1. **Problem Statement:** Given a discrete random variable $X$ with values $0,1,2,3,4,5$ and probabilities $0.05,0.10,0.15,0.30,0.25,0.15$ respectively, find: (i) Probability distributions of $Y_1=5X+2$ and $Y_2=3X+4$. (ii) Expected values $E(Y_1)$ and $E(Y_2)$. (iii) Variances $\mathrm{Var}(Y_1)$ and $\mathrm{Var}(Y_2)$. (iv) Prove $E(5X+2)=5E(X)+2$ and $E(3X+4)=3E(X)+4$. (v) Prove $\mathrm{Var}(5X+2)=25\mathrm{Var}(X)$ and $\mathrm{Var}(3X+4)=9\mathrm{Var}(X)$. 2. **Step (i): Find distributions of $Y_1$ and $Y_2$** - For each $x$, compute $y_1=5x+2$ and $y_2=3x+4$. - Probabilities remain the same since $Y_1$ and $Y_2$ are functions of $X$. Values and probabilities: $$ \begin{aligned} X &: 0,1,2,3,4,5 \\ P(X) &: 0.05,0.10,0.15,0.30,0.25,0.15 \\ Y_1=5X+2 &: 2,7,12,17,22,27 \\ Y_2=3X+4 &: 4,7,10,13,16,19 \end{aligned} $$ So, $$P(Y_1=y_1) = P(X=x)$$ $$P(Y_2=y_2) = P(X=x)$$ 3. **Step (ii): Find $E(Y_1)$ and $E(Y_2)$** Recall $E(Y) = \sum y P(Y=y)$. Calculate $E(X)$ first: $$E(X) = \sum x P(X=x) = 0\times0.05 + 1\times0.10 + 2\times0.15 + 3\times0.30 + 4\times0.25 + 5\times0.15$$ $$= 0 + 0.10 + 0.30 + 0.90 + 1.00 + 0.75 = 3.05$$ Calculate $E(Y_1)$: $$E(Y_1) = \sum y_1 P(Y_1=y_1) = \sum (5x+2) P(X=x) = 5E(X) + 2 \sum P(X=x) = 5\times3.05 + 2\times1 = 15.25 + 2 = 17.25$$ Calculate $E(Y_2)$: $$E(Y_2) = \sum y_2 P(Y_2=y_2) = \sum (3x+4) P(X=x) = 3E(X) + 4 \sum P(X=x) = 3\times3.05 + 4 = 9.15 + 4 = 13.15$$ 4. **Step (iii): Find $\mathrm{Var}(Y_1)$ and $\mathrm{Var}(Y_2)$** Recall variance formula: $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2$$ Calculate $E(X^2)$: $$E(X^2) = \sum x^2 P(X=x) = 0^2\times0.05 + 1^2\times0.10 + 2^2\times0.15 + 3^2\times0.30 + 4^2\times0.25 + 5^2\times0.15$$ $$= 0 + 0.10 + 0.60 + 2.70 + 4.00 + 3.75 = 11.15$$ Calculate $\mathrm{Var}(X)$: $$\mathrm{Var}(X) = 11.15 - (3.05)^2 = 11.15 - 9.3025 = 1.8475$$ Calculate $\mathrm{Var}(Y_1)$: $$\mathrm{Var}(Y_1) = \mathrm{Var}(5X+2) = 5^2 \mathrm{Var}(X) = 25 \times 1.8475 = 46.1875$$ Calculate $\mathrm{Var}(Y_2)$: $$\mathrm{Var}(Y_2) = \mathrm{Var}(3X+4) = 3^2 \mathrm{Var}(X) = 9 \times 1.8475 = 16.6275$$ 5. **Step (iv): Prove $E(aX+b) = aE(X) + b$** By definition: $$E(aX+b) = \sum (a x + b) P(X=x) = a \sum x P(X=x) + b \sum P(X=x) = a E(X) + b \times 1 = a E(X) + b$$ 6. **Step (v): Prove $\mathrm{Var}(aX+b) = a^2 \mathrm{Var}(X)$** Recall: $$\mathrm{Var}(aX+b) = E[(aX+b)^2] - [E(aX+b)]^2$$ $$= E[a^2 X^2 + 2abX + b^2] - [a E(X) + b]^2$$ $$= a^2 E(X^2) + 2ab E(X) + b^2 - (a^2 [E(X)]^2 + 2ab E(X) + b^2)$$ $$= a^2 (E(X^2) - [E(X)]^2) = a^2 \mathrm{Var}(X)$$ **Final answers:** - $P(Y_1 = y_1)$ and $P(Y_2 = y_2)$ as above. - $E(Y_1) = 17.25$, $E(Y_2) = 13.15$. - $\mathrm{Var}(Y_1) = 46.1875$, $\mathrm{Var}(Y_2) = 16.6275$. - Proofs for expectation and variance linearity shown.