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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 deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?
$\mu=\hat{\mu}, \sigma \leq \hat{\sigma}$
$\mu=\hat{\mu}, \sigma \geq \hat{\sigma}$
$\sigma=\hat{\sigma}$
$\mu$ may or may not be equal to $\hat{\mu}$
Solution
(a)
We have,
$\operatorname{Mean}(\mu)=\frac{x_1+x_2+x_3+\ldots+x_n}{n}$
$\mu=\frac{\Sigma x_i}{n}$
Standard deviation $\sigma=\sqrt{\frac{\Sigma x_i^2}{n}-(\mu)^2}$
Mean of other observations
$\operatorname{Mean}\left(\mu^{\prime}\right)=\frac{y_1+y_2+y_3+\ldots+y_{n-1}+y_n}{n}$
$=\frac{0+x_2+x_3+\ldots+x_{n-1}+x_1+x_n}{n}$
$=\frac{\Sigma x_i}{n}=\mu$
$\mu^{\prime} =\mu$
$\mu^{\prime} =\sqrt{\frac{\sum y_i^2}{n}-\left(\mu^{\prime}\right)^2}$
$\sigma^{\prime}=\sqrt{\frac{0+x_2^2+x_3^2+\ldots+x_{n-1}^2}{+\left(x_1+x_n\right)^2}-\mu}$
$\quad \Sigma x_1^2=x_1^2+x_2^2+\ldots+x_n^2$
$\quad \Sigma y_1^2=0+x_2^2+\ldots+x_{n-1}^2+x_1^2+x_n^2+2 x_1 x_n$
$\text { Clearly, } \Sigma y_1^2 \geq \Sigma x_1^2$
$\therefore \quad \sigma^{\prime} \geq \sigma$
Hence, option (a) is correct.
