પાંચ અવલોકનોનો મધ્યક અને વિચરણ અનુક્રમે 5 અને | vedclass

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Question

$\mu  = \frac{{1 + 3 + 8 + x + y}}{5}$

$25 = 12 + x + y \Rightarrow x + y = 13\,\,\,\,\,\,\,\,........\left( 1 \right)$

${\sigma ^2} = \fra

Vedclass · Accepted Answer

$4 : 9$

Vedclass · Answer

$\mu  = \frac{{1 + 3 + 8 + x + y}}{5}$

$25 = 12 + x + y \Rightarrow x + y = 13\,\,\,\,\,\,\,\,........\left( 1 ight)$

${\sigma ^2} = \frac{{\sum {{{\left( {{x_i} - \mu } ight)}^2}} }}{N}$

$9.2 = \frac{{1 + 9 + 64 + {x^2} + {y^2}}}{5} - 25$

$34.2 	imes 5 = 74 + {x^2} + {y^2}$

$171 = 74 + {x^2} + {y^2}$

$97 = {x^2} + {y^2}..........\left( 2 ight)$

${\left( {x + y} ight)^2} = {x^2} + {y^2} + 2xy$

$169 - 97 = 2xy \Rightarrow xy = 36$

$T = 4,9$

So, ratio is $\frac{4}{9}$ or $\frac{9}{4}$

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જો $\sum\limits_{i\, = \,1}^{18} {({x_i}\, - \,\,8)\,\, = \,\,9} $ અને $\,\sum\limits_{i\, = \,1}^{18} {{{({x_i}\, - \,\,8)}^2}\, = \,\,45} ,\,$ હોય, તો $\,{{\text{x}}_{\text{1}}},\,\,{x_2},\,........\,\,{x_{18}}$ નું પ્રમાણિત વિચલન શોધો .