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Question

$\,\because \,\,\frac{1}{n}\,\Sigma {({x_i}\, + \,\,2)^2}\, = \,\,18\,$ અને $\frac{1}{n}\,\Sigma {({x_i}\, - \,\,2)^2}\, = \,\,10$

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Vedclass · Accepted Answer

$3$

Vedclass · Answer

$\,\because \,\,\frac{1}{n}\,\Sigma {({x_i}\, + \,\,2)^2}\, = \,\,18\,$ અને $\frac{1}{n}\,\Sigma {({x_i}\, - \,\,2)^2}\, = \,\,10$

$ \Rightarrow \,\,\Sigma {({x_i}\, + \,\,2)^2}\,\, = \,\,18n\,$ અને $\Sigma {({x_i}\, - \,\,2)^2}\, = \,\,10n$

$ \Rightarrow \,\,\Sigma {({x_i}\, + \,\,2)^2}\, + \,\,\Sigma {({x_i}\, - \,\,2)^2}\, = \,\,28\,\,n\,\,$ અને $\Sigma {({x_i}\, + \,\,2)^2}\, - \,\,\Sigma {({x_i}\, - \,\,2)^2}\,\, = \,\,8\,\,n$

$ \Rightarrow \,\,2\Sigma \,x_i^2\,\, + \,\,8n\,\, = \,\,28\,n$ અને $8\Sigma {x_i}\,\, = \,\,8n$

$\, \Rightarrow \,\,\Sigma \,x_i^2\,\, = \,\,10\,\,n$ અને $\,\Sigma {x_i}\, = \,\,n$

$ \Rightarrow \,\,\frac{{\Sigma x_i^2}}{n}\,\, = \,\,10\,$ અને $\frac{{\Sigma {x_i}}}{n}\,\, = \,\,1\,\,$

$	herefore \,\,\sigma \,\, = \,\,\sqrt {\frac{{\Sigma x_i^2}}{n}\,\, - \,\,{{\left( {\frac{{\Sigma {x_i}}}{n}} ight)}^2}} \,\, = \,\,\sqrt {10\,\, - \,\,{{(1)}^2}} \,\,\, = \,\,3$

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