The mean and standard deviation of a distribu | vedclass

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(C) Given,$n=20$,$\bar{x}=40$,and $\sigma=5$.Sum of weights $\Sigma x = n \bar{x} = 20 	imes 40 = 800$.Variance $\sigma^2 = \frac{\Sigma x^2}{n} - (\

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$26.78$

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(C) Given,$n=20$,$\bar{x}=40$,and $\sigma=5$.Sum of weights $\Sigma x = n \bar{x} = 20 	imes 40 = 800$.Variance $\sigma^2 = \frac{\Sigma x^2}{n} - (\bar{x})^2 = 25$.$\frac{\Sigma x^2}{20} - 40^2 = 25$ $\Rightarrow \frac{\Sigma x^2}{20} = 1625$ $\Rightarrow \Sigma x^2 = 32500$.After excluding two boys with weights $43 \ kg$ and $37 \ kg$:New sum of weights $\Sigma x_{new} = 800 - 43 - 37 = 720$.New number of boys $n_{new} = 18$.New mean $\bar{x}_{new} = \frac{720}{18} = 40$.New sum of squares $\Sigma x_{new}^2 = 32500 - (43)^2 - (37)^2 = 32500 - 1849 - 1369 = 29282$.New variance $\sigma_{new}^2 = \frac{\Sigma x_{new}^2}{n_{new}} - (\bar{x}_{new})^2 = \frac{29282}{18} - (40)^2 = 1626.777... - 1600 = 26.777... \approx 26.78$.

The mean and standard deviation of a distribution of weights of a group of $20$ boys are $40 \ kg$ and $5 \ kg$ respectively. If two boys of weights $43 \ kg$ and $37 \ kg$ are excluded from this group,then the variance of the distribution of weights of the remaining group of boys is

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