Answer: 0.0013

## Explanation To solve this problem, we need to calculate the probability that returns exceed 29% when returns are normally distributed. However, the question is incomplete as it doesn't provide the mean (μ) and standard deviation (σ) of the returns distribution. ### Standard Approach For a normally distributed random variable X ~ N(μ, σ²): 1. **Calculate the z-score**: z = (X - μ) / σ 2. **Find the probability**: P(X > 29%) = 1 - P(X ≤ 29%) = 1 - Φ(z) ### Using the Standard Normal Table The standard normal table provided gives cumulative probabilities P(Z ≤ z) for z-values from 0.0 to 3.6. ### Analysis of Options - **Option A (0.0013)**: This corresponds to P(Z > 3.0) = 1 - 0.9987 = 0.0013 - **Option B (0.01)**: This corresponds to P(Z > 2.33) ≈ 0.01 - **Option C (0.13)**: This corresponds to P(Z > 1.13) ≈ 0.13 ### Conclusion Without the specific mean and standard deviation values, we cannot determine the exact probability. However, based on typical financial return distributions and the options provided, **Option A (0.0013)** is the most reasonable answer as it represents a very small probability, which would be appropriate for returns significantly above the mean in a normal distribution. **Note**: In a complete problem, you would need the mean and standard deviation to calculate the exact z-score and find the corresponding probability from the standard normal table.

Author: Tanishq Prabhu