Let n geq 3. A list of numbers x_1, x_2, ldot | vedclass

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(B) Given the mean $\mu = \frac{1}{n} \sum_{i=1}^{n} x_i$.For the new list,the mean $\hat{\mu} = \frac{1}{n} (y_1 + y_2 + \sum_{j=3}^{n} x_j) = \frac{

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

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(B) Given the mean $\mu = \frac{1}{n} \sum_{i=1}^{n} x_i$.For the new list,the mean $\hat{\mu} = \frac{1}{n} (y_1 + y_2 + \sum_{j=3}^{n} x_j) = \frac{1}{n} (\frac{x_1+x_2}{2} + \frac{x_1+x_2}{2} + \sum_{j=3}^{n} x_j) = \frac{1}{n} \sum_{i=1}^{n} x_i = \mu$.Now,consider the variance $\sigma^2 = \frac{1}{n} \sum x_i^2 - \mu^2$ and $\hat{\sigma}^2 = \frac{1}{n} \sum y_i^2 - \hat{\mu}^2$.Since $\hat{\mu} = \mu$,we compare $\sum x_i^2$ and $\sum y_i^2$.$\sum x_i^2 - \sum y_i^2 = (x_1^2 + x_2^2) - (y_1^2 + y_2^2) = x_1^2 + x_2^2 - 2(\frac{x_1+x_2}{2})^2 = x_1^2 + x_2^2 - \frac{x_1^2 + x_2^2 + 2x_1x_2}{2} = \frac{x_1^2 + x_2^2 - 2x_1x_2}{2} = \frac{(x_1-x_2)^2}{2} \geq 0$.Thus,$\sum x_i^2 \geq \sum y_i^2$,which implies $\sigma^2 \geq \hat{\sigma}^2$,so $\sigma \geq \hat{\sigma}$.

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