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USA Today reported that for all airlines, the number of lost bags was May: 6.02 per 1000 passengers 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
(a) Poisson distribution is suitable because it models rate-based independent events; \( \lambda = 6.0 \) for May. (b) In May, \( P(X = 0) \approx 0.0025 \), \( P(X \geq 3) \approx 0.938 \), \( P(X \geq 6) \approx 0.445 \). (c) In December, \( P(X = 0) \approx 2.26 \times 10^{-6} \), \( P(X \geq 6) \approx 1 \), \( P(X \geq 12) \approx 0.619 \).

Step by step solution

01

Understanding the Poisson Distribution

The Poisson distribution is used for modeling the number of times an event occurs within a fixed interval of time or space. It is suitable for situations where events occur independently at a constant average rate. In this problem, 'lost bags per 1000 passengers' is a perfect fit as a Poisson random variable because losing luggage is a random event happening independently at a certain rate (given as per 1000 passengers).
02

Calculate the Poisson Parameter \( \lambda \) for May

The parameter \( \lambda \) is the average number of occurrences (lost bags) per interval (1000 passengers). For May, this is given as 6.02 lost bags per 1000 passengers. Therefore, \( \lambda = 6.0 \) when rounded to the nearest tenth.
03

Probability of No Bags Lost in May

Use the Poisson probability formula for the number of occurrences \( k = 0 \): \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]Substituting \( \lambda = 6.0 \) and \( k = 0 \): \[ P(X = 0) = \frac{6.0^0 \times e^{-6.0}}{0!} = e^{-6.0} \approx 0.0025 \]
04

Probability of 3 or More Bags Lost in May

Calculate the probabilities for 0, 1, and 2 bags being lost and subtract from 1 for 3 or more:\[ P(X \geq 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \]Using the Poisson formula:\[ P(X = 1) = \frac{6.0^1 \times e^{-6.0}}{1!} \approx 0.0149 \]\[ P(X = 2) = \frac{6.0^2 \times e^{-6.0}}{2!} \approx 0.0448 \]Thus,\[ P(X \geq 3) = 1 - (0.0025 + 0.0149 + 0.0448) \approx 0.938 \]
05

Probability of 6 or More Bags Lost in May

Calculate probabilities for 0 to 5 bags being lost and subtract from 1 for 6 or more:\[ P(X \geq 6) = 1 - \sum_{k=0}^{5} P(X = k) \]This involves computing each probability from 0 through 5 and summing them, and then subtracting from 1. For brevity, the summation can be identified through tables or a Poisson calculator, but it results in:\[ P(X \geq 6) \approx 0.445 \]
06

Calculate Poisson Parameter \( \lambda \) for December

For December, \( \lambda \) is 12.78 when rounded to the nearest tenth, this becomes \( 13.0 \).
07

Probability of No Bags Lost in December

Substitute \( \lambda = 13.0 \) into the Poisson formula for \( k = 0 \):\[ P(X = 0) = \frac{13.0^0 \times e^{-13.0}}{0!} = e^{-13.0} \approx 2.26 \times 10^{-6} \]
08

Probability of 6 or More Bags Lost in December

Calculate probabilities for 0 to 5 bags being lost and subtract from 1:\[ P(X \geq 6) = 1 - \sum_{k=0}^{5} P(X = k) \approx 1 \]The cumulative probability of losing fewer than 6 bags is extremely low, so \( P(X \geq 6) \approx 1 \).
09

Probability of 12 or More Bags Lost in December

Calculate probabilities for 0 to 11 bags being lost and subtract from 1:\[ P(X \geq 12) = 1 - \sum_{k=0}^{11} P(X = k) \]Calculating directly gives:\[ P(X \geq 12) \approx 0.619 \]

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Key Concepts

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

Probability
Probability is a fundamental concept in statistics that measures the likelihood of an event occurring. In the context of the Poisson distribution, we use probability to determine the chances of a specific number of events happening within a defined interval, such as losing a certain number of bags out of 1000 passengers.

The probability of an event happening ranges from 0 to 1, where 0 means the event is impossible, and 1 means it is certain. The Poisson probability can be calculated using the formula: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]where
  • \( X \) is the random variable representing the number of events,
  • \( k \) is the exact number of events we are measuring,
  • \( \lambda \) is the average rate of occurrence,
  • \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
Using this formula, we can find probabilities for different scenarios, such as no lost bags or losing 3 or more bags.
Lambda parameter
In the Poisson distribution, the Lambda parameter \( \lambda \) plays a crucial role. It represents the average number of occurrences in a given interval. For example, in May, the average number of lost bags is 6.02 per 1000 passengers. Thus, \( \lambda = 6.0 \) when rounded to the nearest tenth.

The choice of \( \lambda \) helps model the distribution accurately. It is essential to round \( \lambda \) appropriately to maintain statistical precision and keep calculations straightforward.

In the Poisson formula, \( \lambda \) determines the shape of the distribution, influencing the probabilities of observations. Higher values of \( \lambda \) suggest a greater average number of events, shifting the distribution towards higher probability of more frequent events.

The calculation of \( \lambda \) is straightforward when the context provides sufficient data, as in the lost bags example.
Independent events
Understanding independent events is key when dealing with the Poisson distribution. Independent events are those where the occurrence of one does not affect the probability of the other happening.

In terms of losing bags, whether one passenger loses a bag does not impact the chance of another passenger losing theirs. This independence is crucial because it allows the Poisson distribution to model events accurately.

Each new occurrence is like a fresh start, unaffected by previous outcomes. This condition ensures that the distribution reflects the true chances in similar future scenarios like consecutive days, different months, or intervals.

The property of independence is fundamental because it simplifies the modeling process. It allows events to follow a consistent rate, as reflected in the \( \lambda \), without the added complexity of accounting for interconnected events.
Random variable
A random variable is a fundamental concept in probability and statistics, representing a variable whose possible values are numerical outcomes of a random event.

In the context of the Poisson distribution, the random variable \( X \) denotes the number of events (like lost bags) occurring over a fixed interval (per 1000 passengers).

Random variables can take different forms. In a Poisson distribution, they often measure discrete variables, such as the counts of specific events.

The focus is not just on single outcomes but also on understanding the range and likelihood of outcomes. This is where the Poisson distribution's probabilities help. It gives insights into the most probable number of occurrences, supporting decision-making and predictions.

Such random variables are central in fields like operations research, risk management, and quality control, where understanding event frequency is essential. They help in structuring real-world problems into mathematically manageable models, offering solutions and strategies for better outcomes.

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Most popular questions from this chapter

The Denver Post reported that, on average, a large shopping center has had an incident of shoplifting caught by security once every 3 hours. The shopping center is open from 10 A.M. to 9 P.M. (11 hours). Let \(r\) be the number of shoplifting incidents caught by security in the 11 -hour period during which the center is open. (a) Explain why the Poisson probability distribution would be a good choice for the random variable \(r\). What is \(\lambda ?\) (b) What is the probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security? (c) What is the probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security? (d) What is the probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security?

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Jim is a real estate agent who sells large commercial buildings. Because his commission is so large on a single sale, he does not need to sell many buildings to make a good living. History shows that Jim has a record of selling an average of eight large commercial buildings every 275 days. (a) Explain why a Poisson probability distribution would be a good choice for \(r=\) number of buildings sold in a given time interval. (b) In a 60 -day period, what is the probability that Jim will make no sales? one sale? two or more sales? (c) In a 90 -day period, what is the probability that Jim will make no sales? two sales? three or more sales?

For a binomial experiment, how many outcomes are possible for each trial? What are the possible outcomes?

This problem shows you how to make a better blend of almost anything. Let \(x_{1}, x_{2}, \ldots x_{n}\) be independent random variables with respective variances \(\sigma_{1}^{2}\) \(\sigma_{2}^{2}, \ldots, \sigma_{n}^{2}.\) Let \(c_{1}, c_{2}, \ldots, c_{n}\) be constant weights such that \(0 \leq c_{i} \leq 1\) and \(c_{1}+c_{2}+\ldots+\) \(c_{n}=1 .\) The linear combination \(w=c_{1} x_{1}+c_{2} x_{2}+\ldots+c_{n} x_{n}\) is a random variable with variance $$\sigma_{w}^{2}=c_{1}^{2} \sigma_{1}^{2}+c_{2}^{2} \sigma_{2}^{2}+\cdots+c_{n}^{2} \sigma_{n}^{2}$$ (a) In your own words write a brief explanation regarding the following statement: The variance of \(w\) is a measure of the consistency or variability of performance or outcomes of the random variable \(w\). To get a more consistent performance out of the blend \(w\), choose weights \(c_{i}\) that make \(\sigma_{w}^{2}\) as small as possible. Now the question is how do we choose weights \(c_{i}\) to make \(\sigma_{w}^{2}\) as small as possible? Glad you asked! A lot of mathematics can be used to show the following choice of weights will minimize \(\sigma_{w}^{2}.\) $$c_{i}=\frac{\frac{1}{\sigma_{i}^{2}}}{\left[\frac{1}{\sigma_{i}^{2}}+\frac{1}{\sigma_{2}^{\frac{1}{2}}}+\cdots+\frac{1}{\sigma_{n}^{2}}\right]} \quad \text { for } i=1,2, \cdots, n$$ (Reference: Introduction to Mathematical Statistics, 4th edition, by Paul Hoel.) (b)Two types of epoxy resin are used to make a new blend of superglue. Both resins have about the same mean breaking strength and act independently. The question is how to blend the resins (with the hardener) to get the most consistent breaking strength. Why is this important, and why would this require minimal \(\sigma_{w}^{2} ?\) Hint: We don't want some bonds to be really strong while others are very weak, resulting in inconsistent bonding. Let \(x_{1}\) and \(x_{2}\) be random variables representing breaking strength (lb) of each resin under uniform testing conditions. If \(\sigma_{1}=8\) lb \(\sigma_{2}=12\) lb, show why a blend of about \(69 \%\) resin 1 and \(31 \%\) resin 2 will result in a superglue with smallest \(\sigma_{w}^{2}\) and most consistent bond strength. (c) Use \(c_{1}=0.69\) and \(c_{2}=0.31\) to compute \(\sigma_{w}\) and show that \(\sigma_{w}\) is less than both \(\sigma_{1}\) and \(\sigma_{2}\) The dictionary meaning of the word synergetic is "working together or cooperating for a better overall effect." Write a brief explanation of how the blend \(w=c_{1} x_{1}+c_{2} x_{2}\) has a synergetic effect for the purpose of reducing variance.

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