QJEEM22Mabc2f: JEE Main 2022 Mathematics - Class 11 JEEM22Mabc2f Solution: Let the mean and the variance of 20 observations be 15 and 9 ,…

Given $\sum\limits_{{{i = 1} \over {20}}}^{20} {{x_i} = 15 \Rightarrow \sum\limits_{i = 1}^{20} {{x_i} = 300} }$

and $\sum\limits_{{{i = 1} \over {20}}}^{20} {x_i^2 - {{\left( {\overline x } \right)}^2} = 9 \Rightarrow \sum\limits_{i = 1}^{20} {x_i^2 = 4680} }$

Mean $= {{{{({x_i} + \alpha )}^2} + {{({x_2} + \alpha )}^2}\, + \,.....\, + \,{{({x_{20}} + \alpha )}^2}} \over {20}} = 178$

$\Rightarrow {{\sum\limits_{i = 1}^{20} {x_i^2 + 2\alpha \sum\limits_{i = 1}^{20} {{x_i} + 20{\alpha ^2}} } } \over {20}} = 178$

$\Rightarrow 4680 + 600\alpha + 20{\alpha ^2} = 3560$

$\Rightarrow {\alpha ^2} + 30\alpha + 56 = 0$

$\Rightarrow {\alpha ^2} + 28\alpha + 2\alpha + 56 = 0$

$\Rightarrow (\alpha + 28)(\alpha + 2) = 0$

${\alpha _{\max }} = - 2 \Rightarrow \alpha _{\max }^2 = 4.$