The correct answer is B.

If P/L over some period are normally distributed with mean 12 and standard deviation 18, then the 95% VaR is given by:

$$\begin{aligned} \alpha \text{VaR} &= -\mu_{\text{P/L}} + \sigma_{\text{P/L}} Z_\alpha \\ &= -12 + 18 Z_{0.95} \\ &= -12 + 18 \times 1.645 = 17.61 \end{aligned}$$

Where $z_\alpha$ is the number of standard deviations that correspond to the desired confidence level, $\sigma$ is the standard deviation of the distribution, and $\mu_{\text{P/L}}$ is the mean of the P/L distribution.

Things to Remember

When calculating Value at Risk (VaR) at a 95% confidence level, we typically use a one-tailed approach because we are only concerned with losses beyond a certain threshold (i.e., the worst 5% of outcomes). In a one-tailed setting, the z-score that captures the 5th percentile (the left tail) of a standard normal distribution is approximately -1.645. The negative sign indicates we’re looking at the lower tail (losses).

In contrast, a two-tailed 95% confidence interval (common in hypothesis testing) would split the remaining 5% equally between both tails (2.5% on the left and 2.5% on the right), leading to a z-score of approximately ±1.96. This two-tailed approach is used when deviations on both sides of the mean are relevant. However, for VaR, where only extreme losses (left tail) matter, the one-tailed z-score of -1.645 is appropriate.