MCQ
જો $\mathop \sum \limits_{i = 1}^9 \left( {{x_i} - 5} \right) = 9$ અને $\mathop \sum \limits_{i = 1}^9 {\left( {{x_i} - 5} \right)^2} = 45,$ તો અવલોકનો ${x_1},{x_2},\;.\;.\;.\;,{x_9}$ નું પ્રમાણિત વિચલન . . . . છે.
- A$4$
- ✓$2$
- C$3$
- D$9$
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$
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