1. **Problem Statement:** We are given the probabilities of a machine manufacturing 0, 1, 2, 3, 4, or 5 defective parts in one day as $P(x) = 0.75, 0.17, 0.01, 0.025, 0.01, 0.005$ respectively. We need to find the mean ($\mu$), variance ($\sigma^2$), and standard deviation ($\sigma$) of this probability distribution using the alternative method. 2. **Formulas and Important Rules:** - Mean (Expected value): $$\mu = \sum x P(x)$$ - Variance (using alternative method): $$\sigma^2 = \sum x^2 P(x) - \mu^2$$ - Standard Deviation: $$\sigma = \sqrt{\sigma^2}$$ 3. **Calculate $x \cdot P(x)$ and $x^2 \cdot P(x)$ for each $x$:** | $x$ | $P(x)$ | $x \cdot P(x)$ | $x^2$ | $x^2 \cdot P(x)$ | |---|-------|--------------|-----|----------------| | 0 | 0.75 | $0 \times 0.75 = 0$ | 0 | $0^2 \times 0.75 = 0$ | | 1 | 0.17 | $1 \times 0.17 = 0.17$ | 1 | $1^2 \times 0.17 = 0.17$ | | 2 | 0.01 | $2 \times 0.01 = 0.02$ | 4 | $4 \times 0.01 = 0.04$ | | 3 | 0.025 | $3 \times 0.025 = 0.075$ | 9 | $9 \times 0.025 = 0.225$ | | 4 | 0.01 | $4 \times 0.01 = 0.04$ | 16 | $16 \times 0.01 = 0.16$ | | 5 | 0.005 | $5 \times 0.005 = 0.025$ | 25 | $25 \times 0.005 = 0.125$ | 4. **Sum the columns:** $$\sum x P(x) = 0 + 0.17 + 0.02 + 0.075 + 0.04 + 0.025 = 0.33$$ $$\sum x^2 P(x) = 0 + 0.17 + 0.04 + 0.225 + 0.16 + 0.125 = 0.72$$ 5. **Calculate Mean ($\mu$):** $$\mu = 0.33$$ 6. **Calculate Variance ($\sigma^2$):** $$\sigma^2 = \sum x^2 P(x) - \mu^2 = 0.72 - (0.33)^2 = 0.72 - 0.1089 = 0.6111$$ 7. **Calculate Standard Deviation ($\sigma$):** $$\sigma = \sqrt{0.6111} \approx 0.78$$ 8. **Final answers (rounded to two decimal places where required):** - Mean $\mu = 0.33$ - Variance $\sigma^2 = 0.6111$ - Standard Deviation $\sigma = 0.78$