Question

The distribution below gives the marks obtained by the students of a class in a test. Maximum marks being $80$ . Find the median marks of the students.

Marks Obtained	$below10$	$below20$	$below30$	$below40$	$below50$	$below60$	$below70$	$below80$
No. of students	$15$	$35$	$60$	$80$	$96$	$127$	$190$	$200$

Answer 1

Step 1: Creating cumulative frequency table

First, prepare a cumulative frequency table as below:

Marks Obtained	No. of students (Frequency)	Cumulative Frequency ($cf$)
$0-10$	$15-0=15$	$15$
$10-20$	$35-15=20$	$15+20=35$
$20-30$	$60-35=25$	$35+25=60$
$30-40$	$80-60=20$	$60+20=80$
$40-50$	$96-80=16$	$80+16=96$
$50-60$	$127-96=31$	$96+31=127$
$60-70$	$190-127=63$	$127+63=190$
$70-80$	$200-190=10$	$190+10=200$

Step 2: Calculating the median

The formula for the median is,

$Median=l+\left(\frac{{\displaystyle \frac{\mathrm{n}}{2}-\mathrm{c}}}{\mathrm{f}}\right)\times h$.

Here,

$l$ is the lower limit of the median class,

$n$ is the number of observations,

$c$ is the cumulative frequency of the class preceding the median class,

$f$ is the frequency of the median class, and

$h$ is the class size

Here $n=200$

$\Rightarrow \frac{200}{2}=100$

The observation $100$ lies in the class $50-60$.

So, we have

$l=50,c=96,f=31,h=10$

$$\begin{gathered} \Rightarrow Median=50+\left(\frac{100-96}{31}\right)\times 10 \\ \Rightarrow Median=50+1.29 \\ \therefore Median=51.29 \end{gathered}$$

Hence, the median marks of the students is $51.29$.