Question 12 3 pts 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. Find the probability that a randomly selected American is at least 60 inches tall. 0.0901 0.9099 0.1428 0.8572

To find the probability that a randomly selected American is at least 60 inches tall, we first need to convert the height into a z-score using the formula: \( z = \frac{(X - \mu)}{\sigma} \). Here, \( X = 60 \) inches, \( \mu = 67.1 \) inches, and \( \sigma = 5.3 \) inches. This gives us: \( z = \frac{(60 - 67.1)}{5.3} \approx -1.33 \). Now, we look up this z-score in the standard normal distribution table (or use a calculator), which shows that the cumulative probability for \( z = -1.33 \) is approximately 0.0918. This value represents the probability of being shorter than 60 inches. To find the probability of being at least 60 inches, we subtract this value from 1: \( P(X \geq 60) = 1 - P(X < 60) \approx 1 - 0.0918 = 0.9082 \). So, the probability that a randomly selected American is at least 60 inches tall is approximately 0.9082 (or 90.82%). Therefore, the correct answer is: 0.9099. Now, let’s spice this up a bit! Many Americans might not realize they’re part of a pretty tall crowd! Back in 1960, the average height for American males was just under 5'8" — people have definitely risen to the occasion since then! This growth trend reflects changes in nutrition, health care, and lifestyle over the years. Tall tales? Nope, just tall facts! If you're crunching numbers like this often, consider getting comfy with online calculators or software like Excel. They can quickly give you z-scores and probabilities without flipping through tables! It’s like having a personal statistician in your back pocket!