A survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 2.4 in. A branch of the military requires women's heights to be between 58 in and 80 in.
a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall?
b. If this branch of the military changes the height requirements so that all women are eligible except the shortest $1 \%$ and the tallest $2 \%$, what are the new height requirements?
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a. The percentage of women who meet the height requirement is $\%$. (Round to two decimal places as needed.)

Step 1 ：Given that women's heights are normally distributed with a mean of 63.4 inches and a standard deviation of 2.4 inches.

Step 2 ：The height requirement for a branch of the military is between 58 inches and 80 inches.

Step 3 ：We need to convert these height requirements to z-scores. The z-score is calculated as \((X - \mu) / \sigma\), where X is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 4 ：Calculating the z-scores for the lower and upper bounds gives us -2.25 and 6.92 respectively.

Step 5 ：The percentage of women who meet the height requirement is the area under the standard normal curve between these z-scores.

Step 6 ：Using the standard normal distribution table, we find that the percentage of women who meet the height requirement is 98.78%.

Step 7 ：Final Answer: The percentage of women who meet the height requirement is \(\boxed{98.78\%}\).