Step 1 :Given that the mean sitting knee height for men is 21.2 in. with a standard deviation of 1.2 in., we can calculate the 95th percentile using the formula for the z-score, which is \((X - \mu) / \sigma\), where X is the value we're looking for, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. The z-score for the 95th percentile is approximately 1.645. Rearranging the formula to solve for X gives us \(X = \mu + z * \sigma\). Substituting the given values, we find that the minimum table clearance required to satisfy the requirement of fitting 95% of men is \(\boxed{23.2}\) in.
Step 2 :To determine whether a table clearance that fits 95% of men will also fit all women in the bottom 5%, we need to compare the 95th percentile for men with the 5th percentile for women. The 5th percentile for women can be calculated using the same method as above, but with a z-score of -1.645 (since we're looking at the lower tail of the distribution). Given that the mean sitting knee height for women is 19.1 in. with a standard deviation of 1.1 in., we find that the 5th percentile for women is 17.2905 in. Since the 95th percentile for men is greater than the 5th percentile for women, the statement is true. So, the answer is \(\boxed{B}\).
Step 3 :To find out what percentage of men fit a table with a clearance of 23.6 in., we need to calculate the percentile of 23.6 in the distribution of men's sitting knee heights. This can be done using the z-score formula, and then converting the z-score to a percentile. Given that the z-score for a clearance of 23.6 in. is 2.0000000000000018, we find that the percentile is 97.72498680518208. Therefore, the percentage of men that fit a table with a clearance of 23.6 in is \(\boxed{97.72\%}\).