यदि sum_{i=1}^{9}left(x_{i}-5right)=9 तथा sum | vedclass

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Question

Given $\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)}  = 9 \Rightarrow \sum\limits_{i = 1}^9 {{x_i} = 54\,\,\,.....\left( i \right)} $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Given $\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)}  = 9 \Rightarrow \sum\limits_{i = 1}^9 {{x_i} = 54\,\,\,.....\left( i \right)} $

Also, $\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}}  = 45$

$\sum\limits_{i = 1}^9 {x_i^2}  - 10\sum\limits_{i = 1}^9 {{x_i} + 9\left( {25} \right)}  = 45\,\,\,\,\,...\left( {ii} \right)$

From $(i)$ and $(ii)$ we get,

$\sum\limits_{i = 1}^9 {x_i^2}  = 360$

Since, variance $ = \frac{{\sum {x_i^2} }}{9} - {\left( {\frac{{\sum {{x_i}} }}{9}} \right)^2}$

$ = \frac{{360}}{9} - {\left( {\frac{{54}}{9}} \right)^2} = 40 - 36 = 4$

Standared deviation $ = \sqrt {Variance}  = 2$

व्यास	$33-36$	$37-40$	$41-44$	$45-48$	$49-52$
वृत्तों संख्या	$15$	$17$	$21$	$22$	$25$

यदि $\sum_{i=1}^{9}\left(x_{i}-5\right)=9$ तथा $\sum_{i=1}^{9}\left(x_{i}-5\right)^{2}=45$ है, तो नौ प्रेक्षणों $x_{1}, x_{2}, \ldots . ., x_{9}$ का मानक विचलन है

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