Question

Let the mean and the variance of 20 observations $x_1, x_2,…., x_{20}$ be 15 and 9, respectively. For a ∈ R, if the mean of $(x_1 + α)^2, (x_2 + α)^2,….,(x_{20} + α)^2$ is 178, then the square of the maximum value of $α$ is equal to ___________.

Answer 1

Given, $\begin{array}{l} \displaystyle\sum\limits_{\frac{i=1}{20}}^{20}x_i=15\Rightarrow\ \displaystyle\sum\limits_{i=1}^{20}x_i=300\cdots\left(1\right)\end{array}$
and $\begin{array}{l} \displaystyle\sum\limits_{\frac{i=1}{20}}^{20}x_i^2-\left(\overline{x}\right)^2=9\Rightarrow \displaystyle\sum\limits_{i=1}^{20}x_i^2=4680\cdots\left(2\right)\end{array}$
$\begin{array}{l} \text{Mean}=\frac{\left(x_1+\alpha\right)^2+\left(x_2+\alpha\right)^2+\cdots+\left(x_{20}+\alpha\right)^2}{20} \end{array}$= 178
$\begin{array}{l} \Rightarrow\ \frac{\displaystyle\sum\limits_{i=1}^{20}x_i^2+2\alpha\displaystyle\sum\limits_{i=1}^{20}x_i+20\alpha^2}{20}=178\end{array}$
⇒ 4680 + 600$α$ + 20$α^2$ = 3560
⇒ $α^2$+ 30$α$ + 56 = 0
⇒ $α^2$+ 28$α$ + 2$α$ + 56 = 0
⇒ ($α$ + 28)($α$ + 2) = 0
$α_{max}$= – 2
$\begin{array}{l}\Rightarrow \alpha_{\text{max}}^2=4\end{array}$