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 $◻\mathrm{\%}$. (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 $\boxed{{\displaystyle 51.63\mathrm{\%}}}$.