/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Phoenix is a hub for a large air... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 crucial—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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Roulette is a game of chance that involves spinning a wheel that is divided into 38 equal segments, as shown in the accompanying picture. A metal ball is tossed into the wheel as it is spinning, and the ball eventually lands in one of the 38 segments. Each segment has an associated color. Two segments are green. Half of the other 36 segments are red, and the others are black. When a balanced roulette wheel is spun, the ball is equally likely to land in any one of the 38 segments. a. When a balanced roulette wheel is spun, what is the probability that the ball lands in a red segment? b. In the roulette wheel shown, black and red segments alternate. Suppose instead that all red segments were grouped together and that all black segments were together. Does this increase the probability that the ball will land in a red segment? Explain. c. Suppose that you watch 1000 spins of a roulette wheel and note the color that results from each spin. What would be an indication that the wheel was not balanced?

In an article that appears on the website of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in \(100 .\) And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given - say 99 in 100 - are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty \(T C=\) event that the test result is clean \(D=\) event that the person tested is actually dirty \(C=\) event that the person tested is actually clean a. Using the information in the quote, what are the values of i. \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C)\) iv. \(P(D)\) b. Use the probabilities from Part (a) to construct a "hypothetical 1000 " table. c. What is the value of \(P(T D)\) ? d. Use the table to calculate the probability that a person is clean given that the test result is dirty, \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

A small college has 2,700 students enrolled. Consider the chance experiment of selecting a student at random. For each of the following pairs of events, indicate whether or not you think they are mutually exclusive and explain your reasoning. a. the event that the selected student is a senior and the event that the selected student is majoring in computer science. b. the event that the selected student is female and the event that the selected student is majoring in computer science. c. the event that the selected student's college residence is more than 10 miles from campus and the event that the selected student lives in a college dormitory. d. the event that the selected student is female and the event that the selected student is on the college football team.

Suppose you want to estimate the probability that a randomly selected customer at a particular grocery store will pay by credit card. Over the past 3 months, 80,500 payments were made, and 37,100 of them were by credit card. What is the estimated probability that a randomly selected customer will pay by credit card?

Suppose that \(20 \%\) of all teenage drivers in a certain county received a citation for a moving violation within the past year. Assume in addition that \(80 \%\) of those receiving such a citation attended traffic school so that the citation would not appear on their permanent driving record. Consider the chance experiment that consists of randomly selecting a teenage driver from this county. a. One of the percentages given in the problem specifies an unconditional probability, and the other percentage specifies a conditional probability. Which one is the conditional probability, and how can you tell? b. Suppose that two events \(E\) and \(F\) are defined as follows: \(E=\) selected driver attended traffic school \(F=\) selected driver received such a citation Use probability notation to translate the given information into two probability statements of the form \(P(\ldots)=\) probability value.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.