Answer: 0.016

## Explanation This problem applies the **Central Limit Theorem** to find the probability that a sample mean exceeds a certain value. ### Given: - Population mean (μ) = 42 - Population standard deviation (σ) = 7 - Sample size (n) = 36 - We need P(\bar{X} > 44.5) ### Step 1: Calculate the standard error $$SE = \frac{\sigma}{\sqrt{n}} = \frac{7}{\sqrt{36}} = \frac{7}{6} \approx 1.1667$$ ### Step 2: Calculate the z-score $$z = \frac{\bar{X} - \mu}{SE} = \frac{44.5 - 42}{1.1667} = \frac{2.5}{1.1667} \approx 2.143$$ ### Step 3: Find the probability $$P(\bar{X} > 44.5) = P(Z > 2.143) = 1 - \Phi(2.143)$$ From the standard normal table: - \Phi(2.14) ≈ 0.9838 - \Phi(2.15) ≈ 0.9842 - Interpolating: \Phi(2.143) ≈ 0.9839 Therefore: $$P(\bar{X} > 44.5) = 1 - 0.9839 = 0.0161$$ ### Key Concepts: - **Central Limit Theorem**: For large samples (n ≥ 30), the sampling distribution of the mean is approximately normal - **Standard Error**: Measures the variability of sample means around the population mean - **Z-score**: Standardizes the sample mean to compare with standard normal distribution This matches option **B (0.016)**.

Author: Tanishq Prabhu