If x is a random variable with PMF as follows | vedclass

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(B) The sum of probabilities in a probability distribution must be $1$,i.e.,$\sum P(X = x_i) = 1$. Given $P(X=0) = \frac{5}{16}$,$P(X=1) = \frac{5}{16

Vedclass · Accepted Answer

$1.3125$

Vedclass · Answer

(B) The sum of probabilities in a probability distribution must be $1$,i.e.,$\sum P(X = x_i) = 1$. Given $P(X=0) = \frac{5}{16}$,$P(X=1) = \frac{5}{16}$,$P(X=2) = \frac{2k}{48}$,and $P(X=3) = \frac{1}{4} = \frac{12}{48}$. Summing these: $\frac{5}{16} + \frac{5}{16} + \frac{2k}{48} + \frac{12}{48} = 1$. Converting all to denominator $48$: $\frac{15}{48} + \frac{15}{48} + \frac{2k}{48} + \frac{12}{48} = 1$. $\frac{42 + 2k}{48} = 1 \Rightarrow 42 + 2k = 48 \Rightarrow 2k = 6 \Rightarrow k = 3$. Now,the probability distribution is: $P(X=0) = \frac{15}{48}$,$P(X=1) = \frac{15}{48}$,$P(X=2) = \frac{6}{48} = \frac{1}{8}$,$P(X=3) = \frac{12}{48}$. The expected value $E(X) = \sum x_i P(x_i) = (0 	imes \frac{15}{48}) + (1 	imes \frac{15}{48}) + (2 	imes \frac{6}{48}) + (3 	imes \frac{12}{48})$. $E(X) = 0 + \frac{15}{48} + \frac{12}{48} + \frac{36}{48} = \frac{63}{48} = \frac{21}{16} = 1.3125$.

$X = k$	$-2$	$-1$	$0$	$1$	$2$
$P(X = k)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$

$X = x$	$-1$	$0$	$1$	$2$
$F(X = x)$	$0.3$	$0.7$	$0.8$	$1$

If $x$ is a random variable with $PMF$ as follows: $P(X = x) = \begin{cases} \frac{5}{16}, & x = 0, 1 \\ \frac{kx}{48}, & x = 2 \\ \frac{1}{4}, & x = 3 \end{cases}$ then find $E(x)$.

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