A random variable X takes the values 1, 2, 3 | vedclass

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(A) Given $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4) = k$.Then $P(X=1) = \frac{k}{2}, P(X=2) = \frac{k}{3}, P(X=3) = k, P(X=4) = \frac{k}{5}$.Since $\su

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$\frac{421}{61}$

Vedclass · Answer

(A) Given $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4) = k$.Then $P(X=1) = \frac{k}{2}, P(X=2) = \frac{k}{3}, P(X=3) = k, P(X=4) = \frac{k}{5}$.Since $\sum P(X) = 1$, we have $\frac{k}{2} + \frac{k}{3} + k + \frac{k}{5} = 1$.$\Rightarrow k(\frac{15+10+30+6}{30}) = 1 \Rightarrow k(\frac{61}{30}) = 1 \Rightarrow k = \frac{30}{61}$.The probability distribution is:$x$$1$$2$$3$$4$$P(X=x)$$\frac{15}{61}$$\frac{10}{61}$$\frac{30}{61}$$\frac{6}{61}$Mean $\mu = E(X) = \sum x P(x) = 1(\frac{15}{61}) + 2(\frac{10}{61}) + 3(\frac{30}{61}) + 4(\frac{6}{61}) = \frac{15+20+90+24}{61} = \frac{149}{61}$.$E(X^2) = \sum x^2 P(x) = 1^2(\frac{15}{61}) + 2^2(\frac{10}{61}) + 3^2(\frac{30}{61}) + 4^2(\frac{6}{61}) = \frac{15+40+270+96}{61} = \frac{421}{61}$.Variance $\sigma^2 = E(X^2) - \mu^2$.We need $\sigma^2 + \mu^2 = E(X^2) - \mu^2 + \mu^2 = E(X^2)$.Therefore, $\sigma^2 + \mu^2 = \frac{421}{61}$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x)$	$0.2$	$0.3$	$0.1$	$0.15$	$0.25$

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

$A$ random variable $X$ takes the values $1, 2, 3$ and $4$ such that $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4)$. If $\sigma^2$ is the variance and $\mu$ is the mean of $X$, then $\sigma^2 + \mu^2 =$

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