13.Statistics
medium

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

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

A

$5$

B

$8$

C

$7$

D

$6$

(JEE MAIN-2024)

Solution

$x$ $C$ $2C$ $3C$ $4C$ $5C$ $6C$
$f$ $2$ $1$ $1$ $1$ $1$ $1$

$\bar{x}=\frac{(2+2+3+4+5+6) C}{7}=\frac{22 C}{7}$

$ \operatorname{Var}(\mathrm{x})=\frac{\mathrm{c}^2\left(2+2^2+3^2+4^2+5^2+6^2\right)}{7} $

$ -\left(\frac{22 c}{7}\right)^2 $

$ =\frac{92 c^2}{7}-\mathrm{c}^2 \times \frac{484}{49} $

$ =\frac{(644-484) c^2}{49}=\frac{160 c^2}{49} $

$ 160=\frac{160 \times c^2}{49} \Rightarrow c=7$

Standard 11
Mathematics

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