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	$2$	$4$	$6$

$x_i$	$4$	$8$	$11$	$17$	$20$	$24$	$32$
$f_i$	$3$	$5$	$9$	$5$	$4$	$3$	$1$

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

Similar Questions

If for a distribution $\Sigma(x-5)=3$,$\Sigma(x-5)^{2}=43$ and the total number of items is $18$,find the mean and standard deviation.

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$x_i$ $4$ $8$ $11$ $17$ $20$ $24$ $32$

$f_i$ $3$ $5$ $9$ $5$ $4$ $3$ $1$

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If each observation of a distribution with variance $\sigma^2$ is multiplied by $\lambda$,find the standard deviation of the new observations.

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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$