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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
$4$
$2$
$3$
$9$
Solution
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$
