The mean of 100 observations is 50 and their | vedclass

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(A) Given: $n = 100$,$\bar{x} = 50$,and $\sigma = 5$. We know that the mean $\bar{x} = \frac{\sum x_i}{n} = 50$. Therefore,$\sum x_i = 50 	imes 100 =

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$252500$

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(A) Given: $n = 100$,$\bar{x} = 50$,and $\sigma = 5$. We know that the mean $\bar{x} = \frac{\sum x_i}{n} = 50$. Therefore,$\sum x_i = 50 	imes 100 = 5000$. The formula for standard deviation is $\sigma = \sqrt{\frac{\sum x_i^2}{n} - (\bar{x})^2}$. Substituting the values: $5 = \sqrt{\frac{\sum x_i^2}{100} - (50)^2}$. Squaring both sides: $25 = \frac{\sum x_i^2}{100} - 2500$. $\frac{\sum x_i^2}{100} = 2500 + 25 = 2525$. $\sum x_i^2 = 2525 	imes 100 = 252500$.

Class Interval	$0-10$	$10-20$	$20-30$	$30-40$
Frequency	$2$	$3$	$4$	$1$

$X$	$5$	$6$	$7$	$8$	$10$
Frequency	$3$	$7$	$4$	$2$	$4$

The mean of $100$ observations is $50$ and their standard deviation is $5$. Then,the sum of the squares of all the observations is:

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