Question

Water Taxi Safety Passengers died when a water taxi sank in Baltimore’s Inner Harbor. Men are typically heavier than women and children, so when loading a water taxi, assume a worst-case scenario in which all passengers are men. Assume that weights of men are normally

distributed with a mean of 189 lb and a standard deviation of 39 lb (based on Data Set 1 “Body Data” in Appendix B). The water taxi that sank had a stated capacity of 25 passengers, and the boat was rated for a load limit of 3500 lb.

a. Given that the water taxi that sank was rated for a load limit of 3500 lb, what is the maximum mean weight of the passengers if the boat is filled to the stated capacity of 25 passengers?

b. If the water taxi is filled with 25 randomly selected men, what is the probability that their mean weight exceeds the value from part (a)?

c. After the water taxi sank, the weight assumptions were revised so that the new capacity became 20 passengers. If the water taxi is filled with 20 randomly selected men, what is the probability that their mean weight exceeds 175 lb, which is the maximum mean weight that does not cause the total load to exceed 3500 lb?

d. Is the new capacity of 20 passengers safe?

Short answer

a. The maximum value of the mean weight of passengers that can be allowed if the boat is filled to its capacity is 140 lb.

b. The probability that the mean weight of the men exceeds 140 lb when the capacity of 25 passengers is considered is 0.9999.

c. The probability that the mean weight of the men exceeds 175lbwhen the capacity of 20 passengers is considered is 0.9463.

d. As the probability of overloading with the current number of passengers is very high, it is not safe to travel with the new capacity of 20 passengers.

Step-by-step solution

Given information

The weights of the men are normally distributed with a mean of 189 lb and a standard deviation of 39 lb. The load limit of the water taxi is given to be equal to 3500 lb. The capacity of the water taxi is equal to 25 passengers.

Maximum possible mean weight

a.

The maximum possible mean weight of the passengers is computed as shown below.

$$\begin{gathered} \text{Maximum}\text{mean}\text{weight}=\frac{\text{Maximum}\text{load}\text{limit}}{\text{Number}\text{of}\text{passengers}} \\ =\frac{3500}{25} \\ =140 \end{gathered}$$

Thus, the maximum value of the mean weight of passengers that can be allowed if the boat is filled to its capacity is 140lb.

Sampling distribution of the sample mean

b.

Let $\overline{x}$represent the sample mean weight of the men.

The maximum mean weight of the passengers is considered to be 140 lb.

The sample mean weight of the men follows the normal distribution with a mean of ${\mu }_{\overline{x}}=\mu$and standard deviation of ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

The sample size given is equal to n=25.

The probability that the mean weight of the men is greater than 140 lb is computed using the standard normal table, as shown below.

$$\begin{gathered} P\left(\overline{x}>140\right)=1-P\left(\overline{x}<140\right) \\ =1-P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{140-\mu }{\frac{\sigma }{\sqrt{n}}}\right) \\ =1-P\left(z<\frac{140-189}{\frac{39}{\sqrt{25}}}\right) \end{gathered}$$

$$\begin{gathered} =1-P\left(z<-6.28\right) \\ =1-0.0001 \\ =0.9999 \end{gathered}$$

Therefore, the probability that the mean weight of the men exceeds 140 lb is 0.9999.

c.

The updated maximum mean weight of the passengers is considered to be equal to 175 lb.

The sample mean weight of the men follows the normal distribution with a mean of ${\mu }_{\overline{x}}=\mu$and a standard deviation of ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

The updated sample size given is equal to n=20.

The probability that the mean weight of the men is greater than 175lb is computed using the standard normal table, as shown below.

$$\begin{gathered} P\left(\overline{x}>175\right)=P\left(\overline{x}>175\right) \\ =P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{175-\mu }{\frac{\sigma }{\sqrt{n}}}\right) \\ =1-P\left(z<\frac{175-189}{\frac{39}{\sqrt{20}}}\right) \\ =P\left(z>-1.61\right) \end{gathered}$$

$$\begin{gathered} =P\left(z<1.61\right) \\ =0.9463 \end{gathered}$$

Therefore, the probability that the mean weight of the men exceeds 170lb is 0.9463.

Examining the safety of the current capacity

d.

The probability of exceeding the maximum allowed mean weight of passengers for the capacity of 20 passengers is quite high.

Therefore, the new capacity of 20 passengers is not safe as there is a very high chance of overloading and thus, risking the lives of the passengers.