If the variance of the frequency distribution is 160, then the value of is

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Solution

Q.
If the variance of the frequency distribution

$x$ $c$ $2 c$ $3 c$ $4 c$ $5 c$ $6 c$

$y$ 2 1 1 1 1 1

is 160, then the value of $c \in N$ is [2024]

1 6

2 8

3 5

4 7

Ans.
(4)

We have, $\bar{x} = \frac{2 c + 2 c + 3 c + 4 c + 5 c + 6 c}{2 + 1 + 1 + 1 + 1 + 1}$ $= \frac{22 c}{7}$

$Var (X) = \frac{c^{2} (2 + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2})}{7} - {(\frac{22 c}{7})}^{2}$

$= \frac{92 c^{2}}{7} - \frac{484 c^{2}}{49} = \frac{(644 - 484) c^{2}}{49} = \frac{160 c^{2}}{49}$

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

$\Rightarrow c^{2} = 49$

$\Rightarrow c = 7$ [ $∵ c \in N$ ]

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