1. **State the problem:** We want to find the standard deviation of Ahmed's total winnings after playing the game twice independently. 2. **Given probabilities and winnings:** - Win $40 with probability 0.05 - Win $8 with probability 0.15 - Lose $1 with probability 0.80 3. **Calculate the expected value $E(X)$ of one game:** $$E(X) = 40 \times 0.05 + 8 \times 0.15 + (-1) \times 0.80 = 2 + 1.2 - 0.8 = 2.4$$ 4. **Calculate $E(X^2)$ for one game:** $$E(X^2) = 40^2 \times 0.05 + 8^2 \times 0.15 + (-1)^2 \times 0.80 = 1600 \times 0.05 + 64 \times 0.15 + 1 \times 0.80 = 80 + 9.6 + 0.8 = 90.4$$ 5. **Calculate variance $Var(X)$ for one game:** $$Var(X) = E(X^2) - (E(X))^2 = 90.4 - (2.4)^2 = 90.4 - 5.76 = 84.64$$ 6. **Since Ahmed plays twice independently, total winnings $T = X_1 + X_2$:** - Expected value: $$E(T) = E(X_1) + E(X_2) = 2.4 + 2.4 = 4.8$$ - Variance: $$Var(T) = Var(X_1) + Var(X_2) = 84.64 + 84.64 = 169.28$$ 7. **Calculate standard deviation of total winnings:** $$\sigma_T = \sqrt{Var(T)} = \sqrt{169.28} \approx 13.01$$ **Final answer:** The standard deviation of Ahmed's total winnings from two games is approximately $13.01$.