show-that-the-variance-2-of-a-set-of-observations-x-1-x-2-x-n-with-mean-x-can-be-expressed-in-the-form-2-i-1-n-x-i-2-n-x-2-

Question Number 88592 by Rio Michael last updated on 11/Apr/20

show that the variance δ^{2} of a set of observations x_{1}, x_{2}, \dots x_{n} with mean

\overline{x} can be expressed in the form δ^{2} = \frac{\underset{i = 1}{\sum^{n}} x_{i}^{2}}{n} - \bar{x}^{2}

Answered by mr W last updated on 11/Apr/20

a c c . t o d e f i n i t i o n o f v a r i a n c e :

σ^{2} = \frac{1}{n} \underset{i = 1}{\sum^{n}} {(x_{i} - \overset{―}{x})}^{2}

σ^{2} = \frac{1}{n} \underset{i = 1}{\sum^{n}} (x_{i}^{2} - 2 {\overset{―}{x x}}_{i} + \overset{―}{x^{2}})

σ^{2} = \frac{1}{n} {\underset{i = 1}{\sum^{n}} x_{i}^{2} - 2 \overset{―}{x} \underset{i = 1}{\sum^{n}} x_{i} + \underset{i = 1}{\sum^{n}} \overset{―}{x^{2}}}

σ^{2} = \frac{1}{n} {\underset{i = 1}{\sum^{n}} x_{i}^{2} - 2 \overset{―}{x} (n \overset{―}{x}) + \overset{―}{x^{2}} n}

σ^{2} = \frac{1}{n} {\underset{i = 1}{\sum^{n}} x_{i}^{2} - n \overset{―}{x^{2}}}

σ^{2} = \frac{1}{n} \underset{i = 1}{\sum^{n}} x_{i}^{2} - \overset{―}{x^{2}}

Commented by Rio Michael last updated on 11/Apr/20

thanks sir