The probability function of a discrete random | vedclass

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(D) Given the probability function $P(X=r) = K r^2$ for $r \in \{-2, -1, 0, 1, 2, 3\}$.Since the sum of all probabilities must be $1$,we have:$\sum P(

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$\frac{115}{19}$

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(D) Given the probability function $P(X=r) = K r^2$ for $r \in \{-2, -1, 0, 1, 2, 3\}$.Since the sum of all probabilities must be $1$,we have:$\sum P(X=r) = 1$$K((-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 + 3^2) = 1$$K(4 + 1 + 0 + 1 + 4 + 9) = 1$$19K = 1 \Rightarrow K = \frac{1}{19}$We need to find the sum of the variance $\sigma^2$ and the square of the mean $\mu^2$. We know that $\sigma^2 = E(X^2) - \mu^2$,so $\sigma^2 + \mu^2 = E(X^2)$.$E(X^2) = \sum r^2 P(X=r) = \sum r^2 (K r^2) = K \sum r^4$$E(X^2) = K((-2)^4 + (-1)^4 + 0^4 + 1^4 + 2^4 + 3^4)$$E(X^2) = K(16 + 1 + 0 + 1 + 16 + 81) = K(115)$Substituting $K = \frac{1}{19}$,we get:$E(X^2) = \frac{115}{19}$Thus,$\sigma^2 + \mu^2 = \frac{115}{19}$.

The probability function of a discrete random variable $X$ is given by $P(X=r)=K r^2$,where $r=-2,-1,0,1,2,3$ and $K$ is a constant. The sum of the variance of $X$ and the square of the mean of $X$ is

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