
Answer-first summary for fast verification
Answer: 53
Let’s walk through this carefully and intuitively. ### Step 1: Understand the problem Production capacity is **normally distributed** with: - Mean (μ) = 100 - Standard deviation (σ) = 40 We are asked for the **12th percentile**, meaning **12% of outcomes fall below this value**. --- ### Step 2: Convert percentile to a z‑score In a normal distribution, percentiles are found using **z‑scores**. We want: \[ P(Z \le z) = 0.12 \] From the provided z‑table, the closest probability to 0.12 is **0.1210**, which corresponds to: \[ z \approx -1.17 \] (The negative sign makes sense because the 12th percentile is **below the mean**.) --- ### Step 3: Convert back to the original scale \[ X = \mu + z\sigma = 100 + (-1.17)(40) \] \[ X = 100 - 46.8 \approx 53 \] --- ### Step 4: Evaluate answer choices - **50 (A)**: Too low → corresponds to a much smaller percentile - **56 (B)**: Not far enough below the mean - **53 (C)** ✅ Correct calculation - **54 (D)**: Slightly higher than the true percentile --- ### Key intuition (FRM exam tip) Percentiles below 50% always use **negative z‑scores**, and the distance from the mean is driven by the **standard deviation**. ✅ **Correct answer: C (53)**
Author: Nikitesh Somanthe
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The production capacity of a production entity is normally distributed with a mean of 100 and a standard deviation of 40. Calculate the 12th percentile of the production capacity.
Below is a compact z‑table extract sufficient to answer this question directly.
| z‑score | P(Z ≤ z) |
|---|---|
| −1.28 | 0.1003 |
| −1.25 | 0.1056 |
| −1.20 | 0.1151 |
| −1.17 | 0.1210 |
| −1.15 | 0.1251 |
| −1.10 | 0.1357 |
A
50
B
56
C
53
D
54