Question

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

• A. 28.5
• B. 29
• C. 30
• D. 40

Answer 1

The correct option is A

28.5

$\begin{array}{cc}Marksobtained& Numberofstudents\left(Frequency\right)\\ 20& 6\\ 25& 20\\ 28& 24\\ 29& 28\\ 33& 15\\ 38& 4\\ 42& 2\\ 43& 1\\ Total& 100\end{array}$
As visible in above table, first we arrange the marks in ascending order
and prepare a frequency table as shown above.
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:

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

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

The above table will help us find the ${50}^{th}and{51}^{st}$ terms of distribution.
As 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.