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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}$

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X=x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$0.2$	$0.3$	$0.4$

$X = x$	$-4$	$-2$	$0$	$2$	$4$	$6$	$8$	$10$
$F(X = x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

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:

Similar Questions

The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:
$X$ $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(X=x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$

Then,find the value of $\frac{P[X=-3]}{P[X < 0]}$.

The probability distribution of $X$ is
$\begin{array}{|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & 0.3 & k & 2k & 2k \\ \hline \end{array}$
Find the value of $k$.

$A$ fair six-faced die is rolled $12$ times. The probability that each face turns up exactly twice is equal to:

$A$ random variable $X$ has the following probability distribution:
$X=x$ $1$ $2$ $3$ $4$
$P(X=x)$ $0.1$ $0.2$ $0.3$ $0.4$

The mean and standard deviation of $X$ are respectively:

The cumulative distribution function of a discrete random variable $X$ is given by the following table:
$X = x$ $-4$ $-2$ $0$ $2$ $4$ $6$ $8$ $10$
$F(X = x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$

Then,calculate $\frac{P(X \leqslant 0)}{P(X > 0)}$.

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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: