When women were finally allowed to become pilots of fighter jets, engineers needed to redesign the ejection seats because they had been originally designed for men only. The ejection seats were designed for men weighing between $130 \mathrm{lb}$ and $181 \mathrm{lb}$. Weights of women are now normally distributed with a mean of $172 \mathrm{lb}$ and a standard deviation of $46 \mathrm{lb}$. Complete parts (a) through (c) below.
a. If 1 woman is randomly selected, find the probability that her weight is between $130 \mathrm{lb}$ and $181 \mathrm{lb}$.
The probability is approximately
(Round to four decimal places as needed.)

Step 1 ：Given that the weights of women are normally distributed with a mean of 172lb and a standard deviation of 46lb, we are asked to find the probability that a woman's weight is between 130lb and 181lb.

Step 2 ：To solve this, we first convert the weights into z-scores, which measure how many standard deviations an element is from the mean. The formula for a z-score is: \(z = \frac{X - \mu}{\sigma}\), where X is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：Substituting the given values into the formula, we find the z-scores for the lower and upper bounds of the weight range: \(z_{lower} = \frac{130 - 172}{46} = -0.913\) and \(z_{upper} = \frac{181 - 172}{46} = 0.196\).

Step 4 ：We then use the z-scores to find the corresponding probabilities in the standard normal distribution: \(prob_{lower} = 0.181\) and \(prob_{upper} = 0.578\).

Step 5 ：Finally, we subtract the smaller probability from the larger one to get the probability of the weight range: \(prob_{range} = prob_{upper} - prob_{lower} = 0.578 - 0.181 = 0.397\).

Step 6 ：Thus, the probability that a randomly selected woman's weight is between 130lb and 181lb is approximately \(\boxed{0.3970}\).