Given
\begin{array}{|l|c|c|c|c|}
\hline \mathbf{X} & 0.5 & 1 & 1.5 & 2 \\
\hline \mathbf{P}(\mathbf{X}) & \mathrm{k} & \mathrm{k}^2 & 2 \mathrm{k}^2 & \mathrm{k} \\
\hline
\end{array}
(i) We know that
$$
\begin{aligned}
& \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{P}_{\mathrm{i}}=1 \text {, where } \mathrm{P}_{\mathrm{i}} \geq 0 \\
\Rightarrow & \mathrm{P}_1+\mathrm{P}_2+\mathrm{P}_3+\mathrm{P}_4=1
\end{aligned}
$$
$\Rightarrow \mathrm{k}+\mathrm{k}^2+2 \mathrm{k}^2+\mathrm{k}=1$
$\Rightarrow 3 \mathrm{k}^2+2 \mathrm{k}-1=0$
$\Rightarrow(3 \mathrm{k}-1)(\mathrm{k}+1)=0$
$\Rightarrow \mathrm{k}=1 / 3 \Rightarrow \mathrm{k}=-1$
Since, $\mathrm{k} \geq 0 \Rightarrow \mathrm{k}=1 / 3$
(ii) $\quad$ Mean of the distribution $(\mu)$
$=\mathrm{E}(\mathrm{X})=\sum_{\mathrm{i}=1_{\mathrm{i}}}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}} \mathrm{P}_{\mathrm{i}}$
$=0.5(\mathrm{k})+1\left(\mathrm{k}^2\right)+1.5\left(2 \mathrm{k}^2\right)+2(\mathrm{k})$
$=4 \mathrm{k}^2+2.5 \mathrm{k}$
$=4 \cdot \frac{1}{9}+2.5 \cdot \frac{1}{3}$
$=\frac{4+7.5}{9}=\frac{23}{18}$