Assume that military aircraft use ejection seats designed for men weighing between $147.3 \mathrm{lb}$ and $214 \mathrm{lb}$. If women's weights are normally distributed with a mean of $161.8 \mathrm{lb}$ and a standard deviation of $43.4 \mathrm{lb}$, what percentage of women have weights that are within those limits? Are many women excluded with those specifications? The percentage of women that have weights between those limits is $\square \%$. (Round to two decimal places as needed.)

Step 1 ：We are given that women's weights are normally distributed with a mean of \(161.8 \mathrm{lb}\) and a standard deviation of \(43.4 \mathrm{lb}\).

Step 2 ：We are asked to find the percentage of women whose weights fall within the range of \(147.3 \mathrm{lb}\) to \(214 \mathrm{lb}\).

Step 3 ：To solve this, we can standardize the weights using the z-score formula: \(z = \frac{X - \mu}{\sigma}\), where \(X\) is the value we are standardizing, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 4 ：We calculate two z-scores: one for the lower limit (\(147.3 \mathrm{lb}\)) and one for the upper limit (\(214 \mathrm{lb}\)).

Step 5 ：The z-score for the lower limit is \(-0.334\) and for the upper limit is \(1.203\).

Step 6 ：We then find the probabilities corresponding to these z-scores using the standard normal distribution. The probability for the lower limit is \(0.369\) and for the upper limit is \(0.885\).

Step 7 ：The difference between these probabilities gives us the percentage of women whose weights fall within the given range.

Step 8 ：The percentage of women that have weights between those limits is \(\boxed{51.63 \%}\).