Theory:

The following steps can determine the median of the grouped frequency distribution:
 
1. Find the cumulative frequency distribution and denote \(N\) as the total frequency.
 
2. Find the \(\frac{N}{2}^{th}\) term.
 
3. The class which contains the cumulative frequency \(\frac{N}{2}\) is the median class.
 
4. The median of the class can be determined using the formula:
 
Median \(= l + \frac{(\frac{N}{2} - m)}{f} \times c\)
 
Where \(l\) is the lower limit of the median class,
 
\(m\) is the cumulative frequency of the class preceeding the median class,
 
\(f\) is the frequency of the median class,
 
\(c\) is the width of the median class, and
 
\(N\) is the total frequency.
Example:
Find the median of the following data.
 
Class interval\(100 - 200\)\(200 - 300\)\(300 - 400\)\(400 - 500\)
Frequency\(30\)\(15\)\(26\)\(44\)
 
Solution:
 
Class intervalFrequencyCumulative frequency
\(100 - 200\)\(30\)\(30\)
\(200 - 300\)\(15\)\(45\)
\(300 - 400\)\(26\)\(71\)
\(400 - 500\)\(44\)\(115\)
 
Therefore, the total frequency is \(N = 115\).
 
Median class \(= (\frac{N}{2}^{th})\) value
 
\(= (\frac{115}{2})^{th}\) value
 
\(= (57.5)^{th}\) value
The median of the grouped frequency distribution can be determined using the formula \(l + \frac{(\frac{N}{2} - m)}{f} \times c\)
The value \(57.5\) lies in the class interval \(300 - 400\)
 
Here, \(l = 300\), \(\frac{N}{2} = 57.5\), \(m = 45\), \(f = 26\) and \(c = 100\)
 
Substituting the known values in the above formula, we have:
 
Median \(= 300 + (\frac{57.5 - 45}{26}) \times 100\)
 
\(= 300 + (0.481) \times 100\)
 
\(= 300 + 48.1\)
 
\(= 348.1\)
 
Therefore, the median of the given data is \(348.1\)