Theory:

The formula for finding the arithmetic mean using the direct method is given by:
 
\(\overline X = \frac{\sum f_ix_i}{\sum f_i}\)
 
Where \(i\) varies from \(1\) to \(n\), \(x_i\) is the midpoint of the class interval and \(f_i\) is the frequency.
 
Steps:
 
1. Calculate the midpoint of the class interval and name it as \(x_i\).
 
2. Multiply the midpoints\(x_i\) with the frequency\(f_i\) of each class interval and name it as \(f_ix_i\).
 
3. Find the values \(\sum f_ix_i\) and \(\sum f_i\).
 
4. Divide \(\sum f_ix_i\) by \(\sum f_i\) to determine the mean of the data.
Example:
The following frequency distribution table shows that the number of trees based on the height in metres. Find the average height of the trees.
 
Height (in \(m\))\(30 - 40\)\(40 - 50\)\(50 - 60\)\(60 - 70\)\(70 - 80\)
Number of trees\(124\)\(156\)\(200\)\(10\)\(10\)
 
Solution:
 
Let us form a frequency distribution table.
 
Height
(in \(m\))
Number of trees
(\(f_i\))
Midpoint
(\(x_i\))
\(f_ix_i\)
\(30 - 40\)\(124\)\(35\)\(4340\)
\(40 - 50\)\(156\)\(45\)\(7020\)
\(50 - 60\)\(200\)\(55\)\(11000\)
\(60 - 70\)\(10\)\(65\)\(650\)
\(70 - 80\)\(10\)\(75\)\(750\)
Total\(\sum f_i = 500\) \(\sum f_ix_i = 23760\)
 
Mean \(\overline X = \frac{23760}{500}\) \(= 47.52\)
 
Therefore, the average height of the trees is \(47.52\).