If the variance of the frequency distribution | vedclass

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(C) The total frequency $N = 2 + 1 + 1 + 1 + 1 + 1 = 7$.The mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{2c + 2c + 3c + 4c + 5c + 6c}{7} = \frac{22c

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$7$

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(C) The total frequency $N = 2 + 1 + 1 + 1 + 1 + 1 = 7$.The mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{2c + 2c + 3c + 4c + 5c + 6c}{7} = \frac{22c}{7}$.The variance $\sigma^2 = \frac{\sum f_i x_i^2}{N} - (\bar{x})^2$.$\sum f_i x_i^2 = 2(c)^2 + 1(2c)^2 + 1(3c)^2 + 1(4c)^2 + 1(5c)^2 + 1(6c)^2 = c^2(2 + 4 + 9 + 16 + 25 + 36) = 92c^2$.$\sigma^2 = \frac{92c^2}{7} - \left(\frac{22c}{7}\right)^2 = \frac{92c^2}{7} - \frac{484c^2}{49} = \frac{644c^2 - 484c^2}{49} = \frac{160c^2}{49}$.Given $\sigma^2 = 160$,we have $\frac{160c^2}{49} = 160$.$c^2 = 49 \Rightarrow c = 7$ (since $c \in N$).

$X$	$c$	$2c$	$3c$	$4c$	$5c$	$6c$
$f$	$2$	$1$	$1$	$1$	$1$	$1$

If the variance of the frequency distribution is $160$,then the value of $c \in N$ is

$X$ $c$ $2c$ $3c$ $4c$ $5c$ $6c$

$f$ $2$ $1$ $1$ $1$ $1$ $1$

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If the variance of the frequency distribution is $160$,then the value of $c \in N$ is $X$ $c$ $2c$ $3c$ $4c$ $5c$ $6c$ $f$ $2$ $1$ $1$ $1$ $1$ $1$