Question

Find the missing frequency in the following distribution if N = 100 and median is 32.

Marks	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	Total
No. of students	10	?	25	30	?	10	100

Answer 1

Hint: Assume that number of students having marks in the range 10 – 20 and 40 – 50 are ‘x’ and ‘y’ respectively. Form first linear equation in ‘x’ and ‘y’ by calculating the total number of students from the table and equating it with 100. Now, form a frequency table with three column, having column 1 as marks, column 2 as frequency i.e. number of students and column 3 as cumulative frequency. Cumulative frequency of cf is found by adding the frequencies in each step of progression. Apply the formula: - Median = \[L+\left( \dfrac{\dfrac{N}{2}-cf}{f} \right)\times h\], 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, to calculate the value of x. Substitute this value of x in equation (1) to get the value of y.

Complete step by step answer:
We have been given the following table: -

Marks	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	Total
No. of students	10	?	25	30	?	10	100

Let us assume that, number of students having marks in the range 10 – 20 and 40 – 50 are ‘x’ and ‘y’ respectively. Using the above table let us form a frequency table, having three columns. Column 1 will have marks, column 2 will contain frequency or number of students, column 3 will contain cumulative frequency which is calculated by adding frequencies in each step. Therefore, the frequency table will look like: -

Marks	Frequency (f)	Cumulative frequency (cf)
0 – 10	10	10
10 – 20	x	10 + x
20 – 30	25	10 + x + 25 = 35 + x
30 – 40	30	35 + x + 30 = 65 + x
40 – 50	y	65 + x + y
50 – 60	10	65 + x + y + 10 = 75 + x + y

Now, we have been given that, there are 100 students,
\[\begin{align}
  & \therefore \sum{f}=100 \\
 & \Rightarrow 75+x+y=100 \\
\end{align}\]
\[\Rightarrow x+y=25\] - (1)
We know that, median = \[L+\left( \dfrac{\dfrac{N}{2}-cf}{f} \right)\times h\], where,
L = Lower class containing the median
N = Total number of students
f = frequency of the class containing median
cf = cumulative frequency before the median class
h = class interval = upper limit – lower limit
We have been given that median = 32, which lies in the range 30 – 40. Therefore, 30 – 40 is the median class. So,
L = 30
N = 100
f = 30
cf = 35 + x
h = 10 – 0 = 10
Substituting these values in the formula for median, we get,
\[\begin{align}
  & \Rightarrow 32=30+\left( \dfrac{\dfrac{100}{2}-\left( 35+x \right)}{30} \right)\times 10 \\
 & \Rightarrow 2=\dfrac{50-35-x}{3} \\
\end{align}\]
By cross – multiplication, we get,
\[\Rightarrow 6=15-x\]
\[\begin{align}
  & \Rightarrow x=15-6 \\
 & \Rightarrow x=9 \\
\end{align}\]
Substituting x = 9 in equation (1), we get,
\[\begin{align}
  & \Rightarrow y=25-x \\
 & \Rightarrow y=25-9 \\
 & \Rightarrow y=16 \\
\end{align}\]

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.