1. Problem 15: Given the probability distribution of $X$: $$\begin{array}{c|ccccccc} X & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ p(x) & 0.05 & 0.10 & 0.30 & 0 & 0.30 & 0.15 & 0.10 \end{array}$$ Find (i) $E(X)$, (ii) $E(2X+3)$, (iii) $V(X)$, (iv) $V(2X+3)$. 2. Recall formulas: - Expected value: $E(X) = \sum x_i p(x_i)$ - Variance: $V(X) = E(X^2) - [E(X)]^2$ - For linear transformations: $E(aX+b) = aE(X) + b$, $V(aX+b) = a^2 V(X)$ 3. Calculate $E(X)$: $$E(X) = (-3)(0.05) + (-2)(0.10) + (-1)(0.30) + 0(0) + 1(0.30) + 2(0.15) + 3(0.10)$$ $$= -0.15 - 0.20 - 0.30 + 0 + 0.30 + 0.30 + 0.30 = 0.35$$ 4. Calculate $E(2X+3)$ using linearity: $$E(2X+3) = 2E(X) + 3 = 2(0.35) + 3 = 3.7$$ 5. Calculate $E(X^2)$: $$E(X^2) = (-3)^2(0.05) + (-2)^2(0.10) + (-1)^2(0.30) + 0^2(0) + 1^2(0.30) + 2^2(0.15) + 3^2(0.10)$$ $$= 9(0.05) + 4(0.10) + 1(0.30) + 0 + 1(0.30) + 4(0.15) + 9(0.10)$$ $$= 0.45 + 0.40 + 0.30 + 0 + 0.30 + 0.60 + 0.90 = 2.95$$ 6. Calculate variance $V(X)$: $$V(X) = E(X^2) - [E(X)]^2 = 2.95 - (0.35)^2 = 2.95 - 0.1225 = 2.8275$$ 7. Calculate $V(2X+3)$: $$V(2X+3) = 2^2 V(X) = 4 \times 2.8275 = 11.31$$ --- 8. Problem 16: Given the pdf $$f(x) = \begin{cases} x & 0 \leq x < 1 \\ 2 - x & 1 \leq x < 2 \\ 0 & \text{elsewhere} \end{cases}$$ Find $E(X)$ and $V(X)$. 9. Check pdf validity: $$\int_0^1 x \, dx + \int_1^2 (2 - x) \, dx = \left[ \frac{x^2}{2} \right]_0^1 + \left[ 2x - \frac{x^2}{2} \right]_1^2 = \frac{1}{2} + (4 - 2 - 2 + \frac{1}{2}) = \frac{1}{2} + \frac{1}{2} = 1$$ Valid pdf. 10. Calculate $E(X)$: $$E(X) = \int_0^1 x \cdot x \, dx + \int_1^2 x (2 - x) \, dx = \int_0^1 x^2 \, dx + \int_1^2 (2x - x^2) \, dx$$ $$= \left[ \frac{x^3}{3} \right]_0^1 + \left[ x^2 - \frac{x^3}{3} \right]_1^2 = \frac{1}{3} + \left(4 - \frac{8}{3} - 1 + \frac{1}{3} \right) = \frac{1}{3} + \left(3 - \frac{7}{3} \right) = \frac{1}{3} + \frac{2}{3} = 1$$ 11. Calculate $E(X^2)$: $$E(X^2) = \int_0^1 x^2 \cdot x \, dx + \int_1^2 x^2 (2 - x) \, dx = \int_0^1 x^3 \, dx + \int_1^2 (2x^2 - x^3) \, dx$$ $$= \left[ \frac{x^4}{4} \right]_0^1 + \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_1^2 = \frac{1}{4} + \left( \frac{16}{3} - 4 - \frac{2}{3} + \frac{1}{4} \right) = \frac{1}{4} + \left( \frac{14}{3} - \frac{15}{4} \right)$$ Calculate inside parentheses: $$\frac{14}{3} - \frac{15}{4} = \frac{56}{12} - \frac{45}{12} = \frac{11}{12}$$ So, $$E(X^2) = \frac{1}{4} + \frac{11}{12} = \frac{3}{12} + \frac{11}{12} = \frac{14}{12} = \frac{7}{6} \approx 1.1667$$ 12. Calculate variance: $$V(X) = E(X^2) - [E(X)]^2 = \frac{7}{6} - 1^2 = \frac{7}{6} - 1 = \frac{1}{6} \approx 0.1667$$ --- 13. Problem 17: From 12 items with 5 defective, select 4 at random. Find expected number $E$ of defective items. 14. This is a hypergeometric distribution with parameters $N=12$, $K=5$, $n=4$. Expected value formula: $$E = n \times \frac{K}{N} = 4 \times \frac{5}{12} = \frac{20}{12} = \frac{5}{3} \approx 1.6667$$ --- Final answers: - Problem 15: $E(X) = 0.35$, $E(2X+3) = 3.7$, $V(X) = 2.8275$, $V(2X+3) = 11.31$ - Problem 16: $E(X) = 1$, $V(X) = \frac{1}{6} \approx 0.1667$ - Problem 17: $E = \frac{5}{3} \approx 1.6667$