Chapter 3: Problem 100
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
Step by step solution
Identify the Poisson parameter
Calculate the probability for exactly four arrivals (Part a)
Execute the calculation for Part a
Calculate the probability of at least four arrivals (Part b)
Calculate individual probabilities for Part b
Calculate the cumulative probability for P(X < 4)
Execute the calculation for Part b
Determine expected arrivals for a 45-minute period (Part c)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Calculation
- \( P(X=k) \) 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
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
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%.