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(C) The sum of all probabilities in a probability distribution must be equal to $1$.Therefore,$k^3 + (2k^3 + k) + (4k - 10k^2) + (4k - 1) = 1$.Simplif

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$\frac{23}{27}$

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(C) The sum of all probabilities in a probability distribution must be equal to $1$.Therefore,$k^3 + (2k^3 + k) + (4k - 10k^2) + (4k - 1) = 1$.Simplifying the equation: $3k^3 - 10k^2 + 9k - 2 = 0$.By testing values,we find that $k = \frac{1}{3}$ is a root: $3(\frac{1}{27}) - 10(\frac{1}{9}) + 9(\frac{1}{3}) - 2 = \frac{1}{9} - \frac{10}{9} + 3 - 2 = -1 + 1 = 0$.The probabilities are: $P(-1) = (\frac{1}{3})^3 = \frac{1}{27}$,$P(0) = 2(\frac{1}{27}) + \frac{1}{3} = \frac{11}{27}$,$P(1) = 4(\frac{1}{3}) - 10(\frac{1}{9}) = \frac{12-10}{9} = \frac{2}{9} = \frac{6}{27}$,$P(2) = 4(\frac{1}{3}) - 1 = \frac{1}{3} = \frac{9}{27}$.Sum check: $\frac{1+11+6+9}{27} = \frac{27}{27} = 1$.The mean $E(X) = \sum x_i P(x_i) = (-1)(\frac{1}{27}) + (0)(\frac{11}{27}) + (1)(\frac{6}{27}) + (2)(\frac{9}{27}) = \frac{-1 + 0 + 6 + 18}{27} = \frac{23}{27}$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$0.1$	$0.15$	$0.3$	$0.25$	$k$	$k$

If the probability distribution of a random variable $X$ is as follows,then the mean of $X$ is:
$X = x_i$ $-1$ $0$ $1$ $2$
$P(X = x_i)$ $k^3$ $2k^3 + k$ $4k - 10k^2$ $4k - 1$

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If the probability distribution of a random variable $X$ is as follows,then the mean of $X$ is: $X = x_i$$-1$$0$$1$$2$$P(X = x_i)$$k^3$$2k^3 + k$$4k - 10k^2$$4k - 1$