1. Problem 7(a): Find the mean and variance of $X$ where $f(x) = 2(1-x)$ for $0 \leq x \leq 1$, and $0$ otherwise. 2. Calculate the mean $E(X)$: $$E(X) = \int_0^1 x \cdot 2(1-x) \, dx = 2 \int_0^1 (x - x^2) \, dx = 2 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = 2 \left( \frac{1}{2} - \frac{1}{3} \right) = 2 \times \frac{1}{6} = \frac{1}{3}.$$ 3. Calculate $E(X^2)$: $$E(X^2) = \int_0^1 x^2 \cdot 2(1-x) \, dx = 2 \int_0^1 (x^2 - x^3) \, dx = 2 \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = 2 \left( \frac{1}{3} - \frac{1}{4} \right) = 2 \times \frac{1}{12} = \frac{1}{6}.$$ 4. Variance $Var(X) = E(X^2) - (E(X))^2 = \frac{1}{6} - \left( \frac{1}{3} \right)^2 = \frac{1}{6} - \frac{1}{9} = \frac{1}{18}.$ 5. Problem 7(b): Find the stock quantity to meet demand with 96% certainty. 6. Find $x$ such that $P(X \leq x) = 0.96$. 7. The cumulative distribution function (CDF) is: $$F(x) = \int_0^x 2(1-t) \, dt = 2 \left[ t - \frac{t^2}{2} \right]_0^x = 2 \left( x - \frac{x^2}{2} \right) = 2x - x^2.$$ 8. Set $F(x) = 0.96$: $$2x - x^2 = 0.96 \implies x^2 - 2x + 0.96 = 0.$$ 9. Solve quadratic: $$x = \frac{2 \pm \sqrt{4 - 4 \times 0.96}}{2} = 1 \pm \sqrt{0.04} = 1 \pm 0.2.$$ 10. Since $x$ in $[0,1]$, take $x = 1 - 0.2 = 0.8$. 11. The factory should stock $0.8$ thousand kg = $800$ kg to be 96% certain demand is met. 12. Problem 8(a): Find $k$ for $f(y) = k y (5 - y)$ on $0 \leq y \leq 4$. 13. Use normalization: $$\int_0^4 k y (5 - y) \, dy = 1.$$ 14. Compute integral: $$\int_0^4 y(5-y) \, dy = \int_0^4 (5y - y^2) \, dy = \left[ \frac{5y^2}{2} - \frac{y^3}{3} \right]_0^4 = \frac{5 \times 16}{2} - \frac{64}{3} = 40 - \frac{64}{3} = \frac{120 - 64}{3} = \frac{56}{3}.$$ 15. So: $$k \times \frac{56}{3} = 1 \implies k = \frac{3}{56}.$$ 16. Problem 8(b): Find the cumulative distribution function $F(y)$. 17. For $0 \leq y \leq 4$: $$F(y) = \int_0^y \frac{3}{56} t (5 - t) \, dt = \frac{3}{56} \int_0^y (5t - t^2) \, dt = \frac{3}{56} \left[ \frac{5t^2}{2} - \frac{t^3}{3} \right]_0^y = \frac{3}{56} \left( \frac{5y^2}{2} - \frac{y^3}{3} \right).$$ 18. Problem 8(c): Find $E(Y)$. 19. Compute: $$E(Y) = \int_0^4 y \cdot f(y) \, dy = \int_0^4 y \cdot \frac{3}{56} y (5 - y) \, dy = \frac{3}{56} \int_0^4 y^2 (5 - y) \, dy = \frac{3}{56} \int_0^4 (5y^2 - y^3) \, dy.$$ 20. Evaluate integral: $$\int_0^4 (5y^2 - y^3) \, dy = \left[ \frac{5y^3}{3} - \frac{y^4}{4} \right]_0^4 = \frac{5 \times 64}{3} - \frac{256}{4} = \frac{320}{3} - 64 = \frac{320 - 192}{3} = \frac{128}{3}.$$ 21. So: $$E(Y) = \frac{3}{56} \times \frac{128}{3} = \frac{128}{56} = \frac{32}{14} = \frac{16}{7} \approx 2.2857.$$ 22. Problem 8(d): Find the modal value (mode) of $Y$. 23. The mode is the $y$ that maximizes $f(y) = \frac{3}{56} y (5 - y)$. 24. Differentiate: $$f'(y) = \frac{3}{56} (5 - y - y) = \frac{3}{56} (5 - 2y).$$ 25. Set $f'(y) = 0$: $$5 - 2y = 0 \implies y = \frac{5}{2} = 2.5.$$ 26. Since $f''(y) = \frac{3}{56} (-2) < 0$, $y=2.5$ is a maximum. 27. Problem 8(e): Verify median lies between 2.3 and 2.4. 28. Median $m$ satisfies $F(m) = 0.5$. 29. Using CDF: $$F(m) = \frac{3}{56} \left( \frac{5m^2}{2} - \frac{m^3}{3} \right) = 0.5.$$ 30. Multiply both sides by 56/3: $$\frac{5m^2}{2} - \frac{m^3}{3} = \frac{56}{3} \times 0.5 = \frac{28}{3}.$$ 31. Multiply both sides by 6: $$15 m^2 - 2 m^3 = 56.$$ 32. Rearranged: $$2 m^3 - 15 m^2 + 56 = 0.$$ 33. Test $m=2.3$: $$2(2.3)^3 - 15(2.3)^2 + 56 \approx 2(12.167) - 15(5.29) + 56 = 24.334 - 79.35 + 56 = 1.0 > 0.$$ 34. Test $m=2.4$: $$2(2.4)^3 - 15(2.4)^2 + 56 \approx 2(13.824) - 15(5.76) + 56 = 27.648 - 86.4 + 56 = -2.75 < 0.$$ 35. Since function changes sign between 2.3 and 2.4, median lies in this interval. Final answers: - 7(a) Mean $= \frac{1}{3}$, Variance $= \frac{1}{18}$. - 7(b) Stock quantity $= 800$ kg. - 8(a) $k = \frac{3}{56}$. - 8(b) $F(y) = \frac{3}{56} \left( \frac{5y^2}{2} - \frac{y^3}{3} \right)$ for $0 \leq y \leq 4$. - 8(c) $E(Y) = \frac{16}{7} \approx 2.2857$. - 8(d) Mode $= 2.5$. - 8(e) Median lies between 2.3 and 2.4.