Answer: 0.016

## Explanation This question involves applying the Central Limit Theorem to find the probability that a sample mean exceeds a certain value. **Given:** - Sample size (n) = 36 - Population standard deviation (σ) = 7 - Population mean (μ) = 42 - Sample mean threshold (x̄) = 44.50 **Step 1: Standardize using the Central Limit Theorem** According to the Central Limit Theorem, for large enough sample sizes, the sampling distribution of the sample mean is approximately normal: $$\frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim N(0, 1)$$ **Step 2: Calculate the z-score** $$z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} = \frac{44.50 - 42}{7 / \sqrt{36}} = \frac{2.50}{7 / 6} = \frac{2.50}{1.1667} \approx 2.143$$ **Step 3: Find the probability** We need $P(\bar{X} > 44.50) = P(Z > 2.143)$ From standard normal distribution tables: - $P(Z \leq 2.143) \approx 0.9840$ - Therefore, $P(Z > 2.143) = 1 - 0.9840 = 0.0160$ **Step 4: Interpretation** The probability that the sample mean exceeds 44.50 is approximately 0.016 or 1.6%. **Key Concepts:** 1. **Central Limit Theorem**: For sufficiently large sample sizes (n ≥ 30 is typically considered large enough), the sampling distribution of the sample mean is approximately normal regardless of the population distribution. 2. **Standard Error**: $\sigma/\sqrt{n}$ is the standard deviation of the sampling distribution of the mean. 3. **Z-score**: Measures how many standard errors the sample mean is from the population mean. **Why other options are incorrect:** - **A (0.045)**: This would correspond to approximately $P(Z > 1.70)$ - **C (0.065)**: This would correspond to approximately $P(Z > 1.51)$ - **D (0.042)**: This would correspond to approximately $P(Z > 1.73)$ All these z-scores are lower than our calculated 2.143, resulting in higher probabilities.

Author: Nikitesh Somanthe