If the probability distribution of a random variable $x$ is
$X = x_{1} : - 2 - 10123$
$p (X = x_{1}) : 0.1 k 0.2 2 k 0.3 k$ , then the variance of $x$ is

2.16

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2.8

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\sqrt{2.16}

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\sqrt{2.8}

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Solution

The correct option is A $2.16$
$\sum_{i = - 2}^{3} p_{i} = 1 0.1 + k + 0.2 + 2 k + 0.3 + k = 1 4 k = 1 - 0.6 = 0.4 k = 0.1 E [X] = \sum_{i = - 2}^{3} x_{i} p_{i} = - 0.2 - k + 2 k + 0.6 + 3 k = 4 k + 0.4 = 0.4 + 0.4 = 0.8 E [X^{2}] = \sum_{i = - 2}^{3} x^{2} p_{i} = 0.4 + k + 2 k + 1.2 + 9 k = 2.6 V a r i a n c e = E [X^{2}] - {E [X]}^{2} = 2.6 - {0.8}^{2} = 2.16$

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