To solve this problem, we need to verify that approximately what percent of 18-year-old males have heights between 66.8 inches and 72.4 inches, given the normal distribution parameters: a mean of 69.6 inches and a standard deviation of 1.4 inches.

Step-by-Step Solution:

1. Standardize the values (z-scores):
   Use the z-score formula:

   $z = \frac{x - \mu}{\sigma}$ where:

   • $x$ is the value we are interested in,
   • $\mu$ is the mean (69.6 inches),
   • $\sigma$ is the standard deviation (1.4 inches).

   Let's first calculate the z-scores for 66.8 and 72.4 inches.

   For 66.8 inches: $z_1 = \frac{66.8 - 69.6}{1.4} = \frac{-2.8}{1.4} = -2$

   For 72.4 inches: $z_2 = \frac{72.4 - 69.6}{1.4} = \frac{2.8}{1.4} = 2$

2. Find the probability corresponding to these z-scores: Using a standard normal distribution table, or a calculator, we can find the percentage of the population between z = -2 and z = 2.

   • The probability (or area under the curve) for a z-score of -2 is approximately 0.0228 (or 2.28%).
   • The probability for a z-score of 2 is approximately 0.9772 (or 97.72%).

3. Calculate the percentage of males between these heights: The percentage of males between these z-scores is the difference between the probabilities for z = 2 and z = -2.

   $97.72\% - 2.28\% = 95.44\%$

Thus, approximately 95.44% of 18-year-old males have heights between 66.8 inches and 72.4 inches.

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Would you like further details or have any questions?

Here are 5 related questions:

1. What is the probability that a randomly selected 18-year-old male is shorter than 66.8 inches?
2. How would you calculate the probability that an 18-year-old male is taller than 72.4 inches?
3. What percentage of males are taller than 70 inches?
4. How does changing the standard deviation affect the range of heights for 95.44% of males?
5. If the mean height changes to 70 inches, how would that affect the percentage of males between 66.8 and 72.4 inches?

Tip: The empirical rule (68-95-99.7 rule) tells us that about 95% of data in a normal distribution falls within 2 standard deviations from the mean.