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 sank, th 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 lb . Fir boat is overloaded because the 70 passengers have a mean weight greater than 142 lb . The probability is 10000 . (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 passen (so that their total weight is greater than the maximum capacity of \( 2,610 \mathrm{lb} \) ) The probability is 1 I (Round to four decimal places as needed.)

Below is a step‐by‐step solution in Markdown format with the necessary LaTeX notation. --- ### Part (a) A similar boat is loaded with 70 passengers. We assume that each passenger’s weight is normally distributed with \[ \mu=178.3\text{ lb} \quad \text{and} \quad \sigma=37.6\text{ lb}. \] The boat was originally rated for a mean weight of \(142\) lb per passenger (which corresponds to a total of \[ 70 \times 142 = 9940\text{ lb}). \] Thus, the boat is considered overloaded if the sample mean weight, \(\bar{X}\), of the 70 passengers is greater than \(142\) lb. Since the passenger weights are independent and normally distributed, the sampling distribution of the mean is also normal with mean \[ \mu_{\bar{X}}=178.3\text{ lb} \] and standard error \[ \sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{37.6}{\sqrt{70}}. \] 1. **Compute the standard error:** \[ \sqrt{70}\approx 8.3666,\quad \text{thus} \quad \sigma_{\bar{X}}\approx\frac{37.6}{8.3666}\approx4.496\text{ lb}. \] 2. **Standardize the threshold of \(142\) lb:** We need the \(z\)-score corresponding to \(\bar{X}=142\) lb: \[ z=\frac{142-178.3}{4.496}\approx\frac{-36.3}{4.496}\approx-8.08. \] 3. **Find the probability:** The probability that the boat is overloaded is \[ P(\bar{X}>142)=P(Z>-8.08). \] Since \(z=-8.08\) is far in the left tail of the normal distribution, we have \[ P(Z>-8.08)\approx1.0000. \] Thus, **the probability that the boat is overloaded in part (a) is approximately 1.0000 (to four decimal places).** --- ### Part (b) Now the boat is rated to carry only 15 passengers, and the load limit is changed to \[ 2610\text{ lb}. \] Again the individual weights are assumed to be normally distributed with \[ \mu=178.3\text{ lb} \quad \text{and} \quad \sigma=37.6\text{ lb}. \] The total weight of 15 passengers is given by \[ T=\sum_{i=1}^{15} X_i, \] which follows a normal distribution with - **Mean:** \[ \mu_T=15\times178.3=2674.5\text{ lb}. \] - **Standard deviation:** \[ \sigma_T=\sqrt{15}\times37.6. \] 1. **Calculate \(\sigma_T\):** \[ \sqrt{15}\approx3.873,\quad \text{so} \quad \sigma_T\approx3.873\times37.6\approx145.73\text{ lb}. \] 2. **Determine the \(z\)-score for the total weight limit:** We want the probability of an overload, i.e., when the total weight exceeds \(2610\) lb. Compute the \(z\)-score: \[ z=\frac{2610-2674.5}{145.73}\approx\frac{-64.5}{145.73}\approx-0.4428. \] 3. **Find the probability:** The probability that the boat is overloaded is \[ P(T>2610)=P\left(Z>-0.4428\right). \] Recall that for a normally distributed variable: \[ P(Z>-0.4428)=1-P(Z\le -0.4428)=1-\left[1-P(Z\le0.4428)\right]=P(Z\le0.4428). \] Looking up \(P(Z\le0.4428)\) (or using a calculator), we obtain approximately \[ P(Z\le0.4428)\approx0.6700. \] Thus, **the probability that the boat is overloaded in part (b) is approximately 0.6700 (to four decimal places).** --- ### Final Answers - **(a)** The probability that the boat is overloaded (i.e., the 70 passengers have a mean weight greater than \(142\) lb) is approximately \[ 1.0000. \] - **(b)** The probability that the boat is overloaded (i.e., the total weight of 15 passengers is greater than \(2610\) lb) is approximately \[ 0.6700. \]