Question

If the median of the following frequency distribution is $32.5$, find the missing frequencies.

Class Interval	Frequency
0-10	${f_1}$
10-20	5
20-30	9
30-40	12
40-50	${f_2}$
50-60	3
60-70	2
Total	40

A) 3, 6
B) 4, 9
C) 2, 8
D) None of these

Answer 1

Here we will first find the cumulative frequency table. Then we will find the median class and using this we will find all the unknown values. We will substitute all the values in the formula of median and find our first unknown frequency. Then we will find the second unknown frequency using the total sum of frequency.

Formula Used:
Median $= l + \dfrac{{\dfrac{N}{2} - F}}{f} \times h$, where $l$ is lower limit of median class interval, $F$ is Cumulative Frequency proceeding to the median class frequency,$f$ is Frequency of the class interval to which median belongs, $h$ is Width of the class interval and $N$ is Total Frequency of the distribution.

Complete step by step solution:
First, we will form table of cumulative frequency as below:

Class Interval	Frequency	Cumulative Frequency
0-10	${f_1}$	${f_1}$
10-20	5	$5 + {f_1}$
20-30	9	$14 + {f_1}$
30-40	12	$26 + {f_1}$
40-50	${f_2}$	$26 + f + {f_2}$
50-60	3	$29 + {f_1} + {f_2}$
60-70	2	$31 + {f_1} + {f_2}$
Total	40

We will use the median formula to find the value of ${f_1}$.
Now, it is given that median of the frequency distribution is $32.5$
So Median class will be = 30 - 40
By using the median class we get our values as,
$l = 30,h = 40 - 30 = 10,f = 12$, $N = 40$ and $F = 14 + {f_1}$.
Substituting the above value in the formula Median $= l + \dfrac{{\dfrac{N}{2} - F}}{f} \times h$ we get,
Median $= 30 + \dfrac{{\dfrac{{40}}{2} - \left( {14 + {f_1}} \right)}}{{12}} \times 10$
As median $32.5$ so we get,
$\Rightarrow 32.5 = 30 + \dfrac{{20 - 14 - {f_1}}}{{12}} \times 10$
Subtracting the terms in the numerator, we get
$\Rightarrow 32.5 = 30 + \dfrac{{6 - {f_1}}}{{12}} \times 10$
Subtracting 30 from both sides, we get
$\Rightarrow 32.5 - 30 = \dfrac{{6 - {f_1}}}{6} \times 5$
Simplifying the expression, we get
$\Rightarrow 2.5 \times 6 = 30 - 5{f_1}$
On solving we get,
$\Rightarrow 15 = 30 - 5{f_1}$
Rewriting the terms, we get
$\Rightarrow 5{f_1} = 30 - 15 = 15$
Dividing both sides by 5, we get
$\begin{array}{l} \Rightarrow {f_1} = \dfrac{{15}}{5}\\\ \Rightarrow {f_1} = 3\end{array}$
So, we get ${f_1}$ value as 3.
Next, as we know sum of frequency is 40, so
$31 + {f_1} + {f_2} = 40$
Substituting ${f_1}$ in above equation, we get
$\Rightarrow 31 + 3 + {f_2} = 40$
$\Rightarrow {f_2} = 40 - 31 - 3$
Adding and subtracting the like terms, we get
$\Rightarrow {f_2} = 6$
So we get ${f_2}$ as 6.

Hence, option (A) is correct.

Note:
Median separates the upper half from the lower half of a data set. Median of a data set doesn’t get skewed with a small proportion or a very large proportion and thus it is generally used in robust statistics. If we have a finite set of numbers then the middle term is its median. Before finding the median of a finite set we need to arrange the given set in ether descending order or ascending order.