Question

The median of the following data is $470$. Find the values of $x$ and $y$ if the total frequency is $44$.

Class-Interval	$200-300$	$300-400$	$400-500$	$500-600$	$600-700$
No. of students	$3$	$x$	$20$	$10$	$y$

Answer 1

Step 1 Creating cumulative frequency table:

First, prepare a cumulative frequency table as below:

Class-Interval	No. of students (Frequency)	Cumulative Frequency ($cf$)
$200-300$	$3$	$3$
$300-400$	$x$	$3+x$
$400-500$	$20$	$23+x$
$500-600$	$10$	$33+x$
$600-700$	$y$	$33+x+y$
	$\sum _{}f=44$

Since it is given that the total frequency is $44$,

$$\begin{gathered} 33+x+y=44 \\ x+y=44-33 \\ x+y=11 \end{gathered}$$

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

Given: $Median=470$ , which lies between class $400-500$.

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 $n=44$

$\Rightarrow \frac{44}{2}=22$

The observation $22$ lies in the class $400-500$ .

So, we have

$l=400,c=3+x,f=20,h=100$

$$\begin{gathered} \Rightarrow 470=400+\left(\frac{22-(3+x)}{20}\right)\times 100 \\ \Rightarrow 470=400+\left(\frac{19-x}{20}\right)\times 100 \\ \Rightarrow 470=400+95-5x \\ \Rightarrow 5x=25 \\ \therefore x=5 \end{gathered}$$

Substitute the value of $x$ in $x+y=11$.

$$\begin{gathered} \Rightarrow y=11-5 \\ \therefore y=6 \end{gathered}$$

Hence, the value of $x$ and $y$ is $5$ and $6$ respectively.