Question

If the median of the distribution given below is $27$, find the value of $x$ and $y$.

C.I	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	Total
f	$5$	$x$	$20$	$14$	$y$	$8$	$68$

Answer 1

Hint: In this problem we have given the median of the distribution. So, we will start with constructing the distribution table with cumulative frequency of each class. Here we have given the total value of frequency, hence the cumulative frequency of last class is equal to the total frequency. From this condition we will get an equation involving the variables $x,y$. After constructing the distribution table we will try to find the median of the given data and at the same time we will equate it to given median $27$, then we will get another equation involving the variables $x,y$. By solving the both equations, we will get the value of $x$ and $y$.

Complete step-by-step answer:
Given that,

C.I	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	Total
f	$5$	$x$	$20$	$14$	$y$	$8$	$68$

Now the distribution table with cumulative frequencies is

C.I	f	Cumulative Frequency (C.F)
$0-10$	$5$	$5$
$10-20$	$x$	$5+x$
$20-30$	$20$	$5+x+20=25+x$
$30-40$	$14$	$25+x+14=39+x$
$40-50$	$y$	$39+x+y$
$50-60$	$8$	$39+x+y+8=47+x+y$
	Total $\left( \text{N} \right)=68$

In a distribution table we know that sum of all the frequencies is equal to the cumulative frequency of last class in the distribution table. Then we have
$\text{N}=\text{C}\text{.F of class }50-60$
$\begin{align}
  & \Rightarrow 68=47+x+y \\
 & x+y=21....\left( \text{i} \right) \\
\end{align}$
Given that median of the table is $27$, hence it is lies in the class $20-30$. So, we will treat it as Median Class.
We know that the formula for the median of a distribution is given by
$\text{M}=l+\dfrac{\dfrac{\text{N}}{2}-\text{c}\text{.f}}{f}\times h$
Where
$l$ is the lower limit of the Median class.
$\text{N}$ is the sum of the frequencies in the distribution table.
$\text{c}\text{.f}$ is the value of cumulative frequency of preceding class of Median class.
$f$ is the frequency of the Median Class.
$h$ is the height of the class.
For this problem we already have the Median class as $20-30$. Hence the above values are stated as
$l=20$, $\text{N}=68$, $\text{c}\text{.f}=5+x$, $f=20$, $h=10$ and we also have the value of Median as $27$, then we will have
$\begin{align}
  & \text{M}=l+\dfrac{\dfrac{\text{N}}{2}-\text{c}\text{.f}}{f}\times h \\
 &\Rightarrow 27=20+\dfrac{\dfrac{68}{2}-\left( 5+x \right)}{20}\times 10 \\
 &\Rightarrow 27=20+\dfrac{34-5-x}{2} \\
\end{align}$
Now subtracting $20$ from both sides then we will have
$\begin{align}
  &\Rightarrow 27-20=20+\dfrac{29-x}{2}-20 \\
 &\Rightarrow 7=\dfrac{29-x}{2} \\
\end{align}$
Multiplying $2$on both sides of the equation, then we will get
$\begin{align}
  & 7\times 2=\dfrac{29-x}{2}\times 2 \\
 &\Rightarrow 14=29-x \\
 &\Rightarrow x=29-14 \\
 &\Rightarrow x=15 \\
\end{align}$
So, from the equation $\left( \text{i} \right)$, we will get value of $y$ as
$\begin{align}
  & x+y=21 \\
 &\Rightarrow y=21-15 \\
 &\Rightarrow y=6 \\
\end{align}$
Hence the values of $x$ and $y$ are $15,6$ respectively.

Note: In the problem we have taken the median class from the given median $27$. Students can also take the median class by considering the value $\dfrac{\text{N}}{2}=\dfrac{68}{2}=34$, from this value we can observe that it is lies between $25+x$ and $39+x$,so we can conclude that the Median class is $20-30$. Student may also assume the Median as a variable and with the help of distribution table they will calculate the value of Median in terms of $x$, now they will equate the Median which in terms of $x$ to the given median. But it is the lengthy process and there is a chance of making mistakes.