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(D) Given,the probability distribution of $X$ is defined by $P(X=0)=3C^3$,$P(X=2)=5C-10C^2$,and $P(X=4)=4C-1$.We know that the sum of all probabilitie

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$\frac{128}{81}$

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(D) Given,the probability distribution of $X$ is defined by $P(X=0)=3C^3$,$P(X=2)=5C-10C^2$,and $P(X=4)=4C-1$.We know that the sum of all probabilities in a distribution must be $1$,so $\Sigma P(X)=1$.$3C^3 + (5C-10C^2) + (4C-1) = 1$$3C^3 - 10C^2 + 9C - 2 = 0$Factoring the cubic equation: $(C-1)(3C^2-7C+2) = 0$$(C-1)(3C-1)(C-2) = 0$This gives $C = 1, \frac{1}{3}, 2$.Since probabilities must be between $0$ and $1$,we test the values. If $C=1$,$P(X=4)=4(1)-1=3 > 1$ (impossible). If $C=2$,$P(X=4)=4(2)-1=7 > 1$ (impossible). Thus,$C=\frac{1}{3}$.Substituting $C=\frac{1}{3}$ into the probabilities:$P(X=0) = 3(\frac{1}{3})^3 = \frac{3}{27} = \frac{1}{9}$$P(X=2) = 5(\frac{1}{3}) - 10(\frac{1}{3})^2 = \frac{5}{3} - \frac{10}{9} = \frac{15-10}{9} = \frac{5}{9}$$P(X=4) = 4(\frac{1}{3}) - 1 = \frac{4}{3} - 1 = \frac{1}{3}$The mean $E(X) = \Sigma X P(X) = (0 	imes \frac{1}{9}) + (2 	imes \frac{5}{9}) + (4 	imes \frac{1}{3}) = 0 + \frac{10}{9} + \frac{4}{3} = \frac{10+12}{9} = \frac{22}{9}$.The mean of squares $E(X^2) = \Sigma X^2 P(X) = (0^2 	imes \frac{1}{9}) + (2^2 	imes \frac{5}{9}) + (4^2 	imes \frac{1}{3}) = 0 + \frac{20}{9} + \frac{16}{3} = \frac{20+48}{9} = \frac{68}{9}$.Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{68}{9} - (\frac{22}{9})^2 = \frac{68}{9} - \frac{484}{81} = \frac{612-484}{81} = \frac{128}{81}$.

$X = x$	$0$	$1$	$2$
$P(X = x)$	$4k - 10k^2$	$5k - 1$	$3k^3$

$X = k$	$-2$	$-1$	$0$	$1$	$2$
$P(X = k)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$

If a random variable $X$ has the probability distribution given by $P(X=0)=3C^3$,$P(X=2)=5C-10C^2$ and $P(X=4)=4C-1$,then the variance of that distribution is

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