1. Problem (b): We want to find the sample size $n$ to estimate the percent defective within 2% with 95.45% confidence from a population of 4000. 2. Formula for sample size when estimating a proportion with finite population correction: $$n = \frac{N z^2 p q}{(N-1) E^2 + z^2 p q}$$ where - $N=4000$ (population size), - $z$ is the z-score for confidence level, - $p$ is estimated proportion defective, - $q=1-p$, - $E$ is margin of error (in decimal). 3. For 95.45% confidence, $z \approx 2$ (since 95.45% corresponds to about 2 standard deviations). 4. Since no prior estimate of $p$ is given, use conservative estimate $p=0.5$, $q=0.5$ to maximize sample size. 5. Margin of error $E=0.02$ (2%). 6. Substitute values: $$n = \frac{4000 \times 2^2 \times 0.5 \times 0.5}{(4000-1) \times 0.02^2 + 2^2 \times 0.5 \times 0.5} = \frac{4000 \times 4 \times 0.25}{3999 \times 0.0004 + 4 \times 0.25}$$ 7. Calculate numerator: $$4000 \times 4 \times 0.25 = 4000 \times 1 = 4000$$ 8. Calculate denominator: $$3999 \times 0.0004 = 1.5996$$ $$4 \times 0.25 = 1$$ Sum: $1.5996 + 1 = 2.5996$ 9. Calculate $n$: $$n = \frac{4000}{2.5996} \approx 1538.5$$ 10. Since sample size must be whole number, round up: $$n = 1539$$ **Final answer:** The required sample size is 1539. --- **Slug:** sample size defective **Subject:** statistics **Desmos:** {"latex":"y=0","features":{"intercepts":true,"extrema":true}} **q_count:** 5