JEE Main 2022 — Statistics Question with Solution

Given,

Mean of observation is $15$,

So $\frac{\sum x_i}{20}=15$

$\Rightarrow \sum x_i=15\times 20=300 \cdots \left(\mathrm{i}\right)$

Also variance is given as $9$

So, $\frac{\sum x_i^2}{20}-{\left(\frac{\sum x_i}{20}\right)}^2=9 \cdots \left(\mathrm{ii}\right)$

$\Rightarrow \frac{\sum x_i^2}{20}-{\left(15\right)}^2=9 \cdots \left(\mathrm{ii}\right)$

$\Rightarrow \sum x_i^2=234\times 20=4680$

Given mean of ${\left(x_1+\alpha \right)}^2,{\left(x_2+\alpha \right)}^2,\dots ,{\left(x_{20}+\alpha \right)}^2$ is $178$

So, $\frac{\sum {\left(x_i+\alpha \right)}^2}{20}=178\Rightarrow \sum {\left(x_i+\alpha \right)}^2=3560$

$\Rightarrow \sum x_i^2+2\alpha \sum x_i+\sum {\alpha }^2=3560$

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

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

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

$\alpha =-2,-28$

Square of maximum value of $\alpha$ is $4$