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Question
Show that the two formulae for the standard deviation of ungrouped data
\[\sigma = \sqrt{\frac{1}{n} \sum \left( x_i - X \right)^2_{}}\] and
\[\sigma' = \sqrt{\frac{1}{n} \sum x_i^2 - X^2_{}}\] are equivalent, where \[X = \frac{1}{n}\sum_{} x_i\]
Solution
\[\sigma = \sqrt{\frac{1}{n} \sum \left( x_i - X \right)^2_{}}\]
\[ = \sqrt{\frac{1}{n} \sum \left( x_i^2 - 2 x_i X + X^2 \right)_{}}\]
\[ = \sqrt{\frac{1}{n} \sum x^2 _i- \frac{1}{n}\sum 2 x_i X + \frac{1}{n}\sum X^2}\]
\[ = \sqrt{\frac{1}{n} \sum ^2 x_i - \frac{1}{n} \times 2X\sum x_i + \frac{1}{n} \times X^2 \sum 1}\]
\[ = \sqrt{\frac{1}{n} \sum x^2_i - \frac{1}{n} \times 2X \times nX + \frac{1}{n} \times X^2 \times n} \left( X = \frac{1}{n} \sum x_i \right)\]
\[= \sqrt{\frac{1}{n} \sum x_i^2 - 2 X^2 +_{} X^2}\]
\[ = \sqrt{\frac{1}{n} \sum x_i^2 - X^2_{}}\]
\[ = \sigma'\]
Hence, the formulae
\[\sigma = \sqrt{\frac{1}{n} \sum \left( x_i - X \right)^2_{}}\] and
\[\sigma' = \sqrt{\frac{1}{n} \sum x_i^2 - X^2_{}}\] are equivalent, where
\[X = \frac{1}{n}\sum_{} x_i\].
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