Question

(x1-x)2 = (1 point BONUS) Calculate the computing formula from the definition af sample variance. That is show that

Answer 1

We have,

$\dpi{100} \fn_cm s^{2} = \frac{1}{n-1} \sum (x_{i} - \bar{x})^{2}$

$\dpi{100} \fn_cm = \frac{1}{n-1} \sum (x_{i}^{2} -2 \bar{x}.x_{i} + \bar{x}^{2})$

  $\dpi{100} \fn_cm = \frac{1}{n-1}(\sum x_{i}^{2} - \sum (2 \bar{x}. x_{i}) + \sum \bar{x}^{2})$

$\dpi{100} \fn_cm = \frac{1}{n-1}(\sum x_{i}^{2} - 2 \bar{x}. \sum x_{i} + \bar{x}^{2} \sum 1)$

$\dpi{100} \fn_cm = \frac{1}{n-1}(\sum x_{i}^{2} - 2 \bar{x}. n \bar{x} + n \bar{x}^{2} )$

$\dpi{100} \fn_cm = \frac{1}{n-1}(\sum x_{i}^{2} - n \bar{x}^{2} )$

$\dpi{100} \fn_cm = \frac{1}{n(n-1)} \times ( n \sum x_{i}^{2} - n^{2} \bar{x}^{2} )$

$\dpi{100} \fn_cm = \frac{1}{n(n-1)} \times ( n \sum x_{i}^{2} - (n\bar{x})^{2} )$

$\dpi{100} \fn_cm = \frac{1}{n(n-1)} \times ( n \sum x_{i}^{2} - (\sum x_{i})^{2} )$

$\dpi{100} \fn_cm = \frac{1}{(n-1)} \times ( \sum x_{i}^{2} - \frac{1}{n}(\sum x_{i})^{2} )$