Question

Find the missing frequencies ${f_1}$ and ${f_2}$ in the following distribution table, $N = 100$ and the median is 32.

Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	Total
Frequency	10	${f_1}$	25	30	${f_2}$	10	100

Answer 1

Hint: Form the first linear equation in ${f_1}$ and ${f_2}$ by calculating the total number of students from the table and equating it with 100. Now, form a frequency table with three columns, column 3 as cumulative frequency. Cumulative frequency is found by adding the frequencies in each step of progression. Apply the median formula $L + \left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right)h$ to calculate the value of one frequency. Substitute this value in the equation of frequency to get the value of other missing frequencies.

Formula used: The median formula is,
$L + \left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right)h$
where L = lower class containing the median,
N = total student,
f = frequency of the class containing median,
cf = cumulative frequency before the median class,
h = class interval.

Complete step-by-step solution:
Using the above table let us form a frequency table, having three columns. Column 1 will have class, column 2 will contain frequency, column 3 will contain cumulative frequency which is calculated by adding frequencies in each step. Therefore, the frequency table will look like: -

Age (in years)	Frequency ($f$)	Cumulative Frequency ($cf$)
0 – 10	10	10
10 – 20	${f_1}$	\[10 + {f_1}\]
20 – 30	25	\[35 + {f_1}\]
30 – 40	30	\[65 + {f_1}\]
40 – 50	${f_2}$	\[65 + {f_1} + {f_2}\]
50 – 60	10	\[75 + {f_1} + {f_2}\]

As the total frequency is 100. Then,
$ \Rightarrow \sum {{f_i}} = 100$
Substitute the value of the sum of frequency,
$ \Rightarrow 75 + {f_1} + {f_2} = 100$
Simplify the terms,
$ \Rightarrow {f_1} + {f_2} = 25$..............….. (1)
We have been given that median \[ = 32\], which lies in the range 30 – 40. Therefore, 30 – 40 is the median class. So,
The lower limit of the median class is,
$ \Rightarrow l = 30$
Cumulative frequency of class preceding the median class is,
$ \Rightarrow cf = 35 + {f_1}$
The frequency of the median class is,
$ \Rightarrow f = 30$
The height of the class is,
$ \Rightarrow h = 40 - 30 = 10$
Then the value of the median is given by,
Median $ = l + \left[ {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right]h$
Substitute the values,
$ \Rightarrow 32 = 30 + \left( {\dfrac{{\dfrac{{100}}{2} - \left( {35 + {f_1}} \right)}}{{30}}} \right) \times 10$
Simplify the terms,
\[ \Rightarrow 2 = \dfrac{{50 - 35 - {f_1}}}{3}\]
Again, simplify the terms,
\[ \Rightarrow 6 = 15 - {f_1}\]
Move constant part on one side,
$ \Rightarrow {f_1} = 15 - 6$
Subtract the value on the right side,
$ \Rightarrow {f_1} = 9$
Substitute the value in equation (1),
$ \Rightarrow 9 + {f_2} = 25$
Move constant part on the right side and subtract,
$ \Rightarrow {f_2} = 16$

Hence, the missing frequencies are 9 and 16.

Note: One may note that the class interval (h) is the same for all the rows and therefore we can calculate it by subtracting the lower limit from the upper limit by choosing any one of the rows. In the above solution, we choose row 1, i.e. 10 – 0 = 10. Now, the most important thing is the formula and its terms. We must know about all the terms in the formula and how to calculate it.