1. **State the problem:** Antoine plays a game 50 times per day for 60 days. Each game has a 5% chance to win 40, 15% chance to win 8, and 80% chance to lose 1. Ahmed earns 130 per day for 60 days. We want to find the probability Antoine makes more money than Ahmed over the summer using the Central Limit Theorem (CLT). 2. **Define variables and expectations:** Let $X$ be the amount Antoine wins in one game. Then: $$P(X=40)=0.05, \quad P(X=8)=0.15, \quad P(X=-1)=0.80$$ 3. **Calculate the expected value $E(X)$:** $$E(X) = 40 \times 0.05 + 8 \times 0.15 + (-1) \times 0.80 = 2 + 1.2 - 0.8 = 2.4$$ 4. **Calculate the variance $Var(X)$:** First, find $E(X^2)$: $$E(X^2) = 40^2 \times 0.05 + 8^2 \times 0.15 + (-1)^2 \times 0.80 = 1600 \times 0.05 + 64 \times 0.15 + 1 \times 0.80 = 80 + 9.6 + 0.8 = 90.4$$ Then, $$Var(X) = E(X^2) - (E(X))^2 = 90.4 - (2.4)^2 = 90.4 - 5.76 = 84.64$$ 5. **Sum over 50 games per day:** Let $Y$ be the total winnings in one day: $$Y = \sum_{i=1}^{50} X_i$$ By independence, $$E(Y) = 50 \times 2.4 = 120$$ $$Var(Y) = 50 \times 84.64 = 4232$$ 6. **Sum over 60 days:** Let $S$ be total winnings over 60 days: $$S = \sum_{j=1}^{60} Y_j$$ Then, $$E(S) = 60 \times 120 = 7200$$ $$Var(S) = 60 \times 4232 = 253920$$ 7. **Ahmed's total earnings:** $$A = 130 \times 60 = 7800$$ 8. **Use CLT to approximate $P(S > A)$:** Standardize: $$Z = \frac{S - E(S)}{\sqrt{Var(S)}}$$ We want: $$P(S > 7800) = P\left(Z > \frac{7800 - 7200}{\sqrt{253920}}\right) = P\left(Z > \frac{600}{503.91}\right) = P(Z > 1.19)$$ 9. **Find $P(Z > 1.19)$ using standard normal table:** $$P(Z > 1.19) = 1 - P(Z \leq 1.19) \approx 1 - 0.8830 = 0.1170$$ **Final answer:** The probability Antoine makes more money than Ahmed over the summer is approximately **0.117** (about 11.7%).