Let n geq 3. A list of numbers 0

Question

(A) Given the original list $x_1, x_2, \ldots, x_n$ with mean $\mu = \frac{1}{n} \sum_{i=1}^n x_i$.The new list is $y_1=0, y_2=x_2, \ldots, y_{n-1}=x_

Vedclass · Accepted Answer

$\mu=\hat{\mu}, \sigma \leq \hat{\sigma}$

Vedclass · Answer

(A) Given the original list $x_1, x_2, \ldots, x_n$ with mean $\mu = \frac{1}{n} \sum_{i=1}^n x_i$.The new list is $y_1=0, y_2=x_2, \ldots, y_{n-1}=x_{n-1}, y_n=x_1+x_n$.The mean of the new list is $\hat{\mu} = \frac{1}{n} (0 + x_2 + x_3 + \ldots + x_{n-1} + x_1 + x_n) = \frac{1}{n} \sum_{i=1}^n x_i = \mu$.Now,consider the sum of squares $\sum y_i^2 = 0^2 + x_2^2 + \ldots + x_{n-1}^2 + (x_1+x_n)^2$.Expanding this,$\sum y_i^2 = x_2^2 + \ldots + x_{n-1}^2 + x_1^2 + x_n^2 + 2x_1x_n = \sum_{i=1}^n x_i^2 + 2x_1x_n$.Since $x_1 > 0$ and $x_n > 0$,we have $2x_1x_n > 0$,so $\sum y_i^2 > \sum x_i^2$.The variance is given by $\sigma^2 = \frac{1}{n} \sum x_i^2 - \mu^2$ and $\hat{\sigma}^2 = \frac{1}{n} \sum y_i^2 - \hat{\mu}^2$.Since $\hat{\mu} = \mu$ and $\sum y_i^2 > \sum x_i^2$,it follows that $\hat{\sigma}^2 > \sigma^2$,which implies $\hat{\sigma} > \sigma$.Thus,$\mu = \hat{\mu}$ and $\sigma Therefore,option $A$ is correct.

List-$I$	List-$II$
$(P)$ The mean of the above data is	$(1) 2.5$
$(Q)$ The median of the above data is	$(2) 5$
$(R)$ The mean deviation about the mean of the above data is	$(3) 6$
$(S)$ The mean deviation about the median of the above data is	$(4) 2.7$
	$(5) 2.4$

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