1. **State the problem:** We are given the generating function $$G_X(s) = \frac{s^3}{6} (2 + s^4 + 2s^8)$$ and asked to find $$\text{mean}(3X + 5) + \text{Var}(5X + 7).$$ 2. **Recall formulas and properties:** - The mean of a linear transformation $$aX + b$$ is $$a \mathbb{E}[X] + b$$. - The variance of a linear transformation $$aX + b$$ is $$a^2 \text{Var}(X)$$. - The generating function $$G_X(s)$$ is the probability generating function (PGF) of a discrete random variable $$X$$, defined as $$G_X(s) = \mathbb{E}[s^X].$$ - The first derivative at $$s=1$$ gives the mean: $$G_X'(1) = \mathbb{E}[X]$$. - The second derivative at $$s=1$$ helps find the variance: $$\text{Var}(X) = G_X''(1) + G_X'(1) - (G_X'(1))^2.$$ 3. **Calculate $$G_X(s)$$ explicitly:** $$G_X(s) = \frac{s^3}{6} (2 + s^4 + 2s^8) = \frac{1}{6} (2s^3 + s^{7} + 2s^{11}).$$ 4. **Find the first derivative $$G_X'(s)$$:** $$G_X'(s) = \frac{1}{6} (2 \cdot 3 s^{2} + 7 s^{6} + 2 \cdot 11 s^{10}) = \frac{1}{6} (6 s^{2} + 7 s^{6} + 22 s^{10}).$$ 5. **Evaluate $$G_X'(1)$$ to find $$\mathbb{E}[X]$$:** $$G_X'(1) = \frac{1}{6} (6 + 7 + 22) = \frac{35}{6}.$$ 6. **Find the second derivative $$G_X''(s)$$:** $$G_X''(s) = \frac{1}{6} (6 \cdot 2 s^{1} + 7 \cdot 6 s^{5} + 22 \cdot 10 s^{9}) = \frac{1}{6} (12 s + 42 s^{5} + 220 s^{9}).$$ 7. **Evaluate $$G_X''(1)$$:** $$G_X''(1) = \frac{1}{6} (12 + 42 + 220) = \frac{274}{6} = \frac{137}{3}.$$ 8. **Calculate variance $$\text{Var}(X)$$:** $$\text{Var}(X) = G_X''(1) + G_X'(1) - (G_X'(1))^2 = \frac{137}{3} + \frac{35}{6} - \left(\frac{35}{6}\right)^2.$$ Convert to common denominator 6: $$\frac{137}{3} = \frac{274}{6},$$ so $$\text{Var}(X) = \frac{274}{6} + \frac{35}{6} - \frac{1225}{36} = \frac{309}{6} - \frac{1225}{36} = \frac{1854}{36} - \frac{1225}{36} = \frac{629}{36}.$$ 9. **Calculate $$\text{mean}(3X + 5)$$:** $$3 \mathbb{E}[X] + 5 = 3 \times \frac{35}{6} + 5 = \frac{105}{6} + 5 = \frac{105}{6} + \frac{30}{6} = \frac{135}{6} = \frac{45}{2}.$$ 10. **Calculate $$\text{Var}(5X + 7)$$:** $$5^2 \text{Var}(X) = 25 \times \frac{629}{36} = \frac{15725}{36}.$$ 11. **Sum the results:** $$\text{mean}(3X + 5) + \text{Var}(5X + 7) = \frac{45}{2} + \frac{15725}{36} = \frac{810}{36} + \frac{15725}{36} = \frac{16535}{36}.$$ **Final answer:** $$\boxed{\frac{16535}{36}}.$$