Question

Let the mean and the variance of 20 observations $x_1,x_2,\dots x_{20}$ be 15 and 9 , respectively. For $\alpha \in R$, if the mean of ${\left(x_1+\alpha \right)}^2,{\left(x_2+\alpha \right)}^2,\dots ,{\left(x_{20}+\alpha \right)}^2$ is 178 , then the square of the maximum value of $\alpha$ is equal to

Answer 1

Answer: 4

$\sum x_1=15\times 20=300$...(i)
$\frac{\sum x_1^2}{20}-(15{)}^2=9$...(ii)
$\sum x_1^2=234\times 20=4680$
$\frac{\sum {\left(x_1+\alpha \right)}^2}{20}=178\Rightarrow \sum {\left(x_1+\alpha \right)}^2=3560$
$\Rightarrow \sum x_1^2+2\alpha \sum x_1+\sum {\alpha }^2=3560$
$4680+600\alpha +20{\alpha }^2=3560$
$\Rightarrow {\alpha }^2+30\alpha +56=0$
$\Rightarrow (\alpha +28)(\alpha +2)=0$
$\alpha =-2,-28$
Square of maximum value of $\alpha$ is 4