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(D) The sum of all probabilities in a probability distribution must be equal to $1$.$\sum P(X = x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) =

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$\frac{4}{5}$

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(D) The sum of all probabilities in a probability distribution must be equal to $1$.$\sum P(X = x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$$10k^2 + 9k - 1 = 0$$(10k - 1)(k + 1) = 0$Since $k \geq 0$,we have $k = \frac{1}{10}$.The cumulative distribution function $F(4)$ is defined as $P(X \leq 4)$.$F(4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$$F(4) = 0 + k + 2k + 2k + 3k = 8k$Substituting $k = \frac{1}{10}$:$F(4) = 8 	imes \frac{1}{10} = \frac{8}{10} = \frac{4}{5}$.

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2 + k$

$x = x_{i}$	$0$	$1$	$2$
$P_{i}$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

$A$ random variable $X$ has the following probability distribution:
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$

Then $F(4) = $

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$A$ random variable $X$ has the following probability distribution:$X = x$$0$$1$$2$$3$$4$$5$$6$$7$$P(X = x)$$0$$k$$2k$$2k$$3k$$k^2$$2k^2$$7k^2 + k$Then $F(4) = $