Question

Calculate the median of the following data.

Class-interval	$1-10$	$11-20$	$21-30$	$31-40$	$41-50$	$51-60$	$61-70$	$71-80$	$81-90$	$91-100$
Frequency	$4$	$6$	$7$	$16$	$25$	$21$	$12$	$5$	$3$	$1$

Answer 1

Step 1 Creating cumulative frequency table:

Here the classes are in inclusive form (discontinuous). So, we first convert the classes into exclusive forms (continuous).

$$\begin{gathered} Adjustmentfactor=\frac{1}{2}[lowerlimitofoneclass-upperlimitofpreviousclass] \\ Adjustmentfactor=\frac{1}{2}[11-10] \\ Adjustmentfactor=0.5 \end{gathered}$$

Prepare a cumulative frequency table as below:

Class Interval	Frequency	Cumulative Frequency ($cf$)
$0.5-10.5$	$4$	$4$
$10.5-20.5$	$6$	$10$
$20.5-30.5$	$7$	$17$
$30.5-40.5$	$16$	$33$
$40.5-50.5$	$25$	$58$
$50.5-60.5$	$21$	$79$
$60.5-70.5$	$12$	$91$
$70.5-80.5$	$5$	$96$
$80.5-90.5$	$3$	$99$
$90.5-100.5$	$1$	$100$

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 $n=100$

$\Rightarrow \frac{100}{2}=50$

The observation $50$ lies in the class $40.5-50.5$.

So, we have

$l=40.5,c=33,f=25,h=10$

$$\begin{gathered} \Rightarrow Median=40.5+\left(\frac{50-33}{25}\right)\times 10 \\ \Rightarrow Median=40.5+6.8 \\ \therefore Median=47.3 \end{gathered}$$

Hence, the median is $47.3$.