Question

The median of the following data is 525. Find the values of $x$ and $y$ if the total frequency is 100.

Class Interval	Frequency
0-100	2
100-200	5
200-300	$x$
300-400	12
400-500	17
500-600	20
600-700	$y$
700-800	9
800-900	7
900-1000	4

Answer 1

Hint: We solve this problem by first adding the frequencies of all the classes given and then equate it to 100. Then we get an equation with $x$ and $y$. Then we find the median class from the given value of median. Then we use the formula for median of the grouped frequency, $Median=l+\dfrac{\dfrac{N}{2}-C}{f}\times h$ and substitute the values as required and find the value of $x$. Then we substitute the value of $x$ in the obtained equation with $x$ and $y$ to find the value of $y$.

Complete step by step answer:
We can see that the given frequency distribution is grouped.
Now let us add a column of cumulative frequency to the given frequency table.

Class	${{f}_{i}}$	Cumulative Frequency (less than) ${{F}_{i}}$
0-100	2	2
100-200	5	7
200-300	$x$	7+$x$
300-400	12	19+$x$
400-500	17	36+$x$
500-600	20	56+$x$
600-700	$y$	56+$x+y$
700-800	9	65+$x+y$
800-900	7	72+$x+y$
900-1000	4	76+$x+y$
Total	$\sum{{{f}_{i}}}=76+x+y$

We are given that the total frequency is 100.
So, we get,
$\begin{align}
  & \Rightarrow \sum{{{f}_{i}}}=76+x+y \\
 & \Rightarrow 76+x+y=100 \\
 & \Rightarrow x+y=24..........\left( 1 \right) \\
\end{align}$
Now let us consider the formula for median of the grouped frequency.
$Median=l+\dfrac{\dfrac{N}{2}-C}{f}\times h$
where, $l=$ Lower limit of median class
$h=$ width of the class interval
$f=$ frequency of the median class
$N=$ Sum of all frequencies
$C=$ Cumulative frequency of the class preceding median class

We are given that median of the given data is 525. From the given table, we can see that the median class is 500-600 because given value of median lies in the class 500-600.
So, Median Class = 500-600
So, from the table above we get the values required in the formula as,
$l=500$, $N=100$, $C=36+x$, $h=100$ and $f=20$.
So, substituting them in the formula for median we get,
$\begin{align}
  & \Rightarrow Median=500+\dfrac{\dfrac{100}{2}-\left( 36+x \right)}{20}\times 100 \\
 & \Rightarrow 525=500+\left( 50-\left( 36+x \right) \right)\times 5 \\
 & \Rightarrow 525-500=\left( 50-36-x \right)\times 5 \\
\end{align}$
$\begin{align}
  & \Rightarrow 25=\left( 14-x \right)\times 5 \\
 & \Rightarrow 50=70-5x \\
 & \Rightarrow 5x=70-50 \\
 & \Rightarrow 5x=20 \\
 & \Rightarrow x=4 \\
\end{align}$
Substituting this value in the equation (1) we get,
$\begin{align}
  & \Rightarrow x+y=24 \\
 & \Rightarrow 4+y=24 \\
 & \Rightarrow y=24-4 \\
 & \Rightarrow y=20 \\
\end{align}$
So, we get the values of $x$ and $y$ as 4 and 20 respectively.

Hence the answer is 4 and 20.

Note: There is a possibility of one making a mistake while solving this problem by taking the interpretation of the formula for median in a wrong way as,
$Median=l+\dfrac{\dfrac{N}{2}-C}{f}\times h$
where, $l=$ Lower limit of median class
$h=$ width of the class interval
$f=$ Cumulative frequency of the median class
$N=$ Sum of all frequencies
$C=$ Frequency of the class preceding median class
So, one needs to be careful while applying the formula.