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.)

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### 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).**

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### 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  
  $$
  0.6700.
  $$
**Part (a):** The probability that the boat is overloaded is approximately 1.0000. **Part (b):** The probability that the boat is overloaded is approximately 0.6700.

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