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Lost Bags USA Today reported that for all airlines, the number of lost bags was May: \(6.02\) per 1000 passengers \(\quad\) December: \(12.78\) per 1000 passengers Note: A passenger could lose more than one bag. (a) Let \(r=\) number of bags lost per 1000 passengers in May. Explain why the Poisson distribution would be a good choice for the random variable \(r\). What is \(\lambda\) to the nearest tenth? (b) In the month of May, what is the probability that out of 1000 passengers, no bags are lost? that 3 or more bags are lost? that 6 or more bags are lost? (c) In the month of December, what is the probability that out of 1000 passengers, no bags are lost? that 6 or more bags are lost? that 12 or more bags are lost? (Round \(\lambda\) to the nearest whole number.)

Short Answer

Expert verified
May: (a) Poisson distribution fits, \(\lambda = 6.0\); (b) No bags: \(0.0025\), \(\geq 3\): \(0.938\), \(\geq 6\): \(0.447\). December: \(\lambda = 13\); (c) No bags: \(2.26 \times 10^{-6}\), \(\geq 6\): \(0.998\), \(\geq 12\): \(0.548\).

Step by step solution

01

Identify the Distribution

The number of lost bags is a rare event occurring over a fixed number of 1000 passengers. Since each passenger can lose zero, one, or more bags independently, without affecting others, this matches the properties of the Poisson distribution.
02

Calculate Lambda for May

For the month of May, the average rate of lost bags (or the average number of events per interval) is given as 6.02 per 1000 passengers. Thus, for May, \[ \lambda = 6.02 \] rounded to the nearest tenth, it remains 6.0.
03

Calculate Probability of No Baggage Loss in May

Using the Poisson probability formula: \[ P(X = k) = \frac{{e^{-\lambda} \lambda^k}}{{k!}} \]For no bags lost, \(k = 0\):\[ P(X = 0) = \frac{{e^{-6.0} (6.0)^0}}{{0!}} = e^{-6.0} \approx 0.0025 \]
04

Calculate Probability of 3 or More Bags Lost in May

First, calculate the probability of 0, 1, and 2 lost bags using the Poisson formula. Then:\[ P(X \geq 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \]Calculate these probabilities using \( \lambda = 6.0 \): \[ P(X=1) = \frac{{e^{-6.0} (6.0)^1}}{{1!}} \approx 0.0149 \]\[ P(X=2) = \frac{{e^{-6.0} (6.0)^2}}{{2!}} \approx 0.0448 \]Adding them, \[ P(X \geq 3) = 1 - (0.0025 + 0.0149 + 0.0448) = 0.938 \]
05

Calculate Probability of 6 or More Bags Lost in May

Calculate probability for fewer than 6 lost bags:\[P(X \geq 6) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5))\]Each term can be calculated using the Poisson formula. The combined result is approximately: \[ P(X \geq 6) \approx 0.447 \]
06

Calculate Lambda for December

For the month of December, the average rate of lost bags is given as 12.78 per 1000 passengers. Rounded to the nearest whole number, \( \lambda \approx 13 \).
07

Probability of No Bags Lost in December

Using the Poisson formula for zero events with \( \lambda = 13 \): \[ P(X = 0) = \frac{{e^{-13} (13)^0}}{{0!}} = e^{-13} \approx 2.26 \times 10^{-6} \]
08

Calculate Probability of 6 or More Bags Lost in December

First calculate for fewer than 6 lost bags:\[P(X \geq 6) = 1 - (P(X = 0) + P(X = 1) + ... + P(X = 5))\]By calculating each using \( \lambda = 13 \), the result is: \[ P(X \geq 6) \approx 0.998 \]
09

Calculate Probability of 12 or More Bags Lost in December

For 12 or more bags lost, we find:\[P(X \geq 12) = 1 - (P(X = 0) + P(X = 1) + ... + P(X = 11))\]By calculating each, the combined probability is: \[ P(X \geq 12) \approx 0.548 \]

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Key Concepts

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

Probability Theory
Probability theory is a fundamental branch of mathematics that deals with the analysis of random events. It provides the framework to calculate the likelihood that a given event will occur. In probability theory, different rules and structures help us predict outcomes in various scenarios, including those where events occur randomly, like the number of lost luggage a passenger might experience during a flight.

In our exercise about lost bags, we used probability theory to determine how often such events might happen per 1000 passengers. It helped us compute the chance of losing a certain number of bags, or no bags at all, in a given month. By applying probability theory, we analyzed this statistical information using mathematical functions like the Poisson distribution, which is particularly useful in modeling the frequency of rare, independently occurring events over a fixed period of time or space.
Random Variable
A random variable is a quantity with a numerical outcome, determined by the result of a random phenomenon. In our context of lost airline bags per 1000 passengers, the random variable would represent the number of bags lost. It can take on various integer values based on real-world events.

For example, if we denote the random variable by \( X \), it could be that \( X = 0 \) when no bags are lost, \( X = 1 \) when one bag is lost, and so forth. Each of these is a random outcome, influenced by chance. The Poisson distribution is a good fit for this scenario because the occurrence of lost bags is rare yet possible multiple times over. Thus, defining these events as random variables helps us calculate their probabilities in a structured way.
Lambda Parameter
In probability theory, particularly in the Poisson distribution, the \( \lambda \) parameter is crucial. It indicates the average rate of occurrence of the event of interest over a defined time or space interval, such as our 1000 passengers. This average, noted as \( \lambda \), allows us to predict how often these events, like losing bags, happen.

For instance, in May, \( \lambda = 6.02 \) bags per 1000 passengers. This translates to an expectation that roughly this many bags will be lost given this number of passengers. Rounding \( \lambda \) helps simplify calculations while preserving the essence of the statistical problem.
  • May: \( \lambda = 6.0 \)
  • December: \( \lambda \approx 13 \)
By understanding \( \lambda \), we can plug it into probability equations, such as the Poisson formula, to compute the probability of 0 bags lost, 6 or more bags lost, etc.
Statistical Problem-Solving
Statistical problem-solving involves using statistics to analyze and interpret data and then making informed decisions based on that analysis. It encompasses defining the problem, selecting the appropriate statistical method, and carrying out the computations.

In the case of lost airline bags, statistical problem-solving included the following steps.
  • Identifying the problem: determining how likely it is for passengers to lose their bags during certain months.
  • Choosing a model: using the Poisson distribution fitted well because it deals with counting occurrences of an event over a fixed interval.
  • Calculating probabilities with these parameters: applying \( \lambda \) and the distribution formula to derive the likelihood of different outcomes.
  • Interpreting results: understanding these probabilities helps airlines prepare and mitigate such issues effectively, offering better services and customer satisfaction.
By optimizing these steps, the process becomes efficient, insightful, and based on solid mathematical principles.

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Most popular questions from this chapter

Harper's Index reported that the number of (Orange County, California) convicted drunk drivers whose sentence included a tour of the morgue was 569 , of which only 1 became a repeat offender. (a) Suppose that of 1000 newly convicted drunk drivers, all were required to take a tour of the morgue. Let us assume that the probability of a repeat offender is still \(p=1 / 569 .\) Explain why the Poisson approximation to the binomial would be a good choice for \(r=\) number of repeat offenders out of 1000 convicted drunk drivers who toured the morgue. What is \(\lambda\) to the nearest tenth? (b) What is the probability that \(r=0\) ? (c) What is the probability that \(r>1\) ? (d) What is the probability that \(r>2\) ? (e) What is the probability that \(r>3\) ?

For a binomial experiment, what probability distribution is used to find the probability that the first success will occur on a specified trial?

When using the Poisson distribution, which parameter of the distribution is used in probability computations? What is the symbol used for this parameter?

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Aldrich Ames is a convicted traitor who leaked American secrets to a foreign power. Yet Ames took routine lie detector tests and each time passed them. How can this be done? Recognizing control questions, employing unusual breathing patterns, biting one's tongue at the right time, pressing one's toes hard to the floor, and counting backwards by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination (Source: Lies! Lies!! Lies!!! The Psychology of Deceit, by C. V. Ford, professor of psychiatry, University of Alabama). In fact, it is reported in Professor Ford's book that after only 20 minutes of instruction by "Buzz" Fay (a prison inmate), \(85 \%\) of those trained were able to pass the polygraph examination even when guilty of a crime. Suppose that a random sample of nine students (in a psychology laboratory) are told a "secret" and then given instructions on how to pass the polygraph examination without revealing their knowledge of the secret. What is the probability that (a) all the students are able to pass the polygraph examination? (b) more than half the students are able to pass the polygraph examination? (c) no more than four of the students are able to pass the polygraph examination? (d) all the students fail the polygraph examination?

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