If the mean of 100 observations is 50 and the | vedclass

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(C) Given: $\bar{x} = 50$,$n = 100$,and $\sigma = 5$. We know that $\bar{x} = \frac{\Sigma x_i}{n}$,so $\Sigma x_i = n \cdot \bar{x} = 100 \cdot 50 =

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$252500$

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(C) Given: $\bar{x} = 50$,$n = 100$,and $\sigma = 5$. We know that $\bar{x} = \frac{\Sigma x_i}{n}$,so $\Sigma x_i = n \cdot \bar{x} = 100 \cdot 50 = 5000$. The formula for standard deviation is $\sigma = \sqrt{\frac{\Sigma x_i^2}{n} - (\bar{x})^2}$. Squaring both sides,we get $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$. Substituting the given values: $5^2 = \frac{\Sigma x_i^2}{100} - (50)^2$. $25 = \frac{\Sigma x_i^2}{100} - 2500$. $\frac{\Sigma x_i^2}{100} = 2525$. $\Sigma x_i^2 = 252500$.

Classes	$0-30$	$30-60$	$60-90$	$90-120$	$120-150$	$150-180$	$180-210$
$f_i$	$2$	$3$	$5$	$10$	$3$	$5$	$2$

If the mean of $100$ observations is $50$ and their standard deviation is $5$,then the sum of squares of all observations is

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