Question

Let $n\ge 3$. A list of numbers $x_1,x_2,.....x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers $y_1,y_2,....y_n$ is made as follows : $y_1=\frac{x_1+x_2}{2},y_2=\frac{x_1+x_2}{2}$ and $y_j=x_j$ for $j=3,4,....n$. The mean and the standard deviation of the list are $\hat{\mu }$ and $\hat{\sigma }$. Then which of the following is necessarily true?

• A. $\mu =\hat{\sigma }$ and $\sigma \le \hat{\sigma }$
• B. $\mu =\hat{\mu }$ and $\sigma \ge \hat{\sigma }$
• C. $\sigma =\hat{\sigma }$
• D. $\mu \ne \hat{\mu }$

Answer 1

Answer: (B)

$\mu =\hat{\mu }$ and $\sigma \ge \hat{\sigma }$

$\mu =\frac{x_1+x_2+.......+x_n}{n}\hat{\mu }=\frac{y_1+y_2+.......+y_n}{n}=\frac{\frac{x_1+x_2}{2}+\frac{x_2+x_1}{2}+x_3+...+x_n}{n}\hat{\mu }=\frac{x_1+x_2+.......+x_n}{n}=\mu {\sigma }^2=\sum \frac{x_i^2}{n}-{\mu }^2{\sigma }^2=\frac{x_1^2+x_2^2+....x_n^2}{n}-{\mu }^2{\hat{\sigma }}^2=\frac{y_1^2+y_2^2+....y_n^2}{n}-{\hat{\mu }}^2=\frac{{\left(\frac{x_1+x_2}{2}\right)}^2+{\left(\frac{x_1+x_2}{2}\right)}^2+{x_3}^2......+{x_n}^2}{n}-{\hat{\mu }}^2{\hat{\sigma }}^2=\frac{\frac{x_1^2+x_2^2}{2}+x_1{x}_2+{x_3}^2......+{x_n}^2}{n}-{\hat{\mu }}^2{\sigma }^2-{\hat{\sigma }}^2=\frac{x_1^2+x_2^2}{n}-\left(\frac{x_1^2+x_2^2+2x_1{x}_2}{2n}\right)=\left(\frac{x_1^2+x_2^2-2x_1{x}_2}{2n}\right)=\frac{{(x_1-x_2)}^2}{2n}\ge 0\Rightarrow \sigma \ge \hat{\sigma }\&\mu =\hat{\mu }$

So option $B$ is correct