Theory:

The mean of the ungrouped frequency distribution can be determined using the formula:
\(\overline X = \frac{f_1 x_1 + f_2 x_2 + ... + f_n x_n}{f_1 + f_2 + ... + f_n}\) \(= \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i}\)
Example:
The height(in \(cm\)) of \(20\) students in a classroom are:
 
Height
\(x_i\)
Number of students
\(f_i\)
\(130\)\(1\)
\(135\)\(2\)
\(140\)\(1\)
\(155\)\(2\)
\(163\)\(1\)
\(165\)\(3\)
\(177\)\(2\)
\(189\)\(2\)
\(196\)\(2\)
\(100\)\(4\)
 
Find the mean height of the \(20\) students.
 
Solution:
 
To find the value of \(f_ix_i\), multiply the value of \(x\) and \(f\) of each entry.
 
Consider for the mark \(130\). That is, \(130 \times 1 = 130\)
 
Similarly, for the mark \(135\), we have \(135 \times 2 = 270\) and so on.
 
Tabulating these values, we get:
 
Marks
\(x_i\)
Frequency
\(f_i\)
\(f_ix_i\)
\(130\)\(1\)\(130\)
\(135\)\(2\)\(270\)
\(140\)\(1\)\(140\)
\(155\)\(2\)\(310\)
\(163\)\(1\)\(163\)
\(165\)\(3\)\(495\)
\(177\)\(2\)\(354\)
\(189\)\(2\)\(378\)
\(196\)\(2\)\(392\)
\(100\)\(4\)\(400\)
Total\(\sum f_i = 20\)\(\sum f_ix_i = 3032\)
 
Substituting the known values in the above formula, we get:
 
Mean \(\overline X = \frac{3032}{20}\) \(= 151.6\)
 
Therefore, the mean of the given data is \(151.6\).