When women were finally allowed to become pilots of fighter jets, enginieers needed to redesign the ejection seats because they had been originally designed for men only. The ejection s 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 0.3970 . (Round to four decimal places as needed.) b. If 35 different women are randomly selected, find the probability that their mean weight is between $130 \mathrm{lb}$ and $181 \mathrm{lb}$. The probability is approximately $\square$. (Round to four decimal places as needed.)

Step 1 ：The problem is asking for the probability that the mean weight of 35 randomly selected women is between 130lb and 181lb. This is a problem of normal distribution. We know that the mean weight of women is 172lb and the standard deviation is 46lb.

Step 2 ：We can use the formula for the standard deviation of the mean to calculate the new standard deviation. The formula is \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation of the population and \(n\) is the size of the sample. Substituting the given values, we get \(\sigma_{\bar{x}} = \frac{46}{\sqrt{35}}\), which is approximately 7.7754191435023525.

Step 3 ：We can use the z-score formula to find the z-scores for 130lb and 181lb. The formula is \(z = \frac{x - \mu}{\sigma_{\bar{x}}}\), where \(x\) is the value, \(\mu\) is the mean, and \(\sigma_{\bar{x}}\) is the standard deviation of the mean. Substituting the given values, we get \(z_{130} = \frac{130 - 172}{7.7754191435023525}\), which is approximately -5.401638062830084, and \(z_{181} = \frac{181 - 172}{7.7754191435023525}\), which is approximately 1.1574938706064466.

Step 4 ：We can use the standard normal distribution table to find the probabilities corresponding to these z-scores. The probability corresponding to \(z_{130}\) is approximately 3.301754273860553e-08, and the probability corresponding to \(z_{181}\) is approximately 0.8764646789556032.

Step 5 ：We subtract the probabilities to find the probability that the mean weight is between 130lb and 181lb. The calculation is \(P_{130-181} = P_{181} - P_{130}\), which is approximately 0.8764646459380605.

Step 6 ：The probability that the mean weight of 35 randomly selected women is between 130lb and 181lb is approximately \(\boxed{0.8765}\).