean deviation about the mean $=\frac{1}{N}\sum f_I\times |x_i-\text{Median}|=\frac{1}{40}\times 400=40=10$