1. **Problem statement:** Given a random variable $Y$ with PDF $f(y) = ky^2 e^{-y/4}$ for $y > 0$, find the constant $k$, the CDF $F(y)$, and the probability $P(Y > 8)$. 2. **Find $k$:** The PDF must integrate to 1 over its domain: $$\int_0^\infty ky^2 e^{-y/4} dy = 1$$ 3. Use substitution $t = \frac{y}{4}$, so $y = 4t$, $dy = 4 dt$: $$\int_0^\infty k (4t)^2 e^{-t} 4 dt = 1 \implies k \cdot 16 \cdot 4 \int_0^\infty t^2 e^{-t} dt = 1$$ 4. Simplify constants: $$64 k \int_0^\infty t^2 e^{-t} dt = 1$$ 5. Recall Gamma function: $\int_0^\infty t^{n} e^{-t} dt = n!$ for integer $n$. Here, $n=2$, so $\int_0^\infty t^2 e^{-t} dt = 2! = 2$. 6. Substitute back: $$64 k \times 2 = 1 \implies 128 k = 1 \implies k = \frac{1}{128}$$ 7. **Find the CDF $F(y)$:** $$F(y) = \int_0^y f(t) dt = \int_0^y \frac{1}{128} t^2 e^{-t/4} dt$$ 8. Use substitution $u = \frac{t}{4}$, $t=4u$, $dt=4 du$: $$F(y) = \frac{1}{128} \int_0^{y} t^2 e^{-t/4} dt = \frac{1}{128} \int_0^{y} t^2 e^{-t/4} dt = \frac{1}{128} \int_0^{y/4} (4u)^2 e^{-u} 4 du = \frac{1}{128} \times 16 \times 4 \int_0^{y/4} u^2 e^{-u} du = \frac{64}{128} \int_0^{y/4} u^2 e^{-u} du = \frac{1}{2} \int_0^{y/4} u^2 e^{-u} du$$ 9. The integral $\int_0^a u^2 e^{-u} du$ is the lower incomplete Gamma function $\gamma(3,a)$. 10. Using integration by parts or the known formula: $$\int_0^a u^2 e^{-u} du = 2 - e^{-a} (a^2 + 2a + 2)$$ 11. Therefore, $$F(y) = \frac{1}{2} \left[ 2 - e^{-y/4} \left( \left(\frac{y}{4}\right)^2 + 2 \cdot \frac{y}{4} + 2 \right) \right] = 1 - \frac{1}{2} e^{-y/4} \left( \frac{y^2}{16} + \frac{y}{2} + 2 \right)$$ 12. **Find $P(Y > 8)$:** $$P(Y > 8) = 1 - F(8) = \frac{1}{2} e^{-2} \left( \frac{64}{16} + \frac{8}{2} + 2 \right) = \frac{1}{2} e^{-2} (4 + 4 + 2) = \frac{1}{2} e^{-2} \times 10 = 5 e^{-2}$$ **Final answers:** - $k = \frac{1}{128}$ - $F(y) = 1 - \frac{1}{2} e^{-y/4} \left( \frac{y^2}{16} + \frac{y}{2} + 2 \right)$ - $P(Y > 8) = 5 e^{-2}$