1. **Problem Statement:** We are given that the heights of Filipino young women are normally distributed with mean $\mu = 50.5$ inches and standard deviation $\sigma = 2$ inches. We want to find the fraction of women taller than 69 inches. 2. **Given:** - Mean $\mu = 50.5$ - Standard deviation $\sigma = 2$ - Height threshold $x = 69$ 3. **Required:** Find $P(X > 69)$ where $X$ is the height. 4. **Formula:** To find the probability for a normal distribution, we convert to the standard normal variable $Z$ using: $$Z = \frac{X - \mu}{\sigma}$$ Then use the standard normal distribution table or calculator to find $P(Z > z)$. 5. **Solution:** Calculate $Z$: $$Z = \frac{69 - 50.5}{2} = \frac{18.5}{2} = 9.25$$ 6. **Interpretation:** A $Z$-score of 9.25 is extremely high, far in the tail of the normal distribution. 7. **Probability:** Using standard normal tables or a calculator, $P(Z > 9.25)$ is effectively 0 (practically zero). 8. **Conclusion:** The fraction of Filipino young women taller than 69 inches is nearly zero. 9. **Graph description:** The normal curve is centered at 50.5 with a spread determined by $\sigma=2$. The area to the right of 69 inches (far right tail) is almost zero, indicating very few women exceed this height.