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Phoenix is a hub for a large airline. Suppose that on a particular day, 8,000 passengers arrived in Phoenix on this airline. Phoenix was the final destination for 1,800 of these passengers. The others were all connecting to flights to other cities. On this particular day, several inbound flights were late, and 480 passengers missed their connecting flight. Of these 480 passengers, 75 were delayed overnight and had to spend the night in Phoenix. Consider the chance experiment of choosing a passenger at random from these 8,000 passengers. Calculate the following probabilities: a. the probability that the selected passenger had Phoenix as a final destination. b. the probability that the selected passenger did not have Phoenix as a final destination. c. the probability that the selected passenger was connecting and missed the connecting flight. d. the probability that the selected passenger was a connecting passenger and did not miss the connecting flight. e. the probability that the selected passenger either had Phoenix as a final destination or was delayed overnight in Phoenix. f. An independent customer satisfaction survey is planned. Fifty passengers selected at random from the 8,000 passengers who arrived in Phoenix on the day described above will be contacted for the survey. The airline knows that the survey results will not be favorable if too many people who were delayed overnight are included. Write a few sentences explaining whether or not you think the airline should be worried, using relevant probabilities to support your answer.

Short Answer

Expert verified
a. Probability that the selected passenger had Phoenix as a final destination is 0.225\n b. Probability that the selected passenger did not have Phoenix as a final destination is 0.775\n c. Probability that the selected passenger was connecting and missed the connection is 0.06\n d. Probability that the selected passenger was a connecting passenger and did not miss the connecting flight is 0.715\n e. Probability that the selected passenger either had Phoenix as a final destination or was delayed overnight is 0.234375\n The actual expected number of people who were delayed overnight and are included in the survey is not very high. Thus, the probability is relatively low and the airline should not worry.

Step by step solution

01

Calculate simple probabilities

To find the simple probabilities asked in the first two parts a and b, simply use the formula of probability, which is the ratio of the favourable outcomes to the total outcomes. a. The probability that the selected passenger had Phoenix as a final destination is calculated as \(P(\text{Phoenix as destination}) = \frac{1800}{8000}\).\n b. The probability that the selected passenger did not have Phoenix as a final destination is simply one minus the probability of having Phoenix as a destination \(P(\text{not Phoenix as destination}) = 1 - P(\text{Phoenix as destination}) = 1 - \frac{1800}{8000} = \frac{6200}{8000}\).
02

Calculate compound probabilities

In order to compute compound probabilities, we need to further understand the conditions: c. To compute the probability that the selected passenger was connecting and missed the connecting flight, divide the number of passengers who missed their flight by the total number of passengers \(P(\text{missed connection}) = \frac{480}{8000}\). d. The probability that the selected passenger was a connecting passenger and did not miss the connection is calculated by subtracting the number of passengers who did miss their connection from those who did not have Phoenix as their final destination, all over the total number of passengers \(P(\text{made connection}) = \frac{6200 - 480}{8000}\). e. To calculate the probability that the selected passenger either had Phoenix as a final destination or was delayed overnight, sum up the number of passengers who had Phoenix as a final destination and those who were delayed overnight (as these events are mutually exclusive), then divide by the total number of passengers \(P(\text{Phoenix or delayed overnight}) = \frac{1800 + 75}{8000}\).
03

Analysis based on probabilities

Finally, estimate the chances of the airline including passengers who had bad experiences (delayed overnight) in their customer satisfaction survey. The number of passengers to be surveyed is 50. Multiply this number by the probability that a passenger was delayed overnight to find the expected amount. If this number is significantly high, then the airline might need to worry.

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Key Concepts

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

Understanding Simple Probabilities
Simple probabilities are the foundation of probability theory, reflecting the chance of a single event occurring. These probabilities are calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

For instance, let's consider the scenario where you're picking one passenger from the 8,000 who arrived in Phoenix. To calculate the probability of selecting a passenger whose final destination is Phoenix, we only look at the number of passengers whose final destination is Phoenix (favorable outcomes), which is 1,800, and divide it by the total number of passengers (8,000). This gives us the probability, expressed mathematically as: \[\begin{equation}P(\text{Phoenix as destination}) = \frac{1800}{8000}.\end{equation}\]
Thinking in terms of simple probabilities gives us a straightforward approach to understanding basic outcomes and their likelihoods.
Compound Probabilities Explained
Compound probabilities refer to the likelihood of two or more events occurring together or in succession. These probabilities help in understanding complex scenarios where events are interconnected.

In our exercise, calculating the probability that a passenger missed their connecting flight involves understanding not just one event but the combination of being a connecting passenger and missing the flight. Here's the calculation for reference:\[\begin{equation}P(\text{missed connection}) = \frac{480}{8000}.\end{equation}\]
It’s important to also know the probability of the opposite situation. In this case, finding the likelihood that a passenger made their connection when not Phoenix-bound:\[\begin{equation}P(\text{made connection}) = \frac{6200 - 480}{8000}.\end{equation}\]
Compound probabilities enable us to better assess combined events, adding depth to our statistical analysis.
Statistical Analysis in Real-world Situations
Statistical analysis allows us to interpret data and make predictions about larger populations based on sample information. In the context of our exercise, statistical analysis can help the airline understand how the inclusion of certain passengers in a satisfaction survey might affect the overall results.

The expected number of passengers who were delayed overnight and might be surveyed is calculated using the probability of being delayed overnight and the sample size of the survey. If this expected number is significant compared to the total sample of 50 passengers, the airline may need to be concerned about potentially unfavorable survey results. This statistical approach guides decision-making and risk assessment, ensuring that surveyed groups are representative of the wider population.
The Significance of a Customer Satisfaction Survey
A customer satisfaction survey is a tool used by businesses to gather feedback on their services and identify areas for improvement. In our scenario, the airline's selection of survey participants is crucialβ€”you wouldn’t want a result skewed by an overrepresentation of those who had a negative experience due to delays.

To minimize this risk, an understanding of the probabilities calculated earlier is essential. By considering the probability of selecting someone who was delayed overnight, and by multiplying that by the number of survey participants, the airline can estimate if the sample might lead to biased survey results. Surveys must be designed to be as unbiased as possible to genuinely reflect customer satisfaction, and probability calculations play a key role in achieving that balance.

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Most popular questions from this chapter

A study of how people are using online services for medical consulting is described in the paper "Internet Based Consultation to Transfer Knowledge for Patients Requiring Specialized Care" (British Medical Journal [2003]: \(696-699)\). Patients using a particular online site could request one or both (or neither) of two services: specialist opinion and assessment of pathology results. The paper reported that \(98.7 \%\) of those using the service requested a specialist opinion, \(35.4 \%\) requested the assessment of pathology results, and \(34.7 \%\) requested both a specialist opinion and assessment of pathology results. Consider the following two events: \(S=\) event that a specialist opinion is requested \(A=\) event that an assessment of pathology results is requested a. What are the values of \(P(S), P(A)\), and \(P(S \cap A)\) ? b. Use the given probability information to set up a "hypothetical 1000 " table with columns corresponding to \(S\) and not \(S\) and rows corresponding to \(A\) and \(\operatorname{not} A .\) c. Use the table to find the following probabilities: i. the probability that a request is for neither a specialist opinion nor assessment of pathology results. ii. the probability that a request is for a specialist opinion or an assessment of pathology results.

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The student council for a school of science and math has one representative from each of five academic departments: Biology (B), Chemistry (C), Mathematics (M), Physics (P), and Statistics (S). Two of these students are to be randomly selected for inclusion on a university-wide student committee. a. What are the 10 possible outcomes? b. From the description of the selection process, all outcomes are equally likely. What is the probability of each outcome? c. What is the probability that one of the committee members is the statistics department representative? d. What is the probability that both committee members come from laboratory science departments?

In a small city, approximately \(15 \%\) of those eligible are called for jury duty in any one calendar year. People are selected for jury duty at random from those eligible, and the same individual cannot be called more than once in the same year. What is the probability that an eligible person in this city is selected in both of the next 2 years? All of the next 3 years?

Each time a class meets, the professor selects one student at random to explain the solution to a homework problem. There are 40 students in the class, and no one ever misses class. Luke is one of these students. What is the probability that Luke is selected both of the next two times that the class meets? (Hint: See Example 5.8 )

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