Let's suppose that heights of Americans are normally distributed with a mean of 67.1 inches and a standard deviation of 5.3 inches. If a particular American is shorter than only \( 3 \% \) of all Americans, how tall is that person in inches? 76.1 inches 77.1 inches 57.1 inches 78.1 inches

To determine the height of a person who is shorter than only 3% of all Americans, we need to find the z-score corresponding to the 3rd percentile of the normal distribution. The z-score for the 3rd percentile is approximately -1.88. We can plug this into the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \] Where: - \( z \) is the z-score, - \( X \) is the height, - \( \mu \) is the mean (67.1 inches), and - \( \sigma \) is the standard deviation (5.3 inches). Rearranging the formula to solve for \( X \): \[ X = \mu + z \cdot \sigma \] Substituting in the values: \[ X = 67.1 + (-1.88) \cdot 5.3 \] \[ X = 67.1 - 9.954 \] \[ X \approx 57.146 \] Therefore, the height of a person who is shorter than only 3% of Americans is approximately 57.1 inches. The correct answer is **57.1 inches.**