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When a water taxi sank in Baltimore's Inner Harbor, an investigation revealed that the safe passenger load for the water taxi was 3500 ib It was also noted that the mean weight of a passenger was assumed to be 140 ib. Assume a "worst- case" scenario in which all of the passengers are adult men. Assume that weights of men are normally distributed with a mean of 188.6 Ib and a standard deviation of 38.9 Ib (based on Data Set 1 "Body Data" in Appendix B). a. If one man is randomly selected, find the probability that he weighs less than 174 lb (the new value suggested by the National Transportation and Safety Board). b. With a load limit of 3500 ib, how many male passengers are allowed if we assume a mean weight of 140 lb? c. With a load limit of 3500 ib, how many male passengers are allowed if we assume the updated mean weight of 188.6 lb? d. Why is it necessary to periodically review and revise the number of passengers that are allowed to board?

Step by step solution

01

- Calculate the Probability for Part a

To find the probability that a randomly selected man weighs less than 174 lb, first convert the weight to a z-score using the formula: \[ z = \frac{x - \mu}{\sigma} \]Where: \( x = 174 \text{ lb} \), the weight in question\( \mu = 188.6 \text{ lb} \), the mean weight\( \sigma = 38.9 \text{ lb} \), the standard deviationSo, \[ z = \frac{174 - 188.6}{38.9} = \frac{-14.6}{38.9} \approx -0.375 \]

02

- Find the Respective Probability

Using the z-score table, find the probability corresponding to a z-score of -0.375. This gives approximately 0.3538. So, the probability that one man weighs less than 174 lb is approximately 0.3538.

03

- Maximum Passengers with Mean Weight of 140 lb (Part b)

To find the maximum number of male passengers allowed with a mean weight of 140 lb, use the load limit of 3500 lb:\[ \text{Number of Passengers} = \frac{\text{Load Limit}}{\text{Mean Weight}} = \frac{3500 \text{ lb}}{140 \text{ lb/man}} = 25 \text{ men} \]

04

- Maximum Passengers with Updated Mean Weight of 188.6 lb (Part c)

Using the updated mean weight of 188.6 lb, use the same load limit of 3500 lb:\[ \text{Number of Passengers} = \frac{\text{Load Limit}}{\text{Updated Mean Weight}} = \frac{3500 \text{ lb}}{188.6 \text{ lb/man}} \approx 18.56 \text{ men} \]Since we cannot have a fraction of a person, we round down to 18 men.

05

- Importance of Revising Passenger Limits (Part d)

It's necessary to periodically review and revise the number of passengers allowed to board because averages and distributions of passenger weights can change over time due to changes in population health, lifestyle, and other factors. Failing to adjust limits could compromise safety.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability

Probability helps us understand the likelihood of certain events occurring. In the context of our water taxi problem, we want to find the probability that a randomly selected man weighs less than 174 pounds.

To calculate this, we first convert the weight into a z-score. The formula used is:

\[ z = \frac{x - \mu}{\sigma} \]

Where:

• \(x = 174 \text{ lb}\) (the weight in question)
• \(\mu = 188.6 \text{ lb}\) (the mean weight)
• \(\sigma = 38.9 \text{ lb}\) (the standard deviation)

Using these values, we find that \(z \approx -0.375\).

Next, we use the z-score to find the corresponding probability from a z-score table. A z-score of -0.375 corresponds to a probability of about 0.3538. This means there’s a 35.38% chance that a randomly selected man weighs less than 174 pounds. Understanding these probabilities is crucial, especially in safety assessments.

Z-Score

The z-score is a measure that describes a value's position relative to the mean of a group of values. It tells us how many standard deviations an element is from the mean. For instance, to find out how uncommon a man's weight of 174 pounds is compared to the average, we use:
\[ z = \frac{x - \mu}{\sigma} \]

This formula normalizes the data, making it easier to compare different values within a dataset. For example, in our exercise:

• If \(x = 174\)
• \(\mu = 188.6 \)
• \(\sigma = 38.9 \)

We calculate:
\[ z = \frac{174 - 188.6}{38.9} = \frac{-14.6}{38.9} \approx -0.375 \]

This z-score of -0.375 indicates that a man weighing 174 pounds is 0.375 standard deviations below the mean weight.

Interpreting z-scores allows you to understand how unusual or common a specific value is in the context of the overall data distribution.

Mean and Standard Deviation

The mean is the average value in a dataset. It’s calculated by adding up all the numbers and then dividing by the number of values. For instance, the mean weight of all passengers assumed for safety calculations was 140 pounds initially and 188.6 pounds later. The formula to calculate the mean (\( \mu \)) is:
\[ \mu = \frac{1}{N}\sum_{i=1}^{N}x_{i} \]

Where \(N\) is the number of values and \(x_{i}\) are the values.

Standard deviation (\(\sigma\)) measures the dispersion or spread of the values around the mean. A low standard deviation means the values are close to the mean, while a high standard deviation indicates they are spread out. The formula for standard deviation is:

\[ \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(x_{i} - \mu)^{2}} \]

In our problem:

• Mean weight: 188.6 pounds
• Standard deviation: 38.9 pounds

These measures give us a snapshot of the data's central tendency and variability. This is crucial for safety measures on the water taxi.

Normal Distribution

Normal distribution, often called a bell curve, is a probability distribution that is symmetric about the mean. Most of the data falls near the mean, with fewer values appearing as you move away from the mean. It's characterized by its mean (\(\mu\)) and standard deviation (\(\sigma\)).

In our context, the assumption is that men's weights are normally distributed around a mean of 188.6 pounds with a standard deviation of 38.9 pounds. This allows us to make precise probability calculations.

When dealing with real-life problems—like determining safe passenger loads for a water taxi—this distribution helps predict the likelihood of certain weights and make informed decisions about limits and safety.

To visualize, if we plotted the men's weights, we would see that:

• The peak of the curve would be at the mean (188.6 pounds).
• The spread of the data would be shown by the width of the curve, determined by the standard deviation (38.9 pounds).

This understanding is critical for ensuring that the water taxi can safely handle its passenger load under varying conditions.