1. **Problem Statement:** We have a roulette wheel with numbers 1 to 9 and their respective probabilities given as: | X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |---|---|---|---|---|---|---|---|---|---| | P(X) | 0.10 | 0.10 | 0.10 | 0.10 | 0.10 | 0.10 | 0.10 | 0.10 | 0.20 | We need to find the mean (expected value), variance, and standard deviation of this distribution. 2. **Formulas and Rules:** - Mean (Expected Value): $$\mu = E(X) = \sum x_i P(x_i)$$ - Variance: $$\sigma^2 = Var(X) = E[(X - \mu)^2] = \sum (x_i - \mu)^2 P(x_i)$$ - Standard Deviation: $$\sigma = \sqrt{Var(X)}$$ 3. **Calculate the Mean:** $$\mu = 1(0.10) + 2(0.10) + 3(0.10) + 4(0.10) + 5(0.10) + 6(0.10) + 7(0.10) + 8(0.10) + 9(0.20)$$ $$= 0.10 + 0.20 + 0.30 + 0.40 + 0.50 + 0.60 + 0.70 + 0.80 + 1.80 = 5.40$$ 4. **Calculate the Variance:** First, compute each squared deviation times its probability: $$\sum (x_i - \mu)^2 P(x_i) = \sum (x_i - 5.40)^2 P(x_i)$$ Calculate each term: - For 1: $(1 - 5.40)^2 \times 0.10 = ( -4.40)^2 \times 0.10 = 19.36 \times 0.10 = 1.936$ - For 2: $(2 - 5.40)^2 \times 0.10 = ( -3.40)^2 \times 0.10 = 11.56 \times 0.10 = 1.156$ - For 3: $(3 - 5.40)^2 \times 0.10 = ( -2.40)^2 \times 0.10 = 5.76 \times 0.10 = 0.576$ - For 4: $(4 - 5.40)^2 \times 0.10 = ( -1.40)^2 \times 0.10 = 1.96 \times 0.10 = 0.196$ - For 5: $(5 - 5.40)^2 \times 0.10 = ( -0.40)^2 \times 0.10 = 0.16 \times 0.10 = 0.016$ - For 6: $(6 - 5.40)^2 \times 0.10 = (0.60)^2 \times 0.10 = 0.36 \times 0.10 = 0.036$ - For 7: $(7 - 5.40)^2 \times 0.10 = (1.60)^2 \times 0.10 = 2.56 \times 0.10 = 0.256$ - For 8: $(8 - 5.40)^2 \times 0.10 = (2.60)^2 \times 0.10 = 6.76 \times 0.10 = 0.676$ - For 9: $(9 - 5.40)^2 \times 0.20 = (3.60)^2 \times 0.20 = 12.96 \times 0.20 = 2.592$ Sum all these values: $$1.936 + 1.156 + 0.576 + 0.196 + 0.016 + 0.036 + 0.256 + 0.676 + 2.592 = 7.44$$ 5. **Calculate the Standard Deviation:** $$\sigma = \sqrt{7.44} \approx 2.73$$ **Final answers:** - Mean (Expected Value): $5.40$ - Variance: $7.44$ - Standard Deviation: $2.73$