The mean and standard deviation of 100 observ | vedclass

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(B) Given,$n = 100$,$\bar{x} = 40$,and $\sigma = 5.1$.The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.Substituting the val

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$161701$

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(B) Given,$n = 100$,$\bar{x} = 40$,and $\sigma = 5.1$.The formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.Substituting the values: $(5.1)^2 = \frac{\sum x_i^2}{100} - (40)^2$.$26.01 = \frac{\sum x_i^2}{100} - 1600$.$\sum x_i^2 = (26.01 + 1600) 	imes 100 = 162601$.This sum includes the incorrect observation $50$. To find the correct sum of squares,we subtract the square of the incorrect value and add the square of the correct value:Correct $\sum x_i^2 = 162601 - (50)^2 + (40)^2$.Correct $\sum x_i^2 = 162601 - 2500 + 1600 = 161701$.

$x$	$2$	$6$	$8$	$9$
$f$	$4$	$4$	$\alpha$	$\beta$

The mean and standard deviation of $100$ observations $x_1, x_2, \ldots, x_{100}$ were calculated as $40$ and $5.1$ respectively by a student who took by mistake $50$ instead of $40$ for one observation. Then the correct value of $\sum_{i=1}^{100} x_i^2=$

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