1. **Problem:** Given the mean and standard deviation of family incomes in urban and rural areas, find the probability that the difference in sample means exceeds 5000. Step 1: Identify parameters. - Urban mean $\mu_1 = 10000$, standard deviation $\sigma_1 = 3000$, sample size $n_1 = 50$. - Rural mean $\mu_2 = 4000$, standard deviation $\sigma_2 = 500$, sample size $n_2 = 200$. Step 2: Calculate the mean and standard deviation of the difference of sample means. - Mean difference $\mu_D = \mu_1 - \mu_2 = 10000 - 4000 = 6000$. - Variance of difference $\sigma_D^2 = \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} = \frac{3000^2}{50} + \frac{500^2}{200} = \frac{9,000,000}{50} + \frac{250,000}{200} = 180,000 + 1,250 = 181,250$. - Standard deviation $\sigma_D = \sqrt{181,250} \approx 425.77$. Step 3: Find the probability $P(|\bar{X}_1 - \bar{X}_2| > 5000)$. - This is $P(D > 5000)$ or $P(D < -5000)$. - Standardize: $Z = \frac{D - \mu_D}{\sigma_D}$. - For $D = 5000$, $Z = \frac{5000 - 6000}{425.77} = \frac{-1000}{425.77} \approx -2.35$. - For $D = -5000$, $Z = \frac{-5000 - 6000}{425.77} = \frac{-11000}{425.77} \approx -25.83$ (very extreme). Step 4: Calculate probabilities using standard normal distribution. - $P(D > 5000) = P(Z > -2.35) = 1 - P(Z < -2.35) = 1 - 0.0094 = 0.9906$. - $P(D < -5000) \approx 0$ (negligible). - Total probability $\approx 0.9906$. --- 2. **Problem:** For 100 credit applications with approval probability 0.33, find the probability that more than 40% are approved. Step 1: Define parameters. - Number of trials $n = 100$. - Probability of approval $p = 0.33$. - We want $P(X > 40)$ where $X$ is number approved. Step 2: Use normal approximation to binomial. - Mean $\mu = np = 100 \times 0.33 = 33$. - Variance $\sigma^2 = np(1-p) = 100 \times 0.33 \times 0.67 = 22.11$. - Standard deviation $\sigma = \sqrt{22.11} \approx 4.70$. Step 3: Apply continuity correction and standardize. - $P(X > 40) = P(X \geq 41) \approx P(Y > 40.5)$ where $Y$ is normal. - $Z = \frac{40.5 - 33}{4.70} = \frac{7.5}{4.70} \approx 1.60$. Step 4: Find probability. - $P(Z > 1.60) = 1 - P(Z < 1.60) = 1 - 0.9452 = 0.0548$. --- 3. **Problem:** For layoffs in two companies with proportions 10% and 15%, sample sizes 100 and 200, find probability that difference in sample proportions is less than 5%. Step 1: Define parameters. - $p_1 = 0.10$, $n_1 = 100$. - $p_2 = 0.15$, $n_2 = 200$. - We want $P(\hat{p}_1 - \hat{p}_2 < 0.05)$. Step 2: Calculate mean and standard deviation of difference. - Mean difference $\mu_D = p_1 - p_2 = 0.10 - 0.15 = -0.05$. - Variance $\sigma_D^2 = \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} = \frac{0.10 \times 0.90}{100} + \frac{0.15 \times 0.85}{200} = 0.0009 + 0.0006375 = 0.0015375$. - Standard deviation $\sigma_D = \sqrt{0.0015375} \approx 0.0392$. Step 3: Standardize the variable. - $Z = \frac{0.05 - (-0.05)}{0.0392} = \frac{0.10}{0.0392} \approx 2.55$. Step 4: Find probability. - $P(\hat{p}_1 - \hat{p}_2 < 0.05) = P(Z < 2.55) = 0.9946$. --- **Final answers:** 1. Probability difference in incomes exceeds 5000 is approximately **0.9906**. 2. Probability more than 40% approvals is approximately **0.0548**. 3. Probability difference in layoff percentages is less than 5% is approximately **0.9946**.