Theory:

Let us recall what we learnt in class \(IX\) in the concept of median of ungrouped frequency distribution.
The median of an ungrouped frequency distribution can be determined using the following steps.
 
1. Arrange the given data in ascending or descending order.
 
2. Find the cumulative frequency distribution and denote \(N\) as the total frequency.
 
3. If \(N\) is odd, then median \(= \left(\frac{N + 1}{2} \right)^{th}\) term.
 
4. If \(N\) is even, then median \(= \left(\frac{(\frac{N}{2}^{th}) \text{observation} + (\frac{N}{2}+1)^{th} \text{observation}}{2} \right)\)
Example:
Find the median of the following data.
 
Marks\(30\)\(35\)\(60\)\(92\)\(85\)\(75\)
Number of students\(1\)\(4\)\(6\)\(2\)\(10\)\(7\)
 
Solution:
 
Let us arrange the marks in ascending order and find the cumulative frequency.
 
Marks
Frequency
(\(f\))
Cumulative frequency
(\(cf\))
\(30\)\(1\)\(1\)
\(35\)\(4\)\(1 + 4 = 5\)
\(60\)\(6\)\(5 + 6 = 11\)
\(75\)\(7\)\(11 + 7 = 18\)
\(85\)\(10\)\(18 + 10 = 28\)
\(92\)\(2\)\(28 + 2 = 30\)
 
Therefore, the total frequency is \(N = 30\).
 
Since \(N\) is even, the median can be determined using the formula \(\left(\frac{(\frac{N}{2}^{th}) \text{observation} + (\frac{N}{2}+1)^{th} \text{observation}}{2} \right)\)
 
Substituting the known values, we get:
 
Median \(= \left(\frac{(\frac{30}{2}^{th}) \text{observation} + (\frac{30}{2}+1)^{th} \text{observation}}{2} \right)\)
 
\(= \left(\frac{15^{th} \text{observation} + 16^{th} \text{observation}}{2} \right)\)
 
If the marks of the students are arranged in ascending order, then the \(15^{th}\) and \(16^{th}\) terms would be in the middle value. From the cumulative frequency distribution, let us select the marks near the \(15^{th}\) and \(16^{th}\) terms. Thus, the marks would be \(75\).
 
Substituting these values in the median, we have:
 
Median \(= \frac{75+75}{2}\) \(=75\)
 
Therefore, the median of the given data is \(75\).