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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 \(2.42\).
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