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To solve part (a) of this problem, we need to calculate the percentage of female students with heights shorter than 64 inches, based on the provided normal distribution parameters:

Given:
- Mean height (\(\mu\)) = 64.5 inches
- Standard deviation (\(\sigma\)) = 1.8 inches
- We are given that the area under the normal curve to the left of 64 inches (P(X < 64)) is 0.3903.

This area represents the probability (or proportion) of students with heights shorter than 64 inches.

### Solution:
To find the percentage, we can convert the probability to a percentage:

\[
0.3903 \times 100 = 39.03\%
\]

Thus, **39.03%** of female students are estimated to have heights shorter than 64 inches.

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Would you like a more detailed breakdown, or do you have any further questions?

### Related Questions
1. What is the probability that a female student's height is between 64 and 65 inches?
2. How would the percentage change if the standard deviation were different?
3. Can we calculate the probability for students taller than 66 inches?
4. How does changing the mean height affect the distribution?
5. How can we determine the z-score for a specific height like 63 inches?

### Tip
When dealing with normal distributions, remember that tables or software can often quickly find probabilities using z-scores, which standardize values based on the mean and standard deviation.

The area under the normal curve with parameters μ = 64.5 and σ = 1.8 that lies to the left of 64 is 0.3903. Use this information to estimate the percent of female students who are shorter than 64 inches.

Learn how to calculate the percentage of female college students shorter than 64 inches using a normal distribution with μ = 64.5 and σ = 1.8. This solution covers probability concepts and conversion to percentage, suitable for college-level statistics.

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