Financial Risk Manager Part 1

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.