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Let us look at an example to find standard deviation of ungrouped data by mean method.
Example:
Find the standard deviation of the data \(5.4\), \(8.9\), \(10.1\), \(11.4\) and \(9.2\) by mean method.
Explanation:
Let \(n\) represent the number of values in the data.
\(n\) \(=\) \(5\)
Let \(\overline x\) represent the mean of the given data.
Mean \(\overline x = \frac{\text{Sum of all the observations}}{\text{Total number of observations}}\)
\(\overline x\) \(=\) \(\frac{5.4 + 8.9 + 10.1 + 11.4 + 9.2}{5}\)
\(=\) \(\frac{45}{5}\)
\(=\) \(9\)
Let \(x_{i}\) represent each values in the given data.
\(x_{i}\) | \(d_{i} = x_{i} - \overline x\) \(=\) \(x_{i} - 9\) | \(d_{i}^{2}\) |
\(5.4\) | \(-3.6\) | \(12.96\) |
\(8.9\) | \(-0.1\) | \(0.01\) |
\(10.1\) | \(1.1\) | \(1.21\) |
\(11.4\) | \(2.4\) | \(5.76\) |
\(9.2\) | \(0.2\) | \(0.04\) |
\(\sum d_{i}^{2} = 19.98\) |
The formula to calculate the standard deviation by mean method is given by:
\(\sigma\) \(=\) \(\sqrt{\frac{\sum d_{i}^{2}}{n}}\) where \(d_{i} = x_{i} - \overline{x}\)
Substitute the known values in the above formula.
\(\sigma\) \(=\) \(\sqrt{\frac{19.98}{5}}\)
\(=\) \(\sqrt{3.996}\)
\(=\) \(1.999\)
\(\approx\) \(2\)
Therefore, the standard deviation for the given data is \(2\).
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