Question

Compute the median for each of the following data:

(i)Marks	No. of students	(ii)Marks	No. of students
Less than $10$	$0$	More than $150$	$0$
Less than $30$	$10$	More than $140$	$12$
Less than $50$	$25$	More than $130$	$27$
Less than $70$	$43$	More than $120$	$60$
Less than $90$	$65$	More than $110$	$105$
Less than $110$	$87$	More than $100$	$124$
Less than $130$	$96$	More than $90$	$141$
Less than $150$	$100$	More than $80$	$150$

Answer 1

$\left(i\right)$

Class interval	Frequency	Cumulative frequency
$0-10$	$0$	$0$
$10-30$	$10$	$10$
$30-50$	$15$	$25$
$50-70$	$18$	$43$
$70-90$	$22$	$65$
$90-110$	$22$	$87$
$110-130$	$9$	$96$
$130-150$	$4$	$100$
	$N=100$

We have, $N=100,\frac{N}{2}=50$

The cumulative frequency just greater than $\frac{N}{2}$ is $65$ then median class is $70-90$ such that,

$l=70,h=90-70=20,f=22,cf=43$

$Median=l+\frac{\frac{N}{2}-cf}{f}\times h$

$\Rightarrow$ $Median=70+\frac{50-43}{22}\times 20$

$\Rightarrow$ $Median=70+\frac{7\times 20}{22}$

$\Rightarrow$ $Median=70+6.36$

$\therefore$ $Median=76.36$

$\left(ii\right)$

Class interval	Frequency	Cumulative Frequency
$80-90$	$9$	$9$
$90-100$	$17$	$26$
$100-110$	$19$	$45$
$110-120$	$45$	$90$
$120-130$	$33$	$123$
$130-140$	$15$	$138$
$140-150$	$12$	$150$
$150-160$	$0$	$150$
	$N=150$

We have, $N=150,$ then, $\frac{N}{2}=\frac{150}{2}=75$

The cumulative frequency just more than $\frac{N}{2}$ is $90$ then the median class is $110-120$ such that

$l=70,h=120-110=10,f=45,cf=45$

$Median=l+\frac{\frac{N}{2}-cf}{f}\times h$

$\Rightarrow$ $Median=110+\frac{75-45}{45}\times 10$

$\Rightarrow$ $Median=110+\frac{30\times 10}{45}$

$\Rightarrow$ $Median=110+6.67$

$\therefore$ $Median=116.67$