**Core Concept**
The question is testing the understanding of the normal distribution of hemoglobin levels in pregnant females, specifically the use of the Z-score to determine the percentage of individuals below a certain value.

**Why the Correct Answer is Right**
To solve this question, we need to use the Z-score formula, which is Z = (X - μ) / σ, where X is the value we are interested in (in this case, the lower limit of the normal range), μ is the mean (10.6 gm/dL), and σ is the standard deviation (2 gm/dL). Once we calculate the Z-score, we can use a standard normal distribution table (Z-table) to find the percentage of individuals below that value. In this case, we are looking for the percentage of individuals with hemoglobin levels below 9.6 gm/dL (10.6 - 1 standard deviation). We can calculate the Z-score as follows: Z = (9.6 - 10.6) / 2 = -0.5. Using a Z-table, we find that a Z-score of -0.5 corresponds to a probability of approximately 0.3085. However, since we are looking for the percentage of individuals below 9.6 gm/dL, we need to find the area to the left of the Z-score, which is approximately 0.3085 * 2 = 0.617 (since the area to the left of the mean is 0.5, and the area to the left of the Z-score is half of the area to the left of the mean). This corresponds to a percentage of approximately 61.7%. However, since the question asks for the 5% of females with their hemoglobin level below the given value, we can use the Z-score to find the corresponding value. A Z-score of -1.645 corresponds to a probability of 0.05 (5% of the population). We can calculate the corresponding value as follows: X = μ + (Z * σ) = 10.6 + (-1.645 * 2) = 9.32 gm/dL.

**Why Each Wrong Option is Incorrect**
**Option A:** This option is incorrect because it does not take into account the Z-score or the normal distribution of hemoglobin levels.
**Option B:** This option is incorrect because it does not use the correct formula or the Z-table to determine the percentage of individuals below the given value.
**Option C:** This option is incorrect because it is not a valid Z-score or probability.

**Clinical Pearl / High-Yield Fact**
Remember that the Z-score is a useful tool for determining the percentage of individuals below or above a certain value, but it requires a standard normal distribution table (Z-table) to use effectively.

**Correct Answer: D. 9.32 gm/dL**