Question

The median of the following data is $26$. Find the values of $x$ and $y$ if the total frequency is $80$.

Class-Interval	$0-8$	$8-16$	$16-24$	$24-32$	$32-40$	$40-48$
Frequency	$8$	$10$	$x$	$24$	$y$	$7$

Answer 1

Step 1 Creating cumulative frequency table:

First, prepare a cumulative frequency table as below:

Class-Interval	Frequency	Cumulative Frequency ($cf$)
$0-8$	$8$	$8$
$8-16$	$10$	$18$
$16-24$	$x$	$18+x$
$24-32$	$24$	$42+x$
$32-40$	$y$	$42+x+y$
$40-48$	$7$	$49+x+y$
	$\sum _{}f=80$

Since it is given that the total frequency is $80$,

$$\begin{gathered} 49+x+y=80 \\ x+y=80-49 \\ x+y=31 \end{gathered}$$

Step 2 Find the value of $x$ and $y$:

Given: $Median=26$ , which lies between class $24-32$.

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 $n=80$

$\Rightarrow \frac{80}{2}=40$

The observation $40$ lies in the class $24-32$ .

So, we have

$l=24,c=18+x,f=24,h=8$

$$\begin{gathered} \Rightarrow 26=24+\left(\frac{40-(18+x)}{24}\right)\times 8 \\ \Rightarrow 2=\left(\frac{22-x}{24}\right)\times 8 \\ \Rightarrow 2=\frac{22-x}{3} \\ \Rightarrow 6=22-x \\ \Rightarrow x=22-6 \\ \therefore x=16 \end{gathered}$$

Substitute the value of $x$ in $x+y=31$.

$$\begin{gathered} \Rightarrow y=31-16 \\ \therefore y=15 \end{gathered}$$

Hence, the value of $x$ and $y$ is $16$ and $15$ respectively.