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(B) For a probability distribution,the sum of probabilities is $1$:$a + 2a + (a+b) + 2b + 3b = 1$$4a + 6b = 1$  --- $(i)$The mean $E(X) = \sum X_i P(X

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$\frac{566}{81}$

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(B) For a probability distribution,the sum of probabilities is $1$:$a + 2a + (a+b) + 2b + 3b = 1$$4a + 6b = 1$  --- $(i)$The mean $E(X) = \sum X_i P(X_i) = \frac{46}{9}$:$0(a) + 2(2a) + 4(a+b) + 6(2b) + 8(3b) = \frac{46}{9}$$4a + 4a + 4b + 12b + 24b = \frac{46}{9}$$8a + 40b = \frac{46}{9} \Rightarrow 4a + 20b = \frac{23}{9}$ --- $(ii)$Subtracting $(i)$ from $(ii)$:$(4a + 20b) - (4a + 6b) = \frac{23}{9} - 1$$14b = \frac{14}{9} \Rightarrow b = \frac{1}{9}$Substituting $b = \frac{1}{9}$ in $(i)$:$4a + 6(\frac{1}{9}) = 1 \Rightarrow 4a = 1 - \frac{2}{3} = \frac{1}{3} \Rightarrow a = \frac{1}{12}$Now,$E(X^2) = \sum X_i^2 P(X_i)$:$E(X^2) = 0^2(a) + 2^2(2a) + 4^2(a+b) + 6^2(2b) + 8^2(3b)$$E(X^2) = 8a + 16a + 16b + 72b + 192b = 24a + 280b$Substituting $a = \frac{1}{12}$ and $b = \frac{1}{9}$:$E(X^2) = 24(\frac{1}{12}) + 280(\frac{1}{9}) = 2 + \frac{280}{9} = \frac{298}{9}$Variance $\sigma^2 = E(X^2) - [E(X)]^2 = \frac{298}{9} - (\frac{46}{9})^2$$\sigma^2 = \frac{298}{9} - \frac{2116}{81} = \frac{2682 - 2116}{81} = \frac{566}{81}$

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

$X$	$0$	$1$	$2$
$P(X)$	$0.4$	$0.4$	$0.2$

If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is:

$X$ $0$ $2$ $4$ $6$ $8$

$P(X)$ $a$ $2a$ $a+b$ $2b$ $3b$

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If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is: $X$ $0$ $2$ $4$ $6$ $8$ $P(X)$ $a$ $2a$ $a+b$ $2b$ $3b$