If a random variable X has the following prob | vedclass

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(B) We know that the sum of all probabilities in a distribution is $1$, i.e.,$\Sigma P(X = x_i) = 1$.$	herefore 2k^2 + k + k + k^2 = 1$$\Rightarrow 3

Vedclass · Accepted Answer

$\frac{22}{9}$

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(B) We know that the sum of all probabilities in a distribution is $1$, i.e.,$\Sigma P(X = x_i) = 1$.$	herefore 2k^2 + k + k + k^2 = 1$$\Rightarrow 3k^2 + 2k - 1 = 0$Factoring the quadratic equation: $(3k - 1)(k + 1) = 0$.This gives $k = \frac{1}{3}$ or $k = -1$.Since the probability $P(X = x_i)$ must be non-negative, $k = \frac{1}{3}$ is the only valid solution.The mean of $X$ is given by $E(X) = \Sigma x_i P(X = x_i)$.$E(X) = (1 	imes 2k^2) + (2 	imes k) + (3 	imes k) + (5 	imes k^2)$$E(X) = 2k^2 + 2k + 3k + 5k^2 = 7k^2 + 5k$Substituting $k = \frac{1}{3}$:$E(X) = 7(\frac{1}{3})^2 + 5(\frac{1}{3}) = 7(\frac{1}{9}) + \frac{5}{3} = \frac{7}{9} + \frac{15}{9} = \frac{22}{9}$.

$X = x_i$	$1$	$2$	$3$	$5$
$P(X = x_i)$	$2k^2$	$k$	$k$	$k^2$

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x_i)$	$\alpha$	$\alpha$	$\alpha$	$\beta$	$\beta$	$0.3$

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

If a random variable $X$ has the following probability distribution, then the mean of $X$ is
$X = x_i$ $1$ $2$ $3$ $5$
$P(X = x_i)$ $2k^2$ $k$ $k$ $k^2$

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If a random variable $X$ has the following probability distribution, then the mean of $X$ is$X = x_i$$1$$2$$3$$5$$P(X = x_i)$$2k^2$$k$$k$$k^2$