show that the variance δ^2 of a set of observations x_1 ,x_2 ,...x_n with mean x^_ can be expressed in the form δ^2 = ((Σ_(i=1) ^n x


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

$show that the variance δ^{2} of a set of observations x_{1}, x_{2}, . . . x_{n} with mean$ $\overline{x} can be expressed in the form δ^{2} = \frac{\underset{i = 1}{\sum^{n}} x_{i}^{2}}{n} - \bar{x}^{2}$
	Answered by mr W last updated on 11/Apr/20
	$acc. to definition of variance: σ^2 =(1/n)Σ_(i=1) ^n (x_i −x^(−) )^2 σ^2 =(1/n)Σ_(i=1) ^n (x_i ^2 −2x^(−) x_i +x^2 ^(−) ) σ^2 =(1/n){Σ_(i=1) ^n x_i ^2 −2x^(−) Σ_(i=1) ^n x_i +Σ_(i=1) ^n x^2 ^(−) } σ^2 =(1/n){Σ_(i=1) ^n x_i ^2 −2x^(−) (nx^(−) )+x^2 ^(−) n} σ^2 =(1/n){Σ_(i=1) ^n x_i ^2 −nx^2 ^(−) } σ^2 =(1/n)Σ_(i=1) ^n x_i ^2 −x^2 ^(−)$
	$a c c . t o d e f i n i t i o n o f v a r i a n c e :$ $σ^{2} = \frac{1}{n} \underset{i = 1}{\sum^{n}} {(x_{i} - \overset{―}{x})}^{2}$ $σ^{2} = \frac{1}{n} \underset{i = 1}{\sum^{n}} (x_{i}^{2} - 2 {\overset{―}{x x}}_{i} + \overset{―}{x^{2}})$ $σ^{2} = \frac{1}{n} {\underset{i = 1}{\sum^{n}} x_{i}^{2} - 2 \overset{―}{x} \underset{i = 1}{\sum^{n}} x_{i} + \underset{i = 1}{\sum^{n}} \overset{―}{x^{2}}}$ $σ^{2} = \frac{1}{n} {\underset{i = 1}{\sum^{n}} x_{i}^{2} - 2 \overset{―}{x} (n \overset{―}{x}) + \overset{―}{x^{2}} n}$ $σ^{2} = \frac{1}{n} {\underset{i = 1}{\sum^{n}} x_{i}^{2} - n \overset{―}{x^{2}}}$ $σ^{2} = \frac{1}{n} \underset{i = 1}{\sum^{n}} x_{i}^{2} - \overset{―}{x^{2}}$
	Commented by Rio Michael last updated on 11/Apr/20

	$thanks sir$
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