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 \[X = \frac{1}{n}\sum_{} x_i\]

 

 

Answer 1

\[\sigma = \sqrt{\frac{1}{n} \sum \left( x_i - X \right)^2_{}}\]

\[ = \sqrt{\frac{1}{n} \sum \left( x_i^2 - 2 x_i X + X^2 \right)_{}}\]

\[ = \sqrt{\frac{1}{n} \sum x^2 _i- \frac{1}{n}\sum 2 x_i X + \frac{1}{n}\sum X^2}\]

\[ = \sqrt{\frac{1}{n} \sum ^2 x_i - \frac{1}{n} \times 2X\sum x_i + \frac{1}{n} \times X^2 \sum 1}\]

\[ = \sqrt{\frac{1}{n} \sum  x^2_i - \frac{1}{n} \times 2X \times nX + \frac{1}{n} \times X^2 \times n} \left( X = \frac{1}{n} \sum x_i \right)\]

\[= \sqrt{\frac{1}{n} \sum x_i^2 - 2 X^2 +_{} X^2}\]

\[ = \sqrt{\frac{1}{n} \sum x_i^2 - X^2_{}}\]

\[ = \sigma'\]

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

\[X = \frac{1}{n}\sum_{} x_i\].