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Chapter 3: Problem 86

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of five per hour. a. What is the probability that exactly four arrivals occur during a particular hour? b. What is the probability that at least four people arrive during a particular hour? c. How many people do you expect to arrive during a 45 min period?

Short Answer

Expert verified
(a) 0.1755, (b) 0.4405, (c) 3.75 people expected.

Step by step solution

01

Understanding the Poisson Distribution

A Poisson process models the number of arrivals occurring within a fixed interval of time. The probability of a given number of arrivals is determined by the Poisson distribution, which is defined by the formula: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]where \( \lambda \) is the average rate of arrivals per interval, \( e \) is the base of the natural logarithm, and \( k \) is the number of occurrences.
02

Computing Probability for Exactly Four Arrivals

For part (a), use the Poisson formula with \( \lambda = 5 \) and \( k = 4 \). Substitute into the formula:\[ P(X = 4) = \frac{e^{-5} \times 5^4}{4!} \]Calculate each part:- \( e^{-5} \approx 0.0067 \)- \( 5^4 = 625 \)- \( 4! = 24 \)Now compute the probability:\[ P(X = 4) = \frac{0.0067 \times 625}{24} \approx 0.1755 \]
03

Determining Probability for At Least Four Arrivals

For part (b), we need the probability of at least four arrivals, which is \( P(X \geq 4) \). This can be calculated as:\[ P(X \geq 4) = 1 - P(X < 4) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)) \]Compute each term using the Poisson formula:- \( P(X = 0) = \frac{e^{-5} \times 5^0}{0!} \approx 0.0067 \)- \( P(X = 1) = \frac{e^{-5} \times 5^1}{1!} \approx 0.0337 \)- \( P(X = 2) = \frac{e^{-5} \times 5^2}{2!} \approx 0.0842 \)- \( P(X = 3) = \frac{e^{-5} \times 5^3}{3!} \approx 0.1404 \)Now sum them and subtract from 1:\[ P(X \geq 4) = 1 - (0.0067 + 0.0337 + 0.0842 + 0.1404) \approx 0.4405 \]
04

Calculating Expected Arrivals in 45 Minutes

For part (c), find the expected number of arrivals in 45 minutes. Since 45 minutes is 0.75 of an hour, multiply the hourly rate by this fraction:\[ E = 5 \times 0.75 = 3.75 \]Thus, you expect approximately 3.75 people to arrive in 45 minutes.

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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 and mathematics. It measures the likelihood of an event occurring. In the context of the Poisson distribution, probability helps us understand how likely it is for a certain number of events to happen within a given time frame.
The Poisson distribution formula, \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \), allows us to calculate the probability of \( k \) events occurring in a fixed interval. Here, \( \lambda \) is the average number of events expected to happen, and \( e \) is a constant (approximately 2.71828).
  • To find the probability of exactly four arrivals (from step 2), you substitute \( k = 4 \) and \( \lambda = 5 \) into the formula.
  • The calculations involve the exponential function and factorial, which are core components in computing probabilities in a Poisson process.
This approach allows for determining the exact likelihood of different outcomes, which is crucial in decision-making and predictions.
Expected Value
The expected value is a key concept used to determine what one can anticipate on average in a probabilistic scenario. In a Poisson distribution, the expected value is simply the mean of the distribution, which is represented by \( \lambda \).
  • For the given scenario, the rate of arrivals is five per hour. Thus, \( \lambda = 5 \) per hour is both the mean and the expected value for an hour.
  • To find the expected number of arrivals in a shorter time, such as 45 minutes, you scale \( \lambda \) by that portion of the hour. Hence, \( E = 5 \times 0.75 = 3.75 \).
The expected value provides a simple yet powerful summary. It tells us that, on average, about 3.75 people will arrive in 45 minutes. This helps manage resources and efficiency in operations like those in emergency rooms.
Poisson Process
The Poisson process is a statistical framework used to model random events that occur independently and sporadically over a continuous interval, such as time. It's especially useful in various fields for modeling events like the arrival of calls at a call center or patients at an emergency room.
In the Poisson process, two main characteristics stand out:
  • The average rate, \( \lambda \), which indicates the expected number of occurrences within a specific period.
  • Independence of events, meaning the occurrence of one event doesn't influence the likelihood of another occurring near the same time.

This process is governed by the Poisson distribution formula, which balances event occurrences with probabilities, facilitating strategic planning and effective management of resources.
By applying this model, one can effectively predict probabilistic outcomes for systems experiencing random, independent events, helping anticipate scenarios and prepare accordingly.

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Most popular questions from this chapter

Suppose that only , \(10 \%\) of all computers of a certain type experience CPU failure during the warranty period. Consider a sample of 10,000 computers. a. What are the expected value and standard deviation of the number of computers in the sample that have the defect? b. What is the (approximate) probability that more than 10 sampled computers have the defect? c. What is the (approximate) probability that no sampled computers have the defect?

An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30 , decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects. a. What is the probability that exactly 10 of these are from the second section? b. What is the probability that at least 10 of these are from the second section? c. What is the probability that at least 10 of these are from the same section? d. What are the mean value and standard deviation of the number among these 15 that are from the second section? e. What are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?

If the sample space \(S\) is an infinite set, does this necessarily imply that any rv \(X\) defined from \(\&\) will have an infinite set of possible values? If yes, say why. If no, give an example.

A concrete beam may fail either by shear \((S)\) or flexure \((F)\). Suppose that three failed beams are randomly selected and the type of failure is determined for each one. Let \(X=\) the number of beams among the three selected that failed by shear. List each outcome in the sample space along with the associated value of \(X\).

A very large batch of components has arrived at a distributor. The batch can be characterized as acceptable only if the proportion of defective components is at most .10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most 2 . a. What is the probability that the batch will be accepted when the actual proportion of defectives is .01? .05?.10? 20 ?.25? b. Let \(p\) denote the actual proportion of defectives in the batch. A graph of \(P\) (batch is accepted) as a function of \(p\), with \(p\) on the horizontal axis and \(P\) (batch is accepted) on the vertical axis, is called the operating characteristic curve for the acceptance sampling plan. Use the results of part (a) to sketch this curve for \(0 \leq p \leq 1\). c. Repeat parts (a) and (b) with " 1 " replacing " 2 " in the acceptance sampling plan. d. Repeat parts (a) and (b) with " 15 " replacing " \(10 "\) in the acceptance sampling plan. e. Which of the three sampling plans, that of part (a), (c), or (d), appears most satisfactory, and why?
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