1. **Problem:** Calculate acceptance probability with lot size $N=25$ and 2 nonconforming. (a) Calculate hypergeometric probability $P=\frac{\binom{23}{5}}{\binom{25}{5}}$ with $23$ conforming. Calculate numerator: $\binom{23}{5} = \frac{23!}{5!18!} = 33,649$. Calculate denominator: $\binom{25}{5} = \frac{25!}{5!20!} = 53,130$. So $P = \frac{33,649}{53,130} \approx 0.6336$. (b) Binomial approximation: $p = \frac{2}{25} = 0.08$. Probability zero nonconforming in 5 draws: $P = (1 - p)^5 = (1 - 0.08)^5 = 0.92^5 \approx 0.6591$. Difference: $0.6591 - 0.6336 = 0.0255$, slight overestimate. (c) For $N=150$, $p=\frac{2}{150}=0.0133$ and $n=5$, binomial approximation is better because $n \ll N$ and $p$ is very small. 2. **Problem:** Tensile strength $X \sim N(40,5^2)$, $50,000$ parts, minimum spec $35$ lb. (a) Compute $P(X < 35)$. Standardize: $Z = \frac{35-40}{5} = -1.0$. From standard normal tables: $P(Z < -1.0) \approx 0.1587$. Expected failing: $50,000 \times 0.1587 = 7,935$. (b) Number with $X>48$ lb: $Z = \frac{48-40}{5} = 1.6$. $P(Z>1.6) = 1 - P(Z \leq 1.6) \approx 1 - 0.9452 = 0.0548$. Expected number: $50,000 \times 0.0548 = 2,740$. 3. **Problem:** Battery life $X \sim N(900,35^2)$, find fraction surviving beyond 1000 days. Standardize: $Z = \frac{1000-900}{35} \approx 2.857$. $P(X > 1000) = P(Z > 2.857) = 1 - P(Z \leq 2.857) \approx 1 - 0.9979 = 0.0021$. Fraction surviving is approximately 0.0021. 4. **Problem:** Diameter sample $n=15$, $\bar{x} = 8.2535$, $\sigma=0.002$, test $H_0: \mu=8.25$ vs $H_a: \mu \neq 8.25$, $\alpha=0.05$. (a) Test statistic: $$Z = \frac{8.2535 - 8.25}{0.002/\sqrt{15}} = \frac{0.0035}{0.0005164} \approx 6.77$$ Critical value $z_{0.025} = 1.96$, since $6.77 > 1.96$, reject $H_0$. (b) P-value is almost zero, $< 0.0001$. (c) 95% Confidence interval: $$8.2535 \pm 1.96 \times 0.0005164 = (8.2525, 8.2545)$$ 5. **Problem:** Voltage sample $n=16$, data given, calculated mean $\bar{x} \approx 10.15$, variance $s^2 \approx 0.78$. (a) Test $H_0: \mu=12$ vs two-sided, $\alpha=0.05$. Test statistic: $$t = \frac{10.15 - 12}{\sqrt{0.78}/4} = \frac{-1.85}{0.221} \approx -8.37$$ Degrees freedom $df=15$, critical $t_{0.025,15} \approx 2.131$. Since $|t|=8.37 > 2.131$, reject $H_0$. (b) 95% CI for $\mu$: $$10.15 \pm 2.131 \times 0.221 = (9.62, 10.68)$$ (c) Test $H_0: \sigma^2=11$ vs $\alpha=0.05$ chi-square test: $$\chi^2 = \frac{(15)(0.78)}{11} = 1.064$$ Critical $\chi^2$ values: lower $6.262$, upper $27.488$. Since $1.064 < 6.262$, 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 confidence interval for $\sigma$: $$\sqrt{\frac{15 \times 0.78}{7.261}}=1.13$$ 6. **Problem:** Sample size $n=500$, defective $x=65$, sample fraction $\hat{p} = \frac{65}{500} = 0.13$. (a) Test $H_0: p=0.08$ vs two-sided $\alpha=0.05$. Test statistic: $$z = \frac{0.13 - 0.08}{\sqrt{0.08 \times 0.92 / 500}} = \frac{0.05}{0.0121} \approx 4.13$$ Critical $z_{0.025}=1.96$, reject $H_0$. (b) P-value: $$2 \times P(Z > 4.13) \approx 2 \times 0.000018 = 0.000036$$ (c) 95% upper confidence interval for fraction nonconforming: $$0.13 + 1.645 \times \sqrt{\frac{0.13 \times 0.87}{500}} = 0.1577$$ 7. **Problem:** Paired measurements by 12 inspectors, test mean difference zero at $\alpha=0.01$. Mean difference $\bar{d} = -0.00042$, standard deviation $s_d = 0.0013$, $n=12$. Test statistic: $$t = \frac{-0.00042}{0.0013/\sqrt{12}} = -1.11$$ Critical $t_{0.005, 11} = 3.106$. Since $|t|=1.11 < 3.106$, fail to reject $H_0$, no significant difference. 8. **Problem:** ANOVA for compressive strength at 4 rodding levels with 3 replicates. Means: 10=1500, 15=1586.67, 20=1606.67, 25=1500; overall mean $1548.33$. Calculated $F \approx 10$, critical $F_{3,8,0.05} \approx 4.07$, reject $H_0$. Conclusion: Significant difference in compressive strength due to rodding level.