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 \( 9,940 \mathrm{lb} \) ). for similar boats was changed from 142 lb to 172 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 182.1 lb and a standard o 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 \( 2,580 \mathrm{lb} \). Find the probability that the boat is overloaded because the mea (so that their total weight is greater than the maximum capacity of \( 2,580 \mathrm{lb} \) ). The probability is (Round to four decimal praces as needed.)

Below is a step‐by‐step solution in markdown format. --- ## Problem Recap A boat was originally rated to carry 70 passengers based on each person weighing \(142\) lb. Its load limit was therefore \[ 70(142) = 9\,940\mbox{ lb}. \] Later, similar boats had their assumed mean weight per passenger raised from \(142\) lb to \(172\) lb. In what follows we assume that passenger weights are normally distributed with \[ \mbox{mean } \mu=182.1\mbox{ lb} \] and (for our computations) a standard deviation \(\sigma=29.1\) lb. (This value for \(\sigma\) is typical in such problems; if a different value is given then the following computations can be modified accordingly.) There are two parts: ### Part (a) A similar boat is loaded with 70 passengers. Notice that if every person is assumed to weigh at least \(142\) lb on average (the original rating assumed that!) then the total weight is compared with \[ 70(142)=9\,940\mbox{ lb.} \] Since the true mean weight \(182.1\) lb \((>142\mbox{ lb})\) is much larger than that threshold, the chance of the 70‐passenger load having an average weight less than \(142\) lb is essentially zero. Hence, the boat is virtually certain to be overloaded. Thus, the probability that the load is excessive is \[ \mbox{Probability} \approx 1.0000. \] ### Part (b) Now the boat is later rated to carry only 15 passengers, and the load limit has been changed to \(2\,580\) lb. In this case the overload occurs when the total weight of 15 passengers exceeds \(2\,580\) lb. Equivalently, in terms of the sample mean \(\overline{X}\) we need \[ 15\,\overline{X}>2\,580 \quad\Longleftrightarrow\quad \overline{X}>\frac{2\,580}{15}=172\mbox{ lb}. \] Because the individual weights \(X_i\) are i.i.d. \(N(\mu,\sigma^2)\) with \(\mu=182.1\) and \(\sigma=29.1\), the distribution of the sample mean is \[ \overline{X}\sim N\Bigl(182.1,\Bigl(\frac{29.1}{\sqrt{15}}\Bigr)^2\Bigr). \] The standard error is \[ SE = \frac{29.1}{\sqrt{15}}. \] Calculating the denominator: \[ \sqrt{15}\approx3.873, \quad \mbox{so } SE\approx\frac{29.1}{3.873}\approx7.515. \] We now standardize: \[ Z=\frac{\overline{X}-\mu}{SE}. \] We are interested in \[ P(\overline{X}>172)=P\Bigl(Z>\frac{172-182.1}{7.515}\Bigr). \] Compute the \(z\)–value: \[ 172-182.1=-10.1,\quad z=\frac{-10.1}{7.515}\approx -1.344. \] Thus, \[ P(\overline{X}>172)=P(Z>-1.344). \] Using the symmetry of the normal distribution, \[ P(Z>-1.344)=P(Z<1.344). \] Consulting standard normal tables or a calculator, we find \[ P(Z<1.344)\approx 0.9104. \] So the probability that when 15 people are sampled their total weight (or equivalently, their mean weight) will exceed the load limit is approximately \[ \boxed{0.9104.} \] --- ## Final Answers - **(a)** The probability that the boat is overloaded with 70 passengers is approximately \( 1.0000 \). - **(b)** The probability that the boat is overloaded with 15 passengers is approximately \( 0.9104 \).