પ્રયોગના 5 અલોકનોનો મધ્યક અને વિચરણ અનુ | vedclass

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Question

આપેલ મુજબ $ \bar x\,\, = \,\,{	ext{4,}}\,\,{	ext{n}}\,\, = \,\,{	ext{5}}\,\,$ અને ${\sigma ^{	ext{2}}}\, = \,\,{	ext{5}}{	ext{.2}}\,{\

Vedclass · Accepted Answer

$4, 7$

Vedclass · Answer

આપેલ મુજબ $ \bar x\,\, = \,\,{	ext{4,}}\,\,{	ext{n}}\,\, = \,\,{	ext{5}}\,\,$ અને ${\sigma ^{	ext{2}}}\, = \,\,{	ext{5}}{	ext{.2}}\,{	ext{.}}$ જો બાકીના અવલોકનો ${{	ext{x}}_{	ext{1}}}{	ext{,}}\,\,{{	ext{x}}_{	ext{2}}}{	ext{ }}$ હોય તો 

${\sigma ^{	ext{2}}}\, = \,\,\frac{{\Sigma {{({x_i}\, - \,\,\bar x)}^2}}}{n}\,\,\, = \,\,5.2$

$ \Rightarrow \,\,\frac{{{{({x_1}\, - \,\,4)}^2}\, + \,\,{{({x_2}\, - \,\,4)}^2}\, + \,\,{{(1\,\, - \,\,4)}^2}\, + \,\,{{(2\,\, - \,\,4)}^2}\, + \,\,{{(6\,\, - \,\,4)}^2}}}{5}\,\,\, = \,\,5.2\,\,\,\,\,\,\,\,\,\,\,\,$

$ \Rightarrow \,\,{({x_1}\, - \,\,4)^2}\, + \,\,{({x_2}\, - \,\,4)^2}\, = \,\,9\,\,.......\,(1)$

ઉપરાંત $ \bar x $     $ = \,\,{	ext{4}}\,\,\,\,\, \Rightarrow \,\,\frac{{{{	ext{x}}_{	ext{1}}}\, + \,\,{x_2}\, + \,\,1\,\, + \,\,2\,\, + \,\,6}}{5}\,\,\, = \,\,4\,\,$

$\, \Rightarrow \,\,{x_1}\, + \,\,{x_2}\,\, = \,\,11\,\,..........\,\,(2)$

સમીકરણ ${	ext{ }}{	ext{. (1), (2) }}$ પરથી ${{	ext{x}}_{	ext{1}}}\,,\,\,{x_2}\,\, = \,\,4,\,\,7$

ચલ $( x )$	$x _{1}$	$x _{1}$	$x _{3} \ldots \ldots x _{15}$
આવૃતિ $(f)$	$f _{1}$	$f _{1}$	$f _{3} \ldots f _{15}$

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