The variance sigma^2 of the data is . . . . | vedclass

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(B) First,calculate the mean $\overline{x}$:						$x_i$			$f_i$			$f_ix_i$			$f_ix_i^2$				$0$$3$$0$$0$		$1$$2$$2$$2$		$5$$3$$15$$75$		$6$$2$$12$$72$

Vedclass · Accepted Answer

$29$

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(B) First,calculate the mean $\overline{x}$:						$x_i$			$f_i$			$f_ix_i$			$f_ix_i^2$				$0$$3$$0$$0$		$1$$2$$2$$2$		$5$$3$$15$$75$		$6$$2$$12$$72$		$10$$6$$60$$600$		$12$$3$$36$$432$		$17$$3$$51$$867$		Total$\sum f_i = 22$$\sum f_ix_i = 176$$\sum f_ix_i^2 = 2048$	$\overline{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{176}{22} = 8$Variance $\sigma^2 = \frac{1}{N} \sum f_i x_i^2 - (\overline{x})^2$$\sigma^2 = \frac{2048}{22} - (8)^2$$\sigma^2 = 93.09 - 64 = 29.09$Rounding to the nearest integer,the correct option is $B$.

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

Class	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
$f_i$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

The variance $\sigma^2$ of the data is $ . . . . . .$

$x_i$ $0$ $1$ $5$ $6$ $10$ $12$ $17$

$f_i$ $3$ $2$ $3$ $2$ $6$ $3$ $3$

Similar Questions

Variance of first $n$ natural numbers is $\qquad$ .

If $\sum\limits_{i = 1}^{18} {(x_i - 8) = 9}$ and $\sum\limits_{i = 1}^{18} {(x_i - 8)^2 = 45}$,then the standard deviation of $x_1, x_2, \dots, x_{18}$ is:

The variance of the following continuous frequency distribution is
Class Interval $0-10$ $10-20$ $20-30$ $30-40$
Frequency $2$ $3$ $4$ $1$

Calculate the mean,variance,and standard deviation for the following distribution:

Class $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$

$f_i$ $3$ $7$ $12$ $15$ $8$ $3$ $2$

The variance of the first $10$ natural numbers which are multiples of $3$ is

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The variance $\sigma^2$ of the data is $ . . . . . .$ $x_i$ $0$ $1$ $5$ $6$ $10$ $12$ $17$ $f_i$ $3$ $2$ $3$ $2$ $6$ $3$ $3$