1. **Problem:** Find the mean number of customers arriving at the counter during a 10-minute interval given the probability distribution: $$X = \{0,1,2,3,4,5,6\}$$ $$P(X) = \{0.05, 0.10, 0.20, 0.25, 0.20, 0.15, 0.05\}$$ 2. **Formula for Mean (Expected Value):** $$\mu = E(X) = \sum X \cdot P(X)$$ 3. **Calculate the mean:** $$\mu = 0(0.05) + 1(0.10) + 2(0.20) + 3(0.25) + 4(0.20) + 5(0.15) + 6(0.05)$$ $$= 0 + 0.10 + 0.40 + 0.75 + 0.80 + 0.75 + 0.30 = 3.10$$ 4. **Formula for Variance:** $$\sigma^2 = Var(X) = E(X^2) - [E(X)]^2$$ where $$E(X^2) = \sum X^2 \cdot P(X)$$ 5. **Calculate $E(X^2)$:** $$E(X^2) = 0^2(0.05) + 1^2(0.10) + 2^2(0.20) + 3^2(0.25) + 4^2(0.20) + 5^2(0.15) + 6^2(0.05)$$ $$= 0 + 0.10 + 0.80 + 2.25 + 3.20 + 3.75 + 1.80 = 11.90$$ 6. **Calculate Variance:** $$\sigma^2 = 11.90 - (3.10)^2 = 11.90 - 9.61 = 2.29$$ 7. **Calculate Standard Deviation:** $$\sigma = \sqrt{2.29} \approx 1.51$$ 8. **Expected customers in 2 hours:** Since 2 hours = 120 minutes, and the data is for 10-minute intervals, number of intervals = $\frac{120}{10} = 12$ Expected customers in 2 hours = $12 \times 3.10 = 37.2$ **Final answers:** - Mean number of customers $\mu = 3.10$ - Variance $\sigma^2 = 2.29$ - Standard deviation $\sigma \approx 1.51$ - Expected customers in 2 hours = 37.2