The probability function of a random variable $X$ is given by $P(X=k)=c k^2$, where $c$ is a constant and $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 $\sigma^2+\mu^2=$
3.33
11.8
$\frac{1}{30}$
354
Solution
Given probability function $P(X=k)=C k^2$, where $C$ is a constant and $K \in\{0,1,2,3,4\}$
$\begin{array}{ll}
\because & \sigma^2=E\left(x^2\right)-\mu^2 \\
\Rightarrow & \sigma^2+\mu^2=E\left(x^2\right)=\Sigma\left(x_i^2\right) P\left(x_i\right) \\
= & 0\left(C(0)^2\right)+1\left(C(1)^2\right)+4\left(C(2)^2\right)+9\left(C\left(3^2\right)\right)+16\left(C(4)^2\right) \\
= & C+16 C+81 C+256 C
\end{array}$
$\begin{array}{ll}
\because & \Sigma P\left(x_i\right)=1 \Rightarrow C+4 C+9 C+16 C=1 \\
\Rightarrow & 30 C=1 \Rightarrow C=\frac{1}{30}
\end{array}$
Put the value of $C$, in Eq. (i), we get
$\text { So, } \sigma^2+\mu^2=\frac{354}{30}=11.8 \text {. }$
Hence, option (b) is correct.
