The standard deviation of the following distr | vedclass

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(C) First,we find the midpoints $(x_i)$ for each class interval ($C$.$I$.):For $0-6$,$x_1 = \frac{0+6}{2} = 3$For $6-12$,$x_2 = \frac{6+12}{2} = 9$For

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$2 \sqrt{5}$

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(C) First,we find the midpoints $(x_i)$ for each class interval ($C$.$I$.):For $0-6$,$x_1 = \frac{0+6}{2} = 3$For $6-12$,$x_2 = \frac{6+12}{2} = 9$For $12-18$,$x_3 = \frac{12+18}{2} = 15$Now,calculate $\sum f_i$,$\sum f_i x_i$,and $\sum f_i x_i^2$:$\sum f_i = 2 + 4 + 6 = 12$$\sum f_i x_i = (2 	imes 3) + (4 	imes 9) + (6 	imes 15) = 6 + 36 + 90 = 132$$\sum f_i x_i^2 = (2 	imes 3^2) + (4 	imes 9^2) + (6 	imes 15^2) = (2 	imes 9) + (4 	imes 81) + (6 	imes 225) = 18 + 324 + 1350 = 1692$The variance $V(X)$ is given by:$V(X) = \frac{\sum f_i x_i^2}{\sum f_i} - \left( \frac{\sum f_i x_i}{\sum f_i} ight)^2$$V(X) = \frac{1692}{12} - \left( \frac{132}{12} ight)^2$$V(X) = 141 - (11)^2$$V(X) = 141 - 121 = 20$Standard deviation $\sigma = \sqrt{V(X)} = \sqrt{20} = 2 \sqrt{5}$.

$C$.$I$.	$0$ - $6$	$6$ - $12$	$12$ - $18$
f_i	$2$	$4$	$6$

The standard deviation of the following distribution is:
$C$.$I$. $0$ - $6$ $6$ - $12$ $12$ - $18$
f_i $2$ $4$ $6$

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The standard deviation of the following distribution is:$C$.$I$.$0$ - $6$$6$ - $12$$12$ - $18$f_i$2$$4$$6$