The probability distribution of a random vari | vedclass

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(A) For a probability distribution,the sum of all probabilities must be equal to $1$.$\sum P(X = x_{i}) = 1$$\frac{k^2}{3} + k^2 + \frac{2k^2}{3} + \f

Vedclass · Accepted Answer

$1/3$

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(A) For a probability distribution,the sum of all probabilities must be equal to $1$.$\sum P(X = x_{i}) = 1$$\frac{k^2}{3} + k^2 + \frac{2k^2}{3} + \frac{k}{2} + \frac{k}{2} = 1$$\frac{k^2 + 3k^2 + 2k^2}{3} + k = 1$$\frac{6k^2}{3} + k = 1$$2k^2 + k - 1 = 0$$(2k - 1)(k + 1) = 0$Since $k$ must be positive for probabilities to be valid,$k = 1/2$.The mean $E(X) = \sum x_{i} P(X = x_{i})$$E(X) = (-2) \cdot \frac{k^2}{3} + (-1) \cdot k^2 + (0) \cdot \frac{2k^2}{3} + (1) \cdot \frac{k}{2} + (2) \cdot \frac{k}{2}$$E(X) = -\frac{2k^2}{3} - k^2 + 0 + \frac{k}{2} + k$$E(X) = -\frac{5k^2}{3} + \frac{3k}{2}$Substituting $k = 1/2$:$E(X) = -\frac{5(1/4)}{3} + \frac{3(1/2)}{2}$$E(X) = -\frac{5}{12} + \frac{3}{4} = -\frac{5}{12} + \frac{9}{12} = \frac{4}{12} = 1/3$.

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(a)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

The probability distribution of a random variable $X$ is as follows. Then the mean of $X$ is
$X = x_{i}$ $-2$ $-1$ $0$ $1$ $2$
$P(X = x_{i})$ $k^2 / 3$ $k^2$ $2k^2 / 3$ $k / 2$ $k / 2$

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The probability distribution of a random variable $X$ is as follows. Then the mean of $X$ is $X = x_{i}$$-2$$-1$$0$$1$$2$$P(X = x_{i})$$k^2 / 3$$k^2$$2k^2 / 3$$k / 2$$k / 2$