For a group of 100 students,the mean bar{x}_1 | vedclass

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(C) Given $n=100$,$\bar{x}_1=40$,and $\sigma_1=15$. The sum of observations is $\sum x_i = n 	imes \bar{x}_1 = 100 	imes 40 = 4000$. The corrected s

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$\bar{x}_1=\bar{x}_2 ; \sigma_1 > \sigma_2$

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(C) Given $n=100$,$\bar{x}_1=40$,and $\sigma_1=15$. The sum of observations is $\sum x_i = n 	imes \bar{x}_1 = 100 	imes 40 = 4000$. The corrected sum is $\sum x_i^{\prime} = 4000 - 30 - 60 + 40 + 50 = 4000$. Thus,the corrected mean $\bar{x}_2 = \frac{4000}{100} = 40$. So,$\bar{x}_1 = \bar{x}_2$. Now,$\sigma_1^2 = \frac{\sum x_i^2}{n} - (\bar{x}_1)^2$ $\Rightarrow 225 = \frac{\sum x_i^2}{100} - 1600$ $\Rightarrow \sum x_i^2 = 182500$. The corrected sum of squares is $\sum x_i^{\prime 2} = 182500 - 30^2 - 60^2 + 40^2 + 50^2 = 182500 - 900 - 3600 + 1600 + 2500 = 182100$. The corrected variance is $\sigma_2^2 = \frac{182100}{100} - (40)^2 = 1821 - 1600 = 221$. Since $\sigma_1^2 = 225$ and $\sigma_2^2 = 221$,we have $\sigma_1^2 > \sigma_2^2$,which implies $\sigma_1 > \sigma_2$. Therefore,$\bar{x}_1 = \bar{x}_2$ and $\sigma_1 > \sigma_2$.

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