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(D) Given that the mean $\bar{x} = 15$ and variance $\sigma^{2} = 9$ for $n = 20$ observations.$\sum x_{i} = 15 	imes 20 = 300$$\sigma^{2} = \frac{\s

Vedclass · Accepted Answer

$4$

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(D) Given that the mean $\bar{x} = 15$ and variance $\sigma^{2} = 9$ for $n = 20$ observations.$\sum x_{i} = 15 	imes 20 = 300$$\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} \Rightarrow 9 = \frac{\sum x_{i}^{2}}{20} - 225$$\frac{\sum x_{i}^{2}}{20} = 234 \Rightarrow \sum x_{i}^{2} = 4680$Now,the mean of $(x_{i} + \alpha)^{2}$ is $178$:$\frac{1}{20} \sum (x_{i} + \alpha)^{2} = 178$$\sum (x_{i}^{2} + 2\alpha x_{i} + \alpha^{2}) = 178 	imes 20 = 3560$$\sum x_{i}^{2} + 2\alpha \sum x_{i} + 20\alpha^{2} = 3560$$4680 + 2\alpha(300) + 20\alpha^{2} = 3560$$20\alpha^{2} + 600\alpha + 1120 = 0$Dividing by $20$:$\alpha^{2} + 30\alpha + 56 = 0$$(\alpha + 28)(\alpha + 2) = 0$So,$\alpha = -28$ or $\alpha = -2$.The maximum value of $\alpha$ is $-2$.The square of the maximum value is $(-2)^{2} = 4$.

Let the mean and the variance of $20$ observations $x_{1}, x_{2}, \ldots, x_{20}$ be $15$ and $9$,respectively. For $\alpha \in R$,if the mean of $(x_{1}+\alpha)^{2}, (x_{2}+\alpha)^{2}, \ldots, (x_{20}+\alpha)^{2}$ is $178$,then the square of the maximum value of $\alpha$ is equal to $...........$

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