Question

The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:

Class Interval	Frequency
0 – 100	2
100 – 200	5
200 – 300	${f_1}$
300 – 400	12
400 – 500	17
500 – 600	20
600 – 700	${f_2}$
700 – 800	9
800 – 900	7
900 – 1000	4

Answer 1

We will first make the table of the frequencies and cumulative frequencies. We will make two equations by using the sum of frequencies and median of the given data. The formula for the median is $l + \dfrac{{\dfrac{n}{2} - cf}}{f} \times h$, where $l$ is the lower value of the median class, $n$ is the total number of frequency and $cf$ is the cumulative frequency of previous class, $f$ is the frequency of median class, and $h$ is the width of the class interval.

Complete step-by-step answer:
We are given that the median of the data is 52.5 and the sum of frequency is 100.
First of all, we will make a table of frequency and cumulative frequency.

Class Interval	Frequency	Cumulative Frequency
0 – 100	2	2
100 – 200	5	7
200 – 300	${f_1}$	$7 + {f_1}$
300 – 400	12	$19 + {f_1}$
400 – 500	17	$36 + {f_1}$
500 – 600	20	$56 + {f_1}$
600 – 700	${f_2}$	$56 + {f_1} + {f_2}$
700 – 800	9	$65 + {f_1} + {f_2}$
800 – 900	7	$72 + {f_1} + {f_2}$
900 – 1000	4	$76 + {f_1} + {f_2}$

Here, the sum of frequencies is $76 + {f_1} + {f_2}$ which is also equal to 100. So,
$\Rightarrow 76 + {f_1} + {f_2} = 100$
Move constant part and ${f_1}$ on the right side and subtract,
$\Rightarrow {f_2} = 24 - {f_1}$...........….. (1)
We know that the median is calculated as $l + \dfrac{{\dfrac{n}{2} - cf}}{f} \times h$, where $l$ is the lower value of the median class, $n$ is the total number of frequency and $cf$ is the cumulative frequency of previous class, $f$ is the frequency of median class, and $h$ is the width of the class interval.
Here, the sum of all frequencies is 100 which is equal to $n$.
Then, the value of $\dfrac{n}{2}$ is,
$\Rightarrow \dfrac{{100}}{2} = 50$
Then, the median class is 50 – 60.
The lowest value of the median class is,
$\Rightarrow l = 50$
The width of the class interval is,
$\Rightarrow 60 - 50 = 10$
The frequency of the median class is,
$\Rightarrow f = 20$
The cumulative frequency of the previous class,
$\Rightarrow 36 + {f_1}$
Substitute the values in the median formula is,
$\Rightarrow 52.5 = 50 + \dfrac{{50 - \left( {36 + {f_1}} \right)}}{{20}} \times 10$
Simplify the terms,
$\Rightarrow \dfrac{{14 - {f_1}}}{2} = 2.5$
Cross-multiply the terms,
$\Rightarrow 14 - {f_1} = 5$
Move constant part on one side and variable on another side,
$\Rightarrow {f_1} = 14 - 5$
Subtract the term,
$\therefore {f_1} = 9$
Substitute the values in equation (1),
$\Rightarrow {f_2} = 24 - 9$
Subtract the term,
$\therefore {f_2} = 15$

Hence, the value of missing frequencies is ${f_1} = 9$ , and ${f_2} = 15$.

Note: There is a possibility of one making a mistake while solving this problem by taking the interpretation of the formula for the median in the wrong way. The median is a kind of average where the middle value of the given data. The students must know the formula of the median. They must also know that the cumulative frequency in the formula is the previous median class.