1. **Problem Statement:** Calculate acceptance probability for a lot with $N=25$ items, 2 nonconforming, selecting a sample of 5 without replacement. 2. **Exact hypergeometric probability:** Number conforming $= 25 - 2 = 23$. Probability all 5 selected are conforming: $$P = \frac{\binom{23}{5}}{\binom{25}{5}}$$ Calculate combinations: $$\binom{23}{5} = \frac{23!}{5!18!} = 33649$$ $$\binom{25}{5} = \frac{25!}{5!20!} = 53130$$ So, $$P=\frac{33649}{53130} \approx 0.6336$$ 3. **Binomial approximation:** $p = \frac{2}{25} = 0.08$ is small, and $n=5$ samples. $$P = (1-p)^5 = (1-0.08)^5 = 0.92^5 \approx 0.6591$$ Difference: $0.6591 - 0.6336 = 0.0255$ Approximation slightly overestimates acceptance but is close. 4. For $N=150$, $p=\frac{2}{150}=0.0133$, with $n=5$, binomial approximation is very good due to smaller $p$ and $n \ll N$. --- 5. **Tensile strength:** $X \sim N(40, 5^2)$, 50,000 parts, min spec limit 35 lb. (a) Probability failing: Standardize: $$Z = \frac{35 - 40}{5} = -1.0$$ $$P(Z < -1.0) \approx 0.1587$$ Expected failures: $$50,000 \times 0.1587 = 7935$$ (b) Strength above 48 lb: Standardize: $$Z = \frac{48 - 40}{5} = 1.6$$ $$P(Z > 1.6) = 1 - 0.9452 = 0.0548$$ Expected number: $$50,000 \times 0.0548 = 2740$$ --- 6. **Battery life:** $X \sim N(900, 35^2)$, find fraction surviving beyond 1000 days. Standardize: $$Z = \frac{1000 - 900}{35} \approx 2.857$$ $$P(X > 1000) = P(Z > 2.857) = 1 - 0.9979 = 0.0021$$ Fraction surviving beyond 1000 days is approximately 0.0021. --- 7. **Diameter sample test (n=15):** $\bar{x}=8.2535$, $\sigma=0.002$, test $H_0: \mu=8.25$, $\alpha=0.05$ two-sided. Test statistic: $$Z = \frac{8.2535 - 8.25}{0.002 / \sqrt{15}} = \frac{0.0035}{0.0005164} \approx 6.77$$ Critical value $z_{\alpha/2} = 1.96$, since $6.77 > 1.96$, reject $H_0$. P-value almost zero $< 0.0001$. 95% CI: $$8.2535 \pm 1.96 \times 0.0005164 = (8.2525, 8.2545)$$ --- 8. **Voltage sample (n=16):** Mean $\approx 10.15$, variance $\approx 0.78$, test $H_0: \mu = 12$ two-sided $\alpha=0.05$. T-statistic: $$t = \frac{10.15 - 12}{\sqrt{0.78}/4} = \frac{-1.85}{0.221} \approx -8.37$$ Degrees freedom $=15$, critical $t_{0.025,15} \approx 2.131$, reject $H_0$. 95% CI: $$10.15 \pm 2.131 \times 0.221 = (9.62, 10.68)$$ Test variance $H_0: \sigma^2=11$: $$\chi^2 = \frac{15 \times 0.78}{11} = 1.064$$ Critical chi-square (df=15): lower 6.262, upper 27.488. Since $1.064 < 6.262$, reject $H_0$. 95% CI for $\sigma$: $$\left(\sqrt{\frac{15 \times 0.78}{27.488}}, \sqrt{\frac{15 \times 0.78}{6.262}}\right) = (0.655, 1.363)$$ 95% upper CI limit: $$\sqrt{\frac{15 \times 0.78}{7.261}} = 1.13$$ --- 9. **Defectives sample (n=500, x=65):** Sample fraction: $$\hat{p} = \frac{65}{500} = 0.13$$ Test $H_0: p=0.08$, 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 $H_0$. P-value: $$2 \times P(Z > 4.13) \approx 0.000036$$ 95% upper confidence bound: $$0.13 + 1.645 \times \sqrt{\frac{0.13 \times 0.87}{500}} = 0.1577$$ --- 10. **Paired caliper measurements (n=12):** Differences mean: $$\bar{d} = \frac{-0.005}{12} \approx -0.00042$$ Std deviation $s_d \approx 0.0013$. Test statistic: $$t = \frac{-0.00042}{0.0013/\sqrt{12}} = 1.11$$ Critical $t_{0.005,11} = 3.106$, since $1.11 < 3.106$, fail to reject $H_0$. No significant difference. --- 11. **ANOVA for compressive strength (4 levels, 3 reps each):** Group means: 1500, 1586.67, 1606.67, 1500. Overall mean: $$\bar{x} = \frac{1500 + 1586.67 + 1606.67 + 1500}{4} = 1548.33$$ Calculate SST, SSB, SSE, and F-statistic. Computed $F \approx 10 > F_{3,8,0.05} \approx 4.07$, reject $H_0$. Conclusion: Significant difference due to rodding level.