Question

Find the median of the following data distribution.
$\begin{array}{ccccccccc}\text{Marks obtained}& 20& 29& 28& 33& 42& 38& 43& 25\\ \text{Number of students}& 6& 28& 24& 15& 2& 4& 1& 20\end{array}$

• A. 28.5
• B. 29
• C. 30
• D. 40

Answer 1

The correct option is A

28.5

$\begin{array}{cc}\text{Marks obtained}& \text{Number of students(Frequency)}\\ 20& 6\\ 25& 20\\ 28& 24\\ 29& 28\\ 33& 15\\ 38& 4\\ 42& 2\\ 43& 1\\ Total& 100\end{array}$

As visible in the table above, first we arrange the marks in ascending order and prepare a frequency table.
Then the cumulative frequency table is prepared by adding all the previous frequencies to the frequency of the corresponding class.
From the table, it is inferred that the sum of frequencies is 100.
Here n = 100, which is even.
So, the median will be the average of the $\frac{n}{2}th$ and the$(\frac{n}{2}+1)th$ observations, i.e., the 50th and 51st observations.

To find these observations, we proceed as follows:

$\begin{array}{cc}\text{Marks obtained}& \text{Number of students}\\ 20& 6\\ upto25& 6+20=26\\ upto28& 26+24=50\\ upto29& 50+28=78\\ upto33& 78+15=93\\ upto38& 93+4=97\\ upto42& 97+2=99\\ upto43& 99+1=100\end{array}$

We now add this table of information to the existing frequency table as shown below.
$\begin{array}{ccc}\text{Marks obtained}& \text{Number of students}& \text{Cumulative frequency}\\ 20& 6& 6\\ 25& 20& 26\\ 28& 24& 50\\ 29& 28& 78\\ 33& 15& 93\\ 38& 4& 97\\ 42& 2& 99\\ 43& 1& 100\end{array}$

The above table will help us find the ${50}^{th}and{51}^{st}$ terms of distribution.
The cumulative frequency gives the number of students who have scored less than or equal to a certain mark.
The 50th student, according to cumulative frequency distribution has got 28 marks. The ${51}^{st}$ student has 29 marks.
Thus, the median terms are 28 and 29.
So, the median is the average of these terms = 28.5.