The probability function of a random variable | vedclass

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(B) Given the probability function $P(X=k)=c k^2$,where $c$ is a constant and $k \in\{0,1,2,3,4\}$.Since the sum of probabilities must be $1$,we have

Vedclass · Accepted Answer

$11.8$

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(B) Given the probability function $P(X=k)=c k^2$,where $c$ is a constant and $k \in\{0,1,2,3,4\}$.Since the sum of probabilities must be $1$,we have $\sum_{k=0}^{4} P(X=k) = 1$.$c(0^2) + c(1^2) + c(2^2) + c(3^2) + c(4^2) = 1$$c(0 + 1 + 4 + 9 + 16) = 1$$30c = 1 \implies c = \frac{1}{30}$.We know that the variance $\sigma^2 = E(X^2) - \mu^2$,where $\mu = E(X)$.Therefore,$\sigma^2 + \mu^2 = E(X^2)$.$E(X^2) = \sum_{k=0}^{4} k^2 P(X=k) = \sum_{k=0}^{4} k^2 (c k^2) = c \sum_{k=0}^{4} k^4$.$E(X^2) = c(0^4 + 1^4 + 2^4 + 3^4 + 4^4) = c(0 + 1 + 16 + 81 + 256) = 354c$.Substituting $c = \frac{1}{30}$,we get:$\sigma^2 + \mu^2 = 354 	imes \frac{1}{30} = \frac{354}{30} = 11.8$.Thus,option $(B)$ is correct.

$x$	$-2$	$-1$	$3$	$4$	$6$
$P(X=x)$	$\frac{1}{5}$	$a$	$\frac{1}{3}$	$\frac{1}{5}$	$b$

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(x)$	$k$	$2k$	$3k$	$4k$	$4k$	$3k$	$2k$	$k$	$k$

The probability function of a random variable $X$ is given by $P(X=k)=c k^2$,where $c$ is a constant and $k \in\{0,1,2,3,4\}$. If $\sigma^2$ is the variance of $X$ and $\mu$ is the mean of $X$,then $\sigma^2+\mu^2=$

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