1. **Acceptance Probability, Lot Size N=25, 2 Nonconforming** (a) Probability all 5 selected are conforming (Hypergeometric): Number conforming = $25 - 2 = 23$ $$P = \frac{\binom{23}{5}}{\binom{25}{5}}$$ Calculate numerator and denominator: $$\binom{23}{5} = \frac{23!}{5! \times 18!} = 33,649$$ $$\binom{25}{5} = \frac{25!}{5! \times 20!} = 53,130$$ Therefore: $$P = \frac{33,649}{53,130} \approx 0.6336$$ (b) Binomial approximation: Probability nonconforming in one draw: $$p = \frac{2}{25} = 0.08$$ Probability all 5 conforming: $$P = (1 - p)^5 = (0.92)^5 \approx 0.6591$$ Difference: $$0.6591 - 0.6336 = 0.0255$$ Approximation slightly overestimates probability. (c) For $N=150$ with 2 nonconforming and $n=5$, $p = \frac{2}{150} = 0.0133$, binomial approximation is better because $n \ll N$ and small $p$. 2. **Tensile Strength $X \sim N(40, 5^2)$, 50,000 parts** (a) Number failing below 35 lb: Standardize: $$Z = \frac{35 - 40}{5} = -1.0$$ Probability: $$P(Z < -1.0) \approx 0.1587$$ Expected failures: $$50,000 \times 0.1587 = 7,935$$ (b) Number exceeding 48 lb: $$Z = \frac{48 - 40}{5} = 1.6$$ $$P(Z > 1.6) = 1 - 0.9452 = 0.0548$$ Expected number: $$50,000 \times 0.0548 = 2,740$$ 3. **Battery Life $X \sim N(900, 35^2)$** Fraction surviving beyond 1000 days: $$Z = \frac{1000 - 900}{35} \approx 2.857$$ $$P(X > 1000) = P(Z > 2.857) = 1 - 0.9979 = 0.0021$$ 4. **Diameter Sample, $n=15$, $\bar{x}=8.2535$, $\sigma=0.002$, Test $H_0: \mu=8.25$, $\alpha=0.05$** (a) Test statistic: $$Z = \frac{8.2535 - 8.25}{0.002/\sqrt{15}} = \frac{0.0035}{0.0005164} \approx 6.77$$ Critical value $z_{\alpha/2} = 1.96$, so reject $H_0$. (b) P-value is very small, $<0.0001$. (c) 95% CI: $$8.2535 \pm 1.96 \times 0.0005164 = (8.2525, 8.2545)$$ 5. **Voltage sample, $n=16$** Mean $\approx 10.15$, variance $\approx 0.78$ (a) Test $H_0: \mu=12$ vs two-sided at $\alpha=0.05$: $$t = \frac{10.15 - 12}{\sqrt{0.78}/4} = \frac{-1.85}{0.221} \approx -8.37$$ Critical $t_{0.025,15} \approx 2.131$, reject $H_0$. (b) 95% CI: $$10.15 \pm 2.131 \times 0.221 = (9.62, 10.68)$$ (c) Test $H_0: \sigma^2=11$, $$\chi^2 = \frac{(15)(0.78)}{11} = 1.064$$ Critical values: Lower=6.262, Upper=27.488, so reject $H_0$. (d) 95% CI for $\sigma$: $$\left( \sqrt{\frac{15 \times 0.78}{27.488}}, \sqrt{\frac{15 \times 0.78}{6.262}} \right) = (0.655, 1.363)$$ (e) 95% upper CI for $\sigma$: $$\sqrt{\frac{15 \times 0.78}{7.261}} = 1.13$$ 6. **Sample size $n=500$, defects $x=65$** Sample fraction: $$\hat{p} = \frac{65}{500} = 0.13$$ (a) Test $H_0: p=0.08$: $$z = \frac{0.13 - 0.08}{\sqrt{0.08 \times 0.92 / 500}} = \frac{0.05}{0.0121} \approx 4.13$$ Reject $H_0$ (critical $z=1.96$). (b) P-value: $$2 \times P(Z > 4.13) \approx 2 \times 0.000018 = 0.000036$$ (c) 95% upper CI: $$0.13 + 1.645 \times \sqrt{\frac{0.13 \times 0.87}{500}} = 0.1577$$ 7. **Paired Caliper Measurements, $n=12$** Mean difference: $$\bar{d} = \frac{-0.005}{12} \approx -0.00042$$ Std deviation $s_d \approx 0.0013$ Test statistic: $$t = \frac{-0.00042}{0.0013 / \sqrt{12}} \approx 1.11$$ Critical $t_{0.005,11} = 3.106$, fail to reject $H_0$. 8. **ANOVA for Compressive Strength, 4 levels with 3 replicates** Group means: 1500, 1586.67, 1606.67, 1500 Overall mean: $$1548.33$$ Calculated $F \approx 10$ vs critical $4.07$, reject $H_0$. Significant difference due to rodding level.