Part 3 of 4 HW Score: 50\%, 6 of 12 points Points: 0 of 1 Save A water taxi carries passengers from harbor to another. Assume that weights of passengers are normally distributed with a mean of 195 lb and a standard deviation of 37 lb . The water taxi has a stated capacity of 25 passengers, and the water taxi was rated for a load limit of 3750 lb . Complete parts (a) through (d) below. a. Given that the water taxi was rated for a load limit of 3750 lb , what is the maximum mean weight of the passengers if the water taxi is filled to the stated capacity of 25 passengers? The maximum mean weight is 150 lb . (Type an integer or a decimal. Do not round.) b. If the water taxi is filled with 25 randomly selected passengers, what is the probability that their mean weight exceeds the value from part (a)? The probability is 1 (Round to four decimal places as needed.) c. If the weight assumptions were revised so that the new capacity became 20 passengers and the water taxi is filled with 20 randomly selected passengers, what is the probability that their mean weight exceeds 187.5 lb , which is the maximum mean weight that does not cause the total load to exceed 3750 lb ? The probability is \( \square \) (Round to four decimal places as needed.)

We wish to find the probability that the mean weight of 20 randomly selected passengers exceeds \(187.5\) lb. The individual weights are assumed to be normally distributed with mean \(\mu = 195\) lb and standard deviation \(\sigma = 37\) lb. Since the sample mean \(\bar{X}\) for a sample of \(n=20\) passengers is normally distributed with mean \(\mu\) and standard error \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{37}{\sqrt{20}}, \] we compute the standard error: \[ SE = \frac{37}{\sqrt{20}} \approx \frac{37}{4.4721} \approx 8.277. \] Next, we standardize the value \(187.5\) lb to obtain a \(z\)-score using \[ z = \frac{\bar{x} - \mu}{SE}. \] Substitute the values: \[ z = \frac{187.5 - 195}{8.277} \approx \frac{-7.5}{8.277} \approx -0.906. \] The probability that the mean weight exceeds \(187.5\) lb is \[ P(\bar{X} > 187.5) = P\left(Z > -0.906\right). \] Recall that for a standard normal variable \(Z\), \[ P(Z > -a) = \Phi(a), \] where \(\Phi\) denotes the cumulative distribution function for a standard normal variable. Therefore, \[ P(Z > -0.906) = \Phi(0.906). \] Looking up (or computing) the value of \(\Phi(0.906)\) we find \[ \Phi(0.906) \approx 0.8176. \] Thus, the probability that the mean weight exceeds \(187.5\) lb is approximately \[ 0.8176. \]