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This question involves a normal distribution. We are given:

- Mean (\(\mu\)) = 65 inches
- Standard deviation (\(\sigma\)) = 2 inches
- We need to find the height corresponding to the 99th percentile.

For a normal distribution, the 99th percentile corresponds to a z-score of approximately \(2.33\). The formula for converting a z-score into a height value in a normal distribution is:

\[
X = \mu + z \cdot \sigma
\]

Substituting the known values:

\[
X = 65 + 2.33 \times 2 = 65 + 4.66 = 69.66 \text{ inches}
\]

Rounding to the closest value, the correct answer is closest to **70 inches (D)**.

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:
1. How do you find the z-score for other percentiles in a normal distribution?
2. What is the empirical rule, and how does it apply to this problem?
3. How would the answer change if the standard deviation were larger?
4. What are other ways to interpret the 99th percentile in terms of probability?
5. How would you calculate the area under a normal curve for a given range of values?

**Tip:** Always check the z-score table for precise values when working with percentiles in normal distributions!

The heights of adult women are approximately normally distributed about a mean of 65 inches with a standard deviation of 2 inches. If Rachael is at the 99th percentile in height for adult women, then her height, in inches, is closest to:

Learn how to calculate the height at the 99th percentile for adult women using normal distribution. This problem involves a mean height of 65 inches, a standard deviation of 2 inches, and applies a z-score of 2.33 for the 99th percentile. Suitable for high school students in Grades 10-12.

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