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(B) Step $1$: Find the midpoints $(x_i)$ of each class interval: $x_1 = \frac{0+4}{2} = 2$,$x_2 = \frac{4+8}{2} = 6$,$x_3 = \frac{8+12}{2} = 10$,$x_4

Vedclass · Accepted Answer

$\frac{44}{3}$

Vedclass · Answer

(B) Step $1$: Find the midpoints $(x_i)$ of each class interval: $x_1 = \frac{0+4}{2} = 2$,$x_2 = \frac{4+8}{2} = 6$,$x_3 = \frac{8+12}{2} = 10$,$x_4 = \frac{12+16}{2} = 14$. Step $2$: Calculate the mean $(\bar{x})$: Total frequency $N = \sum f_i = 1+2+2+1 = 6$. Sum of $f_i x_i = (1 	imes 2) + (2 	imes 6) + (2 	imes 10) + (1 	imes 14) = 2 + 12 + 20 + 14 = 48$. Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{48}{6} = 8$. Step $3$: Calculate the variance $(\sigma^2)$: $\sigma^2 = \frac{1}{N} \sum f_i (x_i - \bar{x})^2$. $\sigma^2 = \frac{1}{6} [1(2-8)^2 + 2(6-8)^2 + 2(10-8)^2 + 1(14-8)^2]$. $\sigma^2 = \frac{1}{6} [1(-6)^2 + 2(-2)^2 + 2(2)^2 + 1(6)^2]$. $\sigma^2 = \frac{1}{6} [36 + 8 + 8 + 36] = \frac{88}{6} = \frac{44}{3}$. Thus,the variance is $\frac{44}{3}$.

$x_i$	$92$	$93$	$97$	$98$	$102$	$104$	$109$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

Class Interval	$0$-$5$	$5$-$10$	$10$-$15$	$15$-$20$	$20$-$25$
Frequency	$4$	$1$	$10$	$3$	$2$

Calculate the variance of the following continuous frequency distribution:
Class Interval $0$–$4$ $4$–$8$ $8$–$12$ $12$–$16$
Frequency $(f_i)$ $1$ $2$ $2$ $1$

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Calculate the variance of the following continuous frequency distribution: Class Interval$0$–$4$$4$–$8$$8$–$12$$12$–$16$Frequency $(f_i)$$1$$2$$2$$1$