If mathop sum limits_{i = 1}^9 left( {{x_i} - | 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$

If $\mathop \sum \limits_{i = 1}^9 \left( {{x_i} - 5} \right) = 9$ and $\mathop \sum \limits_{i = 1}^9 {\left( {{x_i} - 5} \right)^2} = 45,$ then the standard deviation of the $9$ items ${x_1},{x_2},\;.\;.\;.\;,{x_9}$ is :

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