Question

If the median of the following frequency distribution is 32.5, then find the values of x and y.

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	Total
Frequency	x	5	9	12	y	3	2	40

Answer 1

Hint: In this question, we first need to make a table including frequency and cumulative frequency. Then choose the median class from that data to get the values of l, h, f, F and substitute these values in the respective formula which on further simplification gives the value of x. Now, on equating total frequency to 40 gives the value of y.
\[M=l+\dfrac{\dfrac{N}{2}-F}{f}\times h\]
Here, cumulative frequency of a class is the sum of the frequency of that class and all frequencies before it.

Complete step by step answer:
Now, from the given data let us write a table including cumulative frequency.

Class Interval	Frequency	Cumulative frequency
0 - 10	x	x
10 - 20	5	x + 5
20 - 30	9	x + 14
30 - 40	12	x + 26
40 - 50	y	x + y + 26
50 - 60	3	x + y + 29
60 - 70	2	x + y + 31
	N = 40

Let us assume the median of the data as M.
As we already know that formula for median of a grouped data is given by
\[M=l+\dfrac{\dfrac{N}{2}-F}{f}\times h\]
Where, l is the lower limit of the median class
             f is the frequency of the median class
             N is the total frequency
             F is the cumulative frequency of the class just before the median class
             h is the length of the median class
Now, from the data above we have
The median class is the class interval of 30 - 40 because the cumulative frequency of this class is just greater than \[\dfrac{N}{2}\] which is 20
\[M=32.5,N=40,l=30,h=10,f=12,F=14+x\]
Let us now substitute these values in the respective formula
\[\Rightarrow M=l+\dfrac{\dfrac{N}{2}-F}{f}\times h\]
Now, on substituting the values we get,
\[\Rightarrow 32.5=30+\dfrac{20-\left( 14+x \right)}{12}\times 10\]
Now, on further simplification we get,
\[\Rightarrow 32.5=30+\dfrac{6-x}{12}\times 10\]
Now, on rearranging the terms we get,
\[\Rightarrow \dfrac{2.5\times 12}{10}=6-x\]
Now, on further simplification we get,
\[\Rightarrow 3=6-x\]
\[\therefore x=3\]
Now, let us find the value of y from the total frequency given
\[\Rightarrow x+5+9+12+y+3+2=40\]
Now, on simplifying it further we get,
\[\Rightarrow x+y+31=40\]
Now, on substituting the value of x and rearranging the terms we get,
\[\begin{align}
  & \Rightarrow 3+y=9 \\
 & \therefore y=6 \\
\end{align}\]

Hence, the values of x and y are 3, 6.

Note: It is important to note that median class is the class whose cumulative frequency is either equal to \[\dfrac{N}{2}\] or is just greater than \[\dfrac{N}{2}\]. So, we cannot choose any random row as a median class which changes the result completely.
It is also to be noted that considering the value of F if we consider the cumulative frequency of the median class instead of the class just before median class then it changes the corresponding value of x and so the value of y.