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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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Verified Answer
The correct answer is:
$\frac{115}{19}$
$P(X=r)=k r^2$, where $r=-2,-1,0,1,2,3$
and $k=$ constant
$$
\begin{aligned}
& \because \quad \sigma^2=E\left(x^2\right)-\mu^2 \\
& \Rightarrow \quad \sigma^2+\mu^2=E\left(x^2\right)=\sum\left(x_i^2\right) P\left(x_i\right) \\
& =4(4 k)+1(k)+0+1(k)+4(4 k)+9(9 k)
\end{aligned}
$$
$$
\begin{aligned}
& \because \quad \sum_{P\left(x_i\right)=1} \\
& \Rightarrow 4 k+k+0+k+4 k+9 k=1 \Rightarrow k=1 / 19
\end{aligned}
$$
From Eq. (i),
$$
\sigma^2+\mu^2=115 / 19
$$
and $k=$ constant
$$
\begin{aligned}
& \because \quad \sigma^2=E\left(x^2\right)-\mu^2 \\
& \Rightarrow \quad \sigma^2+\mu^2=E\left(x^2\right)=\sum\left(x_i^2\right) P\left(x_i\right) \\
& =4(4 k)+1(k)+0+1(k)+4(4 k)+9(9 k)
\end{aligned}
$$
$$
\begin{aligned}
& \because \quad \sum_{P\left(x_i\right)=1} \\
& \Rightarrow 4 k+k+0+k+4 k+9 k=1 \Rightarrow k=1 / 19
\end{aligned}
$$
From Eq. (i),
$$
\sigma^2+\mu^2=115 / 19
$$
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