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.

Similar Questions

The means of two groups of observations $A$ and $B$ are $\bar{x}$ and $\bar{y}$ respectively,and their standard deviations are $2$ and $3$ respectively. In order for group $A$ to be more consistent than group $B$,$\frac{\bar{y}}{\bar{x}} < $

If the coefficients of variation of two distributions are $40$ and $20$ and their variances are $144$ and $64$ respectively,then the mean of their arithmetic means is

The mean of a data set consisting of $20$ observations is $40$. If one observation $53$ was wrongly recorded as $33$,then the correct mean will be

The Harmonic Mean of $3, 7, 8, 10, 14$ is = .........

Vedclass Test Series

Exam Paper Generator

Online Exam Module