If $μ_{2} = 20, μ_{2}^{1} = 276$ for a discrete random variable $X$ , then the mean of the random variable $X$ is

16

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Solution

The correct option is D $16$
$μ_{2} = E (X^{2}) - [E (X)]^{2} . . . . . . (i)$
$μ_{2}^{'} = E (X)^{2} . . . . . (i i)$
Given: $μ_{2} = 20, μ_{2}^{'} = 276$
from equation (ii) $⟹ E (x^{2}) = 276$
and from equation (i), $⟹ 20 = 276 - [E (X)]^{2}$
$⟹ [E (X)]^{2} = 276 - 20 = 250 ⟹ E (X) = 16$
is the mean of random variable.

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