1. **Problem:** Calculate the probability that the difference between the average daily family income in the city and the countryside is more than 5,000. Step 1: Identify given data: - City mean $\mu_1 = 10000$, city standard deviation $\sigma_1 = 3000$, sample size $n_1 = 50$ - Countryside mean $\mu_2 = 4000$, countryside 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 $P(|\bar{X}_1 - \bar{X}_2| > 5000)$. Step 4: Standardize the variable: $$Z = \frac{(\bar{X}_1 - \bar{X}_2) - \mu_D}{\sigma_D}$$ Calculate for $5000$: $$Z = \frac{5000 - 6000}{425.77} = \frac{-1000}{425.77} \approx -2.35$$ Step 5: Since the distribution is normal, the probability that the difference is more than 5000 is: $$P(|\bar{X}_1 - \bar{X}_2| > 5000) = P(Z < -2.35) + P(Z > 2.35) = 2 \times P(Z < -2.35)$$ From standard normal tables, $P(Z < -2.35) \approx 0.0094$ So, $$P = 2 \times 0.0094 = 0.0188$$ --- 2. **Problem:** For 100 credit limits, find the probability that more than 40% of applications are approved, given the approval rate for the first 6 applications is 0.33 and 0.67 for the last week. Step 1: Assume the approval probability $p = 0.33$ (first 6 applications) or $p = 0.67$ (last week). Since the problem is ambiguous, we consider the overall approval rate as $p = 0.33$. Step 2: Number of trials $n = 100$, success probability $p = 0.33$. Step 3: We want $P(\hat{p} > 0.40)$ where $\hat{p}$ is the sample proportion. Step 4: Mean and standard deviation of $\hat{p}$: - Mean $\mu = p = 0.33$ - Standard deviation $\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.33 \times 0.67}{100}} = \sqrt{0.002211} \approx 0.047 $ Step 5: Standardize: $$Z = \frac{0.40 - 0.33}{0.047} \approx 1.49$$ Step 6: Find $P(Z > 1.49)$ from standard normal table: $$P(Z > 1.49) \approx 0.0681$$ So, the probability more than 40% applications are approved is approximately 6.81%. --- 3. **Problem:** Find the probability that the difference in layoff percentages between PT. Karya and PT. Raya is less than 5%. Step 1: Given: - PT. Karya layoff rate $p_1 = 0.10$, sample size $n_1 = 100$ - PT. Raya layoff rate $p_2 = 0.15$, sample size $n_2 = 200$ Step 2: Mean difference: $$\mu_D = p_1 - p_2 = 0.10 - 0.15 = -0.05$$ Step 3: Variance of difference: $$\sigma_D^2 = \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} = \frac{0.10 \times 0.90}{100} + \frac{0.15 \times 0.85}{200} = 0.0009 + 0.0006375 = 0.0015375$$ Standard deviation: $$\sigma_D = \sqrt{0.0015375} \approx 0.0392$$ Step 4: We want $P(|p_1 - p_2| < 0.05)$, which is equivalent to: $$P(-0.05 < p_1 - p_2 < 0.05)$$ Step 5: Standardize the bounds: - Lower bound: $$Z_1 = \frac{-0.05 - (-0.05)}{0.0392} = 0$$ - Upper bound: $$Z_2 = \frac{0.05 - (-0.05)}{0.0392} = \frac{0.10}{0.0392} \approx 2.55$$ Step 6: Probability: $$P(0 < Z < 2.55) = P(Z < 2.55) - P(Z < 0)$$ From standard normal table, $P(Z < 2.55) \approx 0.9946$, $P(Z < 0) = 0.5$ So, $$P = 0.9946 - 0.5 = 0.4946$$ Therefore, the probability that the difference in layoff percentages is less than 5% is approximately 49.46%.