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Let us look at an example to find standard deviation of a grouped data by step deviation method.
Example:
The number of buckets required to fill given volume of water are given below:
 
Volume (in litres)
\(2 - 4\)
\(4 - 6\)
\(6 - 8\)
\(8 - 10\)
\(10 - 12\)
Number of buckets
\(3\)
\(7\)
\(11\)
\(15\)
\(18\)
 
Find its standard deviation  using step deviation method
 
Explanation:
 
Let the assumed mean be \(A = 7\) and the class width \(c = 2\).
 
Volume
Number of buckets
(\(f_{i}\))
Midpoint
(\(x_{i}\))
Deviation
\(d_{i} = \frac{x_{i} - A}{c}\)
\(d_{i}^{2}\)
\(f_{i}d_{i}\)
\(f_{i}d_{i}^{2}\)
\(2 - 4\)
\(3\)
\(3\)
\(-2\)
\(4\)
\(-6\)
\(12\)
\(4 - 6\)
\(7\)
\(5\)
\(-1\)
\(1\)
\(-7\)
\(7\)
\(6 - 8\)
\(11\)
\(7\)
\(0\)
\(0\)
\(0\)
\(0\)
\(8 - 10\)
\(15\)
\(9\)
\(1\)
\(1\)
\(15\)
\(15\)
\(10 - 12\)
\(18\)
\(11\)
\(2\)
\(4\)
\(36\)
\(72\)
 
\(\sum_{i = 1}^{5} f_{i} = 54\)
 
 
 
\(\sum_{i = 1}^{5} f_{i}d_{i} = 38\)
\(\sum_{i = 1}^{5} f_{i}d_{i}^{2} = 106\)
The formula to calculate the standard deviation by step deviation method is given by:
 
\(\sigma = c \times \sqrt{\frac{\sum f_{i} d_{i}^{2}}{N}- \left(\frac{\sum f_{i} d_{i}}{N}\right) ^2}\) where \(N = \sum_{i = 1}^{n} f_{i}\) and \(d_{i} = \frac{x_{i} - A}{c}\).
Substituting the known values in the above formula, we have:
 
\(\sigma = 2 \times \sqrt{\frac{106}{54}- \left(\frac{38}{54}\right) ^2}\)
 
\(= 2 \times \sqrt{1.96 - \left(0.7 \right) ^2}\)
 
\(= 2 \times \sqrt{1.96 - 0.49}\)
 
\(= 2 \times \sqrt{1.47}\)
 
\(= 2 \times 1.212\)
 
\(=\) \(2.424\)
 
\(\approx\) \(2.42\)
 
Therefore, the standard deviation of the given data is \(2.42\).