(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.