A boat capsized and sank in a lake. Based on an assumption of a mean weight of 142 lb , the boat was rated to carry 70 passengers (so the load limit was \( 9,940 \mathrm{lb} \) ). After the boat s for similar boats was changed from 142 lb to 174 lb Complete parts a and b below. a. Assume that a similar boat is loaded with 70 passengers, and assume that the weights of people are normally distributed with a mean of 178.3 lb and a standard deviation of 37.6 boat is overloaded because the 70 passengers have a mean weight greater than 142 lb . The probability is 1.0000 . (Round to four decimal places as needed.) b. The boat was later rated to carry only 15 passengers, and the load limit was changed to \( 2,610 \mathrm{lb} \). Find the probability that the boat is overloaded because the mean weight of the p (so that their total weight is greater than the maximum capacity of \( 2,610 \mathrm{lb} \) ) The probability is (Round to four decimal places as needed )

Given: - The mean weight of a person is 178.3 lb. - The standard deviation of a person's weight is 37.6 lb. - The boat is overloaded if the mean weight of the passengers is greater than 142 lb. a. To find the probability that the boat is overloaded with 70 passengers: - The mean weight of 70 passengers is \( 70 \times 178.3 = 12471 \) lb. - The standard deviation of the total weight of 70 passengers is \( 70 \times 37.6 = 2632 \) lb. - The z-score for the mean weight of 70 passengers is \( \frac{12471 - 142}{2632} \). - The probability that the boat is overloaded is the probability that the z-score is greater than 0. b. To find the probability that the boat is overloaded with 15 passengers: - The mean weight of 15 passengers is \( 15 \times 178.3 = 2680.5 \) lb. - The standard deviation of the total weight of 15 passengers is \( 15 \times 37.6 = 564 \) lb. - The z-score for the mean weight of 15 passengers is \( \frac{2680.5 - 142}{564} \). - The probability that the boat is overloaded is the probability that the z-score is greater than 0. Let's calculate the probabilities. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(70\times 178.3-142\right)}{70}\times 37.6\) - step1: Remove the parentheses: \(\frac{70\times 178.3-142}{70}\times 37.6\) - step2: Multiply the numbers: \(\frac{12481-142}{70}\times 37.6\) - step3: Subtract the numbers: \(\frac{12339}{70}\times 37.6\) - step4: Convert the expressions: \(\frac{12339}{70}\times \frac{188}{5}\) - step5: Multiply the fractions: \(\frac{12339\times 188}{70\times 5}\) - step6: Multiply the terms: \(\frac{2319732}{70\times 5}\) - step7: Multiply the terms: \(\frac{2319732}{350}\) - step8: Reduce the fraction: \(\frac{1159866}{175}\) Calculate or simplify the expression \( (15*178.3 - 142)/15*37.6 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(15\times 178.3-142\right)}{15}\times 37.6\) - step1: Remove the parentheses: \(\frac{15\times 178.3-142}{15}\times 37.6\) - step2: Multiply the numbers: \(\frac{2674.5-142}{15}\times 37.6\) - step3: Subtract the numbers: \(\frac{2532.5}{15}\times 37.6\) - step4: Divide the terms: \(\frac{1013}{6}\times 37.6\) - step5: Convert the expressions: \(\frac{1013}{6}\times \frac{188}{5}\) - step6: Multiply the fractions: \(\frac{1013\times 188}{6\times 5}\) - step7: Multiply the terms: \(\frac{190444}{6\times 5}\) - step8: Multiply the terms: \(\frac{190444}{30}\) - step9: Reduce the fraction: \(\frac{95222}{15}\) a. The probability that the boat is overloaded with 70 passengers is approximately 0.0000. b. The probability that the boat is overloaded with 15 passengers is approximately 0.0000.