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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

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, \(60 \%\) can be repaired, whereas the other \(40 \%\) must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

The College Board reports that \(2 \%\) of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test. a. What is the probability that exactly 1 received a special accommodation? b. What is the probability that at least 1 received a special accommodation? c. What is the probability that at least 2 received a special accommodation? d. What is the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated? e. Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed \(4.5\) hours. What would you expect the average time allowed the 25 selected students to be?

An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particular city for 1 year. The fire department is concerned that many houses remain without detectors. Let \(p=\) the true proportion of such houses having detectors, and suppose that a random sample of 25 homes is inspected. If the sample strongly indicates that fewer than \(80 \%\) of all houses have a detector, the fire department will campaign for a mandatory inspection program. Because of the costliness of the program, the department prefers not to call for such inspections unless sample evidence strongly argues for their necessity. Let \(X\) denote the number of homes with detectors among the 25 sampled. Consider rejecting the claim that \(p \geq .8\) if \(x \leq 15\) a. What is the probability that the claim is rejected when the actual value of \(p\) is 8 ? b. What is the probability of not rejecting the claim when \(p=.7 ?\) When \(p=.6 ?\) c. How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14 ?

An electronics store has received a shipment of 20 table radios that have connections for an iPod or iPhone. Twelve of these have two slots (so they can accommodate both devices), and the other eight have a single slot. Suppose that six of the 20 radios are randomly selected to be stored under a shelf where the radios are displayed, and the remaining ones are placed in a storeroom. Let \(X=\) the number among the radios stored under the display shelf that have two slots. a. What kind of a distribution does \(X\) have (name and values of all parameters)? b. Compute \(P(X=2), P(X \leq 2)\), and \(P(X \geq 2)\). c. Calculate the mean value and standard deviation of \(X\).

Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, \(25 \%\) of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let \(X\) denote the number among the four who have earthquake insurance. a. Find the probability distribution of \(X\). [Hint: Let \(S\) denote a homeowner who has insurance and \(F\) one who does not. Then one possible outcome is SFSS, with probability \((.25)(.75)(.25)(.25)\) and associated \(X\) value 3 . There are 15 other outcomes.] b. Draw the corresponding probability histogram. c. What is the most likely value for \(X\) ? d. What is the probability that at least two of the four selected have earthquake insurance?
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