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The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of \(5 / \mathrm{h}\). 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. Approx. 0.1755 b. Approx. 0.7351 c. 3.75 arrivals expected

Step by step solution

01

Identify the Poisson parameter

The rate parameter (\( \lambda \)) for the Poisson process is given as 5 per hour. For this problem, \( \lambda = 5 \).
02

Calculate the probability for exactly four arrivals (Part a)

Use the Poisson probability formula \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \) where \( k = 4 \). Substitute \( \lambda = 5 \) and \( k = 4 \): \[ P(X=4) = \frac{e^{-5} \times 5^4}{4!} \].Calculate this value to find the probability of exactly four arrivals in one hour.
03

Execute the calculation for Part a

Calculate:\[ P(X=4) = \frac{e^{-5} \times 625}{24} \approx 0.1755 \].Thus, the probability of four arrivals is approximately 0.1755.
04

Calculate the probability of at least four arrivals (Part b)

To find \( P(X \geq 4) \), use the complement rule: \[ P(X \geq 4) = 1 - P(X < 4) \].This requires calculating \( P(X=0) \), \( P(X=1) \), \( P(X=2) \), and \( P(X=3) \) using the Poisson formula and summing these values.
05

Calculate individual probabilities for Part b

Using the Poisson formula:\[ P(X=0) = e^{-5} \times \frac{5^0}{0!} = e^{-5} \approx 0.0067 \]\[ P(X=1) = e^{-5} \times \frac{5^1}{1!} = 5e^{-5} \approx 0.0337 \]\[ P(X=2) = e^{-5} \times \frac{5^2}{2!} = \frac{25e^{-5}}{2} \approx 0.0842 \]\[ P(X=3) = e^{-5} \times \frac{5^3}{3!} = \frac{125e^{-5}}{6} \approx 0.1404 \].
06

Calculate the cumulative probability for P(X < 4)

Sum the individual probabilities: \[ P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \approx 0.0067 + 0.0337 + 0.0842 + 0.1404 = 0.2649 \].
07

Execute the calculation for Part b

Now calculate \( P(X \geq 4) \) using the complement: \[ P(X \geq 4) = 1 - 0.2649 = 0.7351 \].Thus, the probability that at least four people arrive is approximately 0.7351.
08

Determine expected arrivals for a 45-minute period (Part c)

The expected value for a Poisson distribution is \( \lambda \). Since 45 minutes is 0.75 hours, multiply the rate \( \lambda = 5 \) by 0.75: \[ \lambda = 5 \times 0.75 = 3.75 \].The expected number of arrivals in 45 minutes is 3.75.

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Key Concepts

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

Probability Calculation
Probability Calculation in the Poisson process is key to understanding how likely certain events are to occur within a specified period of time. The Poisson distribution helps calculate the probability of a given number of events, say arrivals at an emergency room, happening in a fixed time interval. The formula used is: \[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]where:\
  • \( P(X=k) \) represents the probability of \( k \) events occurring.
  • \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
  • \( \lambda \) is the average number of events (the rate parameter) in the given time.
  • \( k \) is the specified number of occurrences.
  • \( k! \) represents the factorial of \( k \), the product of all positive integers up to \( k \).

In Part a of the exercise, to find the probability of exactly four arrivals, we apply this formula with \( \lambda = 5 \) (since arrivals happen at this average rate per hour) and \( k = 4 \). Calculating this gives approximately 0.1755, meaning there's about a 17.55% chance that exactly four people will arrive.
Expected Value
Expected Value in a Poisson distribution is often very straightforward. It represents the average number we expect when performing a trial or observation over time. The expected value for a Poisson random variable equals its rate parameter (\( \lambda \)).
This tells us how many events, like arrivals, we expect to happen in a period, factoring in the average rate of occurrence.
For calculating how many people to expect in a 45-minute period, consider that 45 minutes is 3/4 of an hour. Thus, multiply the hour rate \( \lambda = 5 \) by 0.75 to get an expected arrival count of 3.75.
This implies that, on average, we would have around 3.75 people arriving during that time. It indicates that even with precise probabilities, real-life results may be slightly non-integers since we are talking about averages over a period.
Complement Rule
The Complement Rule is an efficient way to calculate the probability of "at least" scenarios, which can otherwise involve long computations. In probability theory, the complement of an event is the probability that the event does not happen, and the Complement Rule is a simple subtraction method using the total probability which must sum to 1.

For the problem of finding the probability of at least four arrivals in an hour (Part b in the exercise), directly computing it would require calculating probabilities for all scenarios: four or more arrivals. Alternatively, using the complement, we find probabilities for zero to three arrivals and subtract from one:\[P(X \geq 4) = 1 - P(X < 4)\]This method simplifies the process by finding \( P(X=0) \), \( P(X=1) \), \( P(X=2) \), and \( P(X=3) \) individually and then calculating their sum to find \( P(X < 4) \). The sum gives approximately 0.2649, leading to \( P(X \geq 4) = 1 - 0.2649 = 0.7351 \). Thus, the chance of at least four arrivals in that hour is about 73.51%.

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