Question

Find the values of \['x'\] and \['y'\] in the following data if the median is equal to 31

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	Total
Frequency	5	\['x'\]	6	\['y'\]	6	5	40

Answer 1

Hint: We solve this problem by extending the given table by adding a column of cumulative frequency. The cumulative frequency of a class is the sum of all frequencies up to that class. Then we assume that middle class of unknown frequency as the median class. Then the formula of the median is given as
\[m=l+\left( \dfrac{\dfrac{N}{2}-cf}{f} \right)\times h\]
Where, \['l'\] is the lower interval of the median class, \['N'\] is the total sum of frequencies, \['cf'\] is the cumulative frequency of preceding median class, \['f'\] is the frequency of median class and \['h'\] is the height of the class.
By using this formula we find the required unknown values.

Complete step-by-step answer:
We are given with the data of a grouped data.
Let us extend the given data by adding a column of cumulative frequency
We know that the cumulative frequency of a class is the sum of all frequencies up to that class
By using this definition of cumulative frequency we get

Class	Frequency	Cumulative frequency
0 – 10	5	5
10 – 20	\[x\]	\[5+x\]
20 – 30	6	\[11+x\]
30 – 40	\[y\]	\[11+x+y\]
40 – 50	6	\[17+x+y\]
50 – 60	5	\[22+x+y\]

We are given that the total frequency as 40
Let us assume that the total sum of frequencies as
\[\Rightarrow N=40\]
Here, from the table we can see that the total sum of frequencies as
\[\Rightarrow N=22+x+y\]
By substituting the value of \['N'\] in above equation we get
\[\begin{align}
  & \Rightarrow 22+x+y=40 \\
 & \Rightarrow x+y=18......equation(i) \\
\end{align}\]
We know that the middle class of unknown frequency will be the median class in this type of question.
We are given that the median of given data as 31
Let us assume that the median as
\[\Rightarrow m=31\]
Now, let us assume that the 30 – 40 is the median class.
We know that the formula of the median is given as
\[m=l+\left( \dfrac{\dfrac{N}{2}-cf}{f} \right)\times h\]
Where, \['l'\] is the lower interval of the median class, \['N'\] is the total sum of frequencies, \['cf'\] is the cumulative frequency of preceding median class, \['f'\] is the frequency of median class and \['h'\] is the height of the class.
Now, let us take all the required values from the table then we get
\[\begin{align}
  & \Rightarrow l=30 \\
 & \Rightarrow N=40 \\
 & \Rightarrow cf=11+x \\
 & \Rightarrow f=y \\
 & \Rightarrow h=10 \\
\end{align}\]
Now, by substituting all the required values in the median formula we get\[\Rightarrow 31=30+\left( \dfrac{\dfrac{40}{2}-\left( 11+x \right)}{y} \right)\times 10\]
Now, by cross multiplying the terms from LHS to RHS we get
\[\begin{align}
  & \Rightarrow y=\left( 20-11-x \right)\times 10 \\
 & \Rightarrow 10x+y=90.....equation(ii) \\
\end{align}\]
Now, let us subtract equation (i) from equation (ii) then we get
\[\begin{align}
  & \Rightarrow \left( 10x+y \right)-\left( x+y \right)=90-18 \\
 & \Rightarrow 9x=72 \\
 & \Rightarrow x=8 \\
\end{align}\]
Now, by substituting the value of \['x'\] in equation (i) we get
\[\begin{align}
  & \Rightarrow 8+y=18 \\
 & \Rightarrow y=10 \\
\end{align}\]
Therefore the values of \['x'\] and \['y'\] are 8 and 10 respectively.

Note: Students may make mistakes in considering the median class.
The median class is very important for calculating the median of a grouped data.
When there are unknown values in the frequencies then we consider the middle class as the median class.
Here there are two middle classes : 20 – 30 and 30 – 40.
We consider 30 – 40 as the median class because it has an unknown frequency.
But students may consider the class 20 – 30 as the median class which gives the wrong answer because we need to consider the class having the unknown frequency as the median class.