To determine the new height requirements for men that exclude the tallest 50% and the shortest 5%, we need to use the concept of percentiles and the normal distribution.

Step-by-step Solution:

1. Normal Distribution for Men's Heights:
   • Mean ($\mu$): 67.6 inches
   • Standard deviation ($\sigma$): 3.1 inches

(a) Find the height that excludes the tallest 50% of men:

The tallest 50% of men would be at the 50th percentile, which corresponds to the mean in a normal distribution.

• The 50th percentile is exactly at the mean, so the height requirement to exclude the tallest 50% is: $\text{50th percentile} = \mu = 67.6 \text{ inches}.$

(b) Find the height that excludes the shortest 5% of men:

We need to find the height corresponding to the 5th percentile of the distribution. Using standard normal distribution tables or a calculator, the z-score for the 5th percentile is approximately -1.645 (negative because it's below the mean).

We now use the z-score formula to find the corresponding height: $z = \frac{X - \mu}{\sigma}$ Where:

• $z = -1.645$
• $\mu = 67.6$
• $\sigma = 3.1$

Solving for $X$ (the height): $-1.645 = \frac{X - 67.6}{3.1}$ $X - 67.6 = -1.645 \times 3.1$ $X - 67.6 = -5.0995$ $X = 67.6 - 5.0995 = 62.5005 \text{ inches}.$

Conclusion:

• The minimum height requirement (excluding the shortest 5% of men) is approximately 62.5 inches.
• The maximum height requirement (excluding the tallest 50% of men) is 67.6 inches.

Final Answer:

The new height requirements will be:

• Minimum height: 62.5 inches (to exclude the shortest 5%)
• Maximum height: 67.6 inches (to exclude the tallest 50%)

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Would you like further details on how to calculate percentiles using z-scores, or any other related information? Here are a few more questions you might want to consider:

1. How do you calculate z-scores for different percentiles?
2. What is the standard normal distribution, and how is it used?
3. How would the results change if the standard deviation of men's heights increased?
4. How does changing the mean affect the height requirements?
5. Could the same approach be applied to women's height requirements?

Tip: For normally distributed data, the mean always represents the 50th percentile, and z-scores help in finding specific percentiles based on standard deviations.