We have,

$$\begin{array}{ll} & \operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2 \\ \text { where, } & E(X)=\mu=\sum_\limits{i=1}^n x_i P_i(x i) \\ \text { and } & E\left(X^2\right)=\sum_\limits{i=1}^n x_i^2 P\left(x_i\right) \end{array}$$

$$\begin{array}{lrl} \therefore & E(X)=0+0.25+0.6+0.6+0.60=2.05 \\ & E\left(X^2\right)=0+0.25+1.2+1.8+2.40=5.65 \end{array}$$

$$\begin{aligned} \text{(i)}\quad V\left(\frac{X}{2}\right) & =\frac{1}{4} V(X)=\frac{1}{4}\left[5.65-(2.05)^2\right] \\ & =\frac{1}{4}[5.65-4.2025]=\frac{1}{4} \times 1.4475=0.361875 \end{aligned}$$

(ii) $V(X)=1.4475$

We have,

$$\begin{aligned} &\text { (i) Since, }\\ &\sum_{i=1}^n P_i=1, i=1,2, \ldots, n \text { and } P_i \geq 0\\ &\begin{aligned} \therefore \quad& k+\frac{k}{2}+\frac{k}{4}+\frac{k}{8} =1 \\ \Rightarrow \quad & 8 k+4 k+2 k+k =8 \\ \therefore \quad & k =\frac{8}{15} \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { (ii) } P(X \leq 2)=P(0)+P(1)+P(2) & =k+\frac{k}{2}+\frac{k}{4} \\ & =\frac{(4 k+2 k+k)}{4}=\frac{7 k}{4}=\frac{7}{4} \cdot \frac{8}{15}=\frac{14}{15} \end{aligned}$$

(iii) $P(X \leq 2)+P(X>2)=\frac{14}{15}+\frac{1}{15}=1$

We have,

We know that, standard deviation of $X=\sqrt{\operatorname{Var} X}$

where,

$$\begin{aligned} \operatorname{Var} X & =E\left(X^2\right)-[E(X)]^2 \\ & =\sum_{i=1}^n x_i^2 P\left(x_i\right)-\left[\sum_{i=1}^n x_i P_i\right]^2 \end{aligned}$$

$$\begin{aligned} \therefore\quad\operatorname{Var} X & =[0.8+4.5+4.8]-[0.4+1.5+1.2]^2 \\ & =10.1-(3.1)^2=10.1-9.61=0.49 \end{aligned}$$

$\therefore$ Standard deviation of $X=\sqrt{\operatorname{Var} X}=\sqrt{0.49}=0.7$

Since, $X=$ Number of fours seen

On tossing two die, $X=0,1,2$.

Also, $$P_{(4)}=\frac{1}{10} \text { and } P_{(\text {not } 4)}=\frac{9}{10}$$

So,

$$\begin{aligned} & P(X=0)=P_{(\text {not } 4)} \cdot P_{(\text {not } 4)}=\frac{9}{10} \cdot \frac{9}{10}=\frac{81}{100} \\ & P(X=1)=P_{(\text {not } 4)} \cdot P_{(4)}+P_{(4)} \cdot P_{(\text {not } 4)}=\frac{9}{10} \cdot \frac{1}{10}+\frac{1}{10} \cdot \frac{9}{10}=\frac{18}{100} \\ & P(X=2)=P_{(4)} \cdot P_{(4)}=\frac{1}{10} \cdot \frac{1}{10}=\frac{1}{100} \end{aligned}$$

Thus, we get following table

$$\begin{aligned} \therefore \quad \operatorname{Var}(X) & =E\left(X^2\right)-[E(X)]^2=\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =\left[0+\frac{18}{100}+\frac{4}{100}\right]-\left[0+\frac{18}{100}+\frac{2}{100}\right]^2 \\ & =\frac{22}{100}-\left(\frac{20}{100}\right)^2=\frac{11}{50}-\frac{1}{25} \\ & =\frac{11-2}{50}=\frac{9}{50}=\frac{18}{100}=0.18 \end{aligned}$$

We have, $X=$ number of twos seen

So, on throwing a die three times, we will have $X=0,1,2,3$.

$$\begin{aligned} & \therefore \quad P(X=0)=P_{(\text {(not 2) }} \cdot P_{(\text {(not 2) }} \cdot P_{(\text {(not 2) }}=\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6}=\frac{125}{216} \\ & P(X=1)=P_{(\text {(not 2) }} \cdot P_{(\text {not 2) }} \cdot P_{(2)}+P_{(\text {not 2) }} \cdot P_{(2)} \cdot P_{(\text {not 2) }}+P_{(2)} \cdot P_{(\text {not 2) }} \cdot P_{(\text {not 2) }} \\ & =\frac{5}{6} \cdot \frac{5}{6} \frac{1}{6}+\frac{5}{6} \cdot \frac{1}{6} \cdot \frac{5}{6}+\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6}=\frac{25}{36} \cdot \frac{3}{6}=\frac{25}{72} \\ & P(X=2)=P_{(\text {not 2) }} \cdot P_{(2)} \cdot P_{(2)}+P_{(2)} \cdot P_{(2)} \cdot P_{(\text {not 2) }}+P_{(2)} \cdot P_{(\text {not 2) }}+P_{(2)} \\ & =\frac{5}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6}+\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \\ & =\frac{1}{36} \cdot\left[\frac{15}{6}\right]=\frac{15}{216} \\ & P(X=3)=P_{(2)} \cdot P_{(2)} \cdot P_{(2)}=\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}=\frac{1}{216} \end{aligned}$$

We know that, $E(X)=\Sigma X P(X)=0 \cdot \frac{125}{216}+1 \cdot \frac{25}{72}+2 \cdot \frac{15}{216}+3 \cdot \frac{1}{216}$

$$=\frac{75+30+3}{216}=\frac{108}{216}=\frac{1}{2}$$