A water taxi carries passengers from harbor to another. Assume that weights of passengers are normally distributed with a mean of $186 \mathrm{lb}$ and a standard deviation of $44 \mathrm{lb}$. The water taxi has a stated capacity of 25 passengers, and the water taxi was rated for a load limit of $3500 \mathrm{lb}$. Complete parts (a) through (d) below.
The maximum mean weight is $140 \mathrm{lb}$
(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 0.9999
(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 $175 \mathrm{lb}$, which is the maximum mean weight that does not cause the total load to exceed $3500 \mathrm{lb}$ ?
The probability is
(Round to four decimal places as needed.)

Step 1 ：Given that the water taxi has a load limit of 3500 lb and can carry 25 passengers, we can calculate the maximum mean weight that the taxi can carry without exceeding its load limit by dividing the total load limit by the number of passengers. This gives us \(\frac{3500}{25} = 140\) lb.

Step 2 ：Next, we want to find the probability that the mean weight of 25 randomly selected passengers exceeds this maximum mean weight. We can use the z-score formula, which is \(z = \frac{x - \mu}{\sigma}\), where x is the value we're interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. In this case, x is the maximum mean weight of 140 lb, \(\mu\) is the given mean weight of 186 lb, and \(\sigma\) is the given standard deviation of 44 lb. However, since we're dealing with a sample of 25 passengers, the standard deviation of the sample mean is \(\sigma/\sqrt{n}\), where n is the sample size. So, \(\sigma = \frac{44}{\sqrt{25}} = 8.8\) lb. Substituting these values into the z-score formula, we get \(z = \frac{140 - 186}{8.8} = -5.23\). The probability that the z-score is less than -5.23 is approximately 0.9999.

Step 3 ：For the third question, we revise the weight assumptions so that the new capacity is 20 passengers. We want to find the probability that the mean weight of 20 randomly selected passengers exceeds 175 lb, which is the new maximum mean weight that does not cause the total load to exceed 3500 lb. We use the same process as in the second question, but with the new capacity and maximum mean weight. The standard deviation of the sample mean is now \(\sigma = \frac{44}{\sqrt{20}} = 9.84\) lb. Substituting these values into the z-score formula, we get \(z = \frac{175 - 186}{9.84} = -1.12\). The probability that the z-score is less than -1.12 is approximately 0.8682.

Step 4 ：Final Answer: The maximum mean weight is \(\boxed{140}\) lb. The probability that the mean weight of 25 randomly selected passengers exceeds this value is approximately \(\boxed{0.9999}\). If the weight assumptions were revised so that the new capacity became 20 passengers, the probability that their mean weight exceeds 175 lb is approximately \(\boxed{0.8682}\).