1. **Problem Statement:** We have two probability distributions for the number of aces served by Michelle and Amanda in each tennis set. We need to find the expected value (mean), variance, and standard deviation for each player and determine who has greater variation. 2. **Formulas:** - Expected value (mean): $$E(X) = \sum x_i p_i$$ where $x_i$ are values and $p_i$ their probabilities. - Variance: $$Var(X) = E(X^2) - [E(X)]^2$$ where $$E(X^2) = \sum x_i^2 p_i$$ - Standard deviation: $$\sigma = \sqrt{Var(X)}$$ 3. **Calculate Expected Value for Michelle:** $$E(X) = 0(0.1) + 1(0.15) + 2(0.45) + 3(0.25) + 4(0.05)$$ $$= 0 + 0.15 + 0.9 + 0.75 + 0.2 = 2.0$$ 4. **Calculate Expected Value for Amanda:** $$E(X) = 0(0.2) + 1(0.1) + 2(0.35) + 3(0.2) + 4(0.15)$$ $$= 0 + 0.1 + 0.7 + 0.6 + 0.6 = 2.0$$ 5. **Calculate $E(X^2)$ for Michelle:** $$E(X^2) = 0^2(0.1) + 1^2(0.15) + 2^2(0.45) + 3^2(0.25) + 4^2(0.05)$$ $$= 0 + 0.15 + 1.8 + 2.25 + 0.8 = 4.999999999999999$$ 6. **Calculate Variance for Michelle:** $$Var(X) = E(X^2) - [E(X)]^2 = 5.0 - 2.0^2 = 5.0 - 4.0 = 1.0$$ 7. **Calculate Standard Deviation for Michelle:** $$\sigma = \sqrt{1.0} = 1.0$$ 8. **Calculate $E(X^2)$ for Amanda:** $$E(X^2) = 0^2(0.2) + 1^2(0.1) + 2^2(0.35) + 3^2(0.2) + 4^2(0.15)$$ $$= 0 + 0.1 + 1.4 + 1.8 + 2.4 = 5.7$$ 9. **Calculate Variance for Amanda:** $$Var(X) = 5.7 - 2.0^2 = 5.7 - 4.0 = 1.7$$ 10. **Calculate Standard Deviation for Amanda:** $$\sigma = \sqrt{1.7} \approx 1.303$$ 11. **Conclusion:** Both players have an expected average of 2 aces per set. Amanda has a greater variance (1.7) and standard deviation (~1.303) than Michelle (variance 1.0, std dev 1.0), so Amanda's number of aces varies more.