1. **State the problem:** Shelia has a probability distribution for the number of defective parts $X$ found in a day with probabilities given as: $$P(X=0)=0.85, P(X=1)=0.10, P(X=2)=0.04, P(X=3)=0.01$$ 2. **Check the total probability:** The sum of all probabilities must equal 1: $$0.85 + 0.10 + 0.04 + 0.01 = 1.00$$ This confirms the distribution is valid. 3. **Calculate the expected value (mean) $E(X)$:** The formula for expected value is: $$E(X) = \sum X \cdot P(X)$$ Calculate: $$E(X) = 0 \times 0.85 + 1 \times 0.10 + 2 \times 0.04 + 3 \times 0.01$$ $$= 0 + 0.10 + 0.08 + 0.03 = 0.21$$ 4. **Calculate the variance $Var(X)$:** Variance formula: $$Var(X) = E(X^2) - [E(X)]^2$$ First find $E(X^2)$: $$E(X^2) = 0^2 \times 0.85 + 1^2 \times 0.10 + 2^2 \times 0.04 + 3^2 \times 0.01$$ $$= 0 + 0.10 + 0.16 + 0.09 = 0.35$$ Now variance: $$Var(X) = 0.35 - (0.21)^2 = 0.35 - 0.0441 = 0.3059$$ 5. **Calculate the standard deviation $\sigma$:** $$\sigma = \sqrt{Var(X)} = \sqrt{0.3059} \approx 0.553$$ **Final answers:** - Expected number of defective parts per day: $E(X) = 0.21$ - Variance of defective parts: $Var(X) = 0.3059$ - Standard deviation: $\sigma \approx 0.553$