If the probability mass function (p.m.f.) of | vedclass

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(B) The variance of a random variable $X$ is given by $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$. First,we calculate $E(X) = \sum_{x=1}^{10} x P(X=x)

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$\frac{33}{4}$

Vedclass · Answer

(B) The variance of a random variable $X$ is given by $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$. First,we calculate $E(X) = \sum_{x=1}^{10} x P(X=x) = \frac{1}{10} (1 + 2 + 3 + \ldots + 10) = \frac{1}{10} 	imes \frac{10 	imes 11}{2} = 5.5$. Next,we calculate $E(X^2) = \sum_{x=1}^{10} x^2 P(X=x) = \frac{1}{10} (1^2 + 2^2 + 3^2 + \ldots + 10^2) = \frac{1}{10} 	imes \frac{10 	imes 11 	imes 21}{6} = \frac{231}{6} = 38.5$. Finally,$\operatorname{Var}(X) = 38.5 - (5.5)^2 = 38.5 - 30.25 = 8.25$. Since $8.25 = \frac{33}{4}$,the correct option is $B$.

$X$	$0$	$1$	$2$
$P(X)$	$\frac{25}{36}$	$k$	$\frac{1}{36}$

If the probability mass function (p.m.f.) of a random variable $X$ is $P(X=x) = \frac{1}{10}$ for $x = 1, 2, 3, \ldots, 10$,and $0$ otherwise,then $\operatorname{Var}(X)$ is equal to:

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