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 all Filipino young women who are taller than 69 inches. 2. **Given:** - Mean $\mu = 50.5$ inches - Standard deviation $\sigma = 2$ inches - Height threshold $x = 69$ inches 3. **Required:** Find the fraction (probability) $P(X > 69)$ where $X$ is the height. 4. **Formula:** For a normal distribution, the standardized $z$-score is given by: $$ z = \frac{x - \mu}{\sigma} $$ The fraction of values greater than $x$ is: $$ P(X > x) = 1 - P(Z \leq z) = 1 - \Phi(z) $$ where $\Phi(z)$ is the cumulative distribution function (CDF) of the standard normal distribution. 5. **Solution:** Calculate the $z$-score: $$ z = \frac{69 - 50.5}{2} = \frac{18.5}{2} = 9.25 $$ A $z$-score of 9.25 is extremely high, meaning 69 inches is far above the mean. 6. **Finding the probability:** The CDF value $\Phi(9.25)$ is practically 1 because 9.25 is far in the right tail. Therefore, $$ P(X > 69) = 1 - \Phi(9.25) \approx 1 - 1 = 0 $$ 7. **Conclusion:** The fraction of Filipino young women taller than 69 inches is effectively 0, meaning it is extremely rare or almost nonexistent. 8. **Normal Curve Sketch Explanation:** The normal curve is centered at 50.5 inches with a spread determined by $\sigma=2$. The area to the right of 69 inches (far right tail) is nearly zero, indicating very few women exceed this height.