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Let us look at an example to find standard deviation of a grouped data by mean method.
Example:
Case 1: Discrete data
Find the standard deviation of the following data by mean method.
\(x\) | \(1\) | \(2\) | \(3\) | \(4\) |
\(f\) | \(4\) | \(8\) | \(12\) | \(16\) |
Explanation:
We know that the formula for finding the arithmetic mean using direct method is given by:
\(\overline x = \frac{\sum_{i = 1}^{n} f_{i} x_{i}}{\sum_{i = 1}^{n} f_{i}}\)
Where \(x_{i}\) is the midpoint of the class interval and \(f_{i}\) is the frequency.
Let us form a frequency distribution table.
\(x_{i}\) | \(f_{i}\) | \(f_{i} x_{i}\) |
\(1\) | \(4\) | \(4\) |
\(2\) | \(8\) | \(16\) |
\(3\) | \(12\) | \(36\) |
\(4\) | \(16\) | \(64\) |
\(\sum_{i = 1}^{4} f_{i} = 40\) | \(\sum_{i = 1}^{4} f_{i} x_{i} = 120\) |
Substituting the known values in the above formula, we get:
Mean \(\overline x = \frac{120}{40}\) \(=\) \(3\)
Thus, the meanof the given data is
Now, let us find the standard deviation for the given data.
\(x_{i}\) | \(f_{i}\) | \(d_{i} = x_{i} - \overline x\) \(=\) \(x_{i} - 3\) | \(d_{i}^{2}\) | \(f_{i}d_{i}^{2}\) |
\(1\) | \(4\) | \(-2\) | \(4\) | \(16\) |
\(2\) | \(8\) | \(-1\) | \(1\) | \(8\) |
\(3\) | \(12\) | \(0\) | \(0\) | \(0\) |
\(4\) | \(16\) | \(1\) | \(1\) | \(16\) |
\(\sum_{i = 1}^{4} f_{i} = 40\) | \(\sum_{i = 1}^{4} f_{i} d_{i}^{2} = 40\) |
The formula to calculate the standard deviation by mean method is given by:
\(\sigma\) \(=\) \(\sqrt{\frac{\sum f_{i} d_{i}^{2}}{N}}\) where \(N = \sum_{i = 1}^{n} f_{i}\)
Substitute the required values in the above formula.
\(\sigma\) \(=\) \(\sqrt{\frac{40}{40}}\)
\(=\) \(\sqrt{1}\)
\(=\) \(1\)
Therefore, the standard deviation of the given data is \(1\).
Case 2: Continuous data
Find the standard deviation of the following data by mean method.
\(x\) | \(0 - 2\) | \(2 - 4\) | \(4 - 6\) | \(6 - 8\) |
\(f\) | \(4\) | \(8\) | \(12\) | \(16\) |
Explanation:
We know that the formula for finding the arithmetic mean using direct method is given by:
\(\overline x = \frac{\sum_{i = 1}^{n} f_{i} x_{i}}{\sum_{i = 1}^{n} f_{i}}\)
Where \(x_{i}\) is the midpoint of the class interval and \(f_{i}\) is the frequency.
Let us form a frequency distribution table.
Class interval | Frequency (\(f_{i}\)) | Midpoint (\(x_{i}\)) | \(f_{i} x_{i}\) |
\(0 - 2\) | \(4\) | \(1\) | \(4\) |
\(2 - 4\) | \(8\) | \(3\) | \(24\) |
\(4 - 6\) | \(12\) | \(5\) | \(60\) |
\(6 - 8\) | \(16\) | \(7\) | \(112\) |
\(\sum_{i = 1}^{4} f_{i} = 40\) | \(\sum_{i = 1}^{4} f_{i} x_{i} = 200\) |
Substituting the known values in the above formula, we get:
Mean \(\overline x = \frac{200}{40}\) \(=\) \(5\)
Thus, the mean of the given data is \(5\)
Now, let us find the standard deviation for the given data.
\(x_{i}\) | \(f_{i}\) | \(d_{i} = x_{i} - \overline x\) \(=\) \(x_{i} - 5\) | \(d_{i}^{2}\) | \(f_{i}d_{i}^{2}\) |
\(1\) | \(4\) | \(-4\) | \(16\) | \(64\) |
\(3\) | \(8\) | \(-2\) | \(4\) | \(32\) |
\(5\) | (12\) | \(0\) | \(0\) | \(0\) |
\(7\) | \(16\) | \(2\) | \(4\) | \(64\) |
\(\sum_{i = 1}^{4} f_{i} = 40\) | \(\sum_{i = 1}^{4} f_{i} d_{i}^{2} = 160\) |
The formula to calculate the standard deviation by mean method is given by:
\(\sigma\) \(=\) \(\sqrt{\frac{\sum f_{i} d_{i}^{2}}{N}}\) where \(N = \sum_{i = 1}^{n} f_{i}\)
Substitute the required values in the above formula.
\(\sigma\) \(=\) \(\sqrt{\frac{160}{40}}\)
\(=\) \(\sqrt{4}\)
\(=\) \(2\)
Therefore, the standard deviation of the given data is \(2\).
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