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Let $n \geq 3$. A list of numbers $0 < x_{1} < x_{2} < \ldots < x_{n}$ has mean $\mu$ and standard deviatiion $\sigma$. A new list of numbers is made as follows : $y_{1}=0, y_{2}=x_{2}, \ldots, y_{n-1}=x_{n-1}, y_{n}=x_{1}+x_{n}$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?
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Verified Answer
The correct answer is:
$\mu=\hat{\mu}, \sigma \leq \hat{\sigma}$
$\hat{\mu}=\frac{0+x_{2}+x_{3} \ldots \ldots x_{n-1}+\left(x_{1}+x_{n}\right)}{n}=\mu$
$\hat{\sigma}^{2}=\frac{1}{n} \Sigma y_{i}^{2}-\hat{\mu}^{2}$
$=\frac{1}{n}\left[0^{2}+x_{2}^{2}+x_{3}^{2} \ldots\left(x_{n-1}\right)^{2}+\left(x_{1}+x_{n}\right)^{2}\right]-\mu^{2}$
$\hat{\sigma}^{2}-\sigma^{2}=\frac{2 x_{1} x_{n}}{n}>0$
so $\hat{\sigma}>\sigma$
$\hat{\sigma}^{2}=\frac{1}{n} \Sigma y_{i}^{2}-\hat{\mu}^{2}$
$=\frac{1}{n}\left[0^{2}+x_{2}^{2}+x_{3}^{2} \ldots\left(x_{n-1}\right)^{2}+\left(x_{1}+x_{n}\right)^{2}\right]-\mu^{2}$
$\hat{\sigma}^{2}-\sigma^{2}=\frac{2 x_{1} x_{n}}{n}>0$
so $\hat{\sigma}>\sigma$
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