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(D) Given the probability distribution table:$X = x$$0$$1$$2$$3$$P(X = x)$$K$$K + \frac{1}{7}$$2K$$\frac{2}{5}$Since the sum of all probabilities must

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$\frac{67}{35}$

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(D) Given the probability distribution table:$X = x$$0$$1$$2$$3$$P(X = x)$$K$$K + \frac{1}{7}$$2K$$\frac{2}{5}$Since the sum of all probabilities must be $1$,we have:$K + (K + \frac{1}{7}) + 2K + \frac{2}{5} = 1$$4K + \frac{5 + 14}{35} = 1$$4K + \frac{19}{35} = 1$$4K = 1 - \frac{19}{35} = \frac{16}{35}$$K = \frac{4}{35}$Now,substituting $K$ back into the table:$P(X=0) = \frac{4}{35}$$P(X=1) = \frac{4}{35} + \frac{5}{35} = \frac{9}{35}$$P(X=2) = 2 	imes \frac{4}{35} = \frac{8}{35}$$P(X=3) = \frac{2}{5} = \frac{14}{35}$The mean $\mu = E(X) = \sum x_i P(x_i)$:$\mu = 0 	imes \frac{4}{35} + 1 	imes \frac{9}{35} + 2 	imes \frac{8}{35} + 3 	imes \frac{14}{35}$$\mu = 0 + \frac{9}{35} + \frac{16}{35} + \frac{42}{35}$$\mu = \frac{67}{35}$

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

For the random variable $X$ with the probability distribution given by the table:
$X = x$ $0$ $1$ $2$ $3$
$P(X = x)$ $K$ $K + \frac{1}{7}$ $2K$ $\frac{2}{5}$

The mean of $X$ is:

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$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$K$	$K + \frac{1}{7}$	$2K$	$\frac{2}{5}$

For the random variable $X$ with the probability distribution given by the table:$X = x$$0$$1$$2$$3$$P(X = x)$$K$$K + \frac{1}{7}$$2K$$\frac{2}{5}$The mean of $X$ is: