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 \(\boxed{0.3970}\).