1. **Problem Statement:** We analyze the quality control data of light bulbs using various probability distributions and statistical methods. --- ### Part 1: Discrete Random Variables and Distributions **Question 1:** Probability exactly 5 out of 100 bulbs are defective (Binomial distribution). - Parameters: $n=100$, $p=0.03$, $k=5$. - Formula: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ - Calculation: $$P(X=5) = \binom{100}{5} (0.03)^5 (0.97)^{95}$$ **Question 2:** Probability fewer than 3 bulbs defective. - Calculate $P(X<3) = P(X=0) + P(X=1) + P(X=2)$ using the binomial formula. **Question 3:** Probability first defective bulb appears on 4th inspection (Geometric distribution). - $p=0.03$, $k=4$. - Formula: $$P(X=k) = (1-p)^{k-1} p$$ - Calculation: $$P(X=4) = (0.97)^3 \times 0.03$$ **Question 4:** Probability of selecting 3 defective bulbs in a sub-sample of 10 from 100 bulbs with 10 defective (Hypergeometric distribution). - Population size $N=100$, defective $K=10$, sample size $n=10$, successes $k=3$. - Formula: $$P(X=k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$$ **Question 5:** Probability of exactly 5 defects in an hour given average rate 3/hour (Poisson distribution). - $\\lambda=3$, $k=5$. - Formula: $$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$$ --- ### Part 2: Continuous Random Variables and Distributions **Question 1:** Probability bulb lasts longer than 1500 hours (Normal distribution). - Mean $\mu=1200$, SD $\sigma=200$. - Calculate $P(X>1500) = 1 - P(X \leq 1500)$. - Convert to z-score: $$z = \frac{1500 - 1200}{200} = 1.5$$ - Use standard normal table: $P(Z>1.5)$. **Question 2:** Find lifetime at which 5% bulbs fail. - Find $x$ such that $P(X \leq x) = 0.05$. - Find $z$ for 5th percentile: $z \approx -1.645$. - Calculate: $$x = \mu + z \sigma = 1200 + (-1.645)(200) = 871$$ hours. **Question 3:** Convert raw score 1500 hours to z-score. - $$z = \frac{1500 - 1200}{200} = 1.5$$ - Interpretation: 1500 hours is 1.5 standard deviations above the mean. **Question 4:** Normal approximation to binomial for at least 10 defective bulbs. - $n=100$, $p=0.03$, mean $\mu = np = 3$, SD $\sigma = \sqrt{np(1-p)} = \sqrt{3 \times 0.97} \approx 1.705$. - Use continuity correction: $P(X \geq 10) \approx P(Y > 9.5)$. - Convert to z: $$z = \frac{9.5 - 3}{1.705} \approx 3.80$$. - Probability is $P(Z > 3.80)$, very small. **Question 5:** Probability next defect occurs within 20 minutes (Exponential distribution). - Mean time $\theta=30$ minutes, rate $\lambda = \frac{1}{30}$. - Formula: $$P(T \leq t) = 1 - e^{-\lambda t}$$ - Calculation: $$P(T \leq 20) = 1 - e^{-\frac{20}{30}} = 1 - e^{-0.6667} \approx 0.487$$. --- ### Part 3: Central Limit Theorem and Hypothesis Testing **Question 1:** Standard deviation of sample mean for samples of size 36. - Population SD $\sigma=200$. - Sample size $n=36$. - Standard error: $$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{200}{6} = 33.33$$ hours. **Question 2:** Probability average lifespan of sample of 36 bulbs is less than 1150 hours. - Mean $\mu=1200$, SE $33.33$. - Calculate $P(\bar{X} < 1150)$. - Convert to z: $$z = \frac{1150 - 1200}{33.33} = -1.5$$. - Probability: $P(Z < -1.5)$. **Question 3:** Hypothesis test for claim less than 4% defective at 5% significance. - Null hypothesis $H_0: p = 0.04$, alternative $H_a: p < 0.04$. - Sample proportion $\hat{p} = \frac{5}{100} = 0.05$ (assuming 5 defective found). - Standard error: $$SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.04 \times 0.96}{100}} = 0.0196$$. - Test statistic: $$z = \frac{\hat{p} - p}{SE} = \frac{0.05 - 0.04}{0.0196} = 0.51$$. - Since $z=0.51$ is not less than critical value $-1.645$, fail to reject $H_0$. **Question 4:** Standard error with finite population correction (FPC). - Population $N=10000$, sample $n=100$, SD $\sigma=200$. - FPC: $$SE = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}} = 20 \times \sqrt{\frac{9900}{9999}} \approx 19.9$$ hours. --- **Final answers:** - Binomial exact 5 defective: $P(X=5) = \binom{100}{5} (0.03)^5 (0.97)^{95}$. - Binomial fewer than 3 defective: sum of $P(X=0), P(X=1), P(X=2)$. - Geometric first defective on 4th: $(0.97)^3 \times 0.03$. - Hypergeometric 3 defective in 10: $\frac{\binom{10}{3} \binom{90}{7}}{\binom{100}{10}}$. - Poisson exactly 5 defects: $\frac{e^{-3} 3^5}{5!}$. - Normal $P(X>1500) = P(Z>1.5)$. - 5% failure lifetime: 871 hours. - Z-score for 1500 hours: 1.5. - Normal approx binomial $P(X \geq 10) \approx P(Z>3.8)$. - Exponential $P(T \leq 20) \approx 0.487$. - CLT SE for n=36: 33.33 hours. - $P(\bar{X} < 1150) = P(Z < -1.5)$. - Hypothesis test: fail to reject $H_0$. - FPC SE: approx 19.9 hours.