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

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

$\begin{array}{cc}\text{Marks obtained}& \text{Number of students}\\ 20& 6\\ \text{upto}25& 6+20=26\\ \text{upto}28& 26+24=50\\ \text{upto}29& 50+28=78\\ \text{upto}33& 78+15=93\\ \text{upto}38& 93+4=97\\ \text{upto}42& 97+2=99\\ \text{upto}43& 99+1=100\end{array}$

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

The above table will help us find
the ${50}^{th}\text{and}{51}^{st}$ 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.