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).