Patients arrive at a clinic following the Poisson process, with the mean rate of 10 customers per hour. The inter-arrival time of a customer follows:
Exponential distribution
Understanding Inter-Arrival Times in a Poisson Process
The question describes a scenario where patients arrive at a clinic following a Poisson process. A Poisson process is a widely used model in probability theory to describe the occurrence of random events in a continuous time interval or spatial region.
A key characteristic given is the mean rate of arrival, which is 10 customers per hour. In a Poisson process, events occur independently of each other, and the average rate of occurrence is constant over time.
Relationship between Poisson Process and Inter-Arrival Time Distribution
A fundamental property links the Poisson process to the time between consecutive events. If the number of events in a fixed interval of time follows a Poisson distribution, then the time elapsed between any two consecutive events (known as the inter-arrival time) follows an Exponential distribution.
This relationship is a cornerstone of queuing theory and reliability analysis. The rate parameter of the exponential distribution is the same as the rate parameter ($\lambda$) of the Poisson process.
Why Exponential Distribution is Correct
Given that the patient arrivals follow a Poisson process with a mean rate $\lambda = 10$ customers per hour, the time between one patient's arrival and the next patient's arrival (the inter-arrival time) is described by an exponential distribution with the rate parameter $\lambda = 10$ per hour.
The probability density function (PDF) of this exponential distribution is $\text{f(t)} = \lambda e^{-\lambda t}$ for $t \ge 0$, where $t$ is the inter-arrival time.
Analyzing the Given Options
Let's examine why the other options are not the correct distributions for the inter-arrival time in a continuous-time Poisson process:
- Geometric distribution: This is a discrete probability distribution. It models the number of Bernoulli trials needed to get the first success. Inter-arrival time is a continuous variable (time), not a discrete count of trials.
- Poisson distribution: This distribution models the number of events that occur in a fixed interval of time or space, not the time between events. The number of customers arriving in one hour follows a Poisson distribution, but the time between arrivals follows a different distribution.
- Gamma distribution: The Gamma distribution is a continuous distribution that generalizes the exponential distribution. The sum of $k$ independent exponential random variables, each with rate $\lambda$, follows a Gamma distribution with shape parameter $k$ and rate parameter $\lambda$. While related, the inter-arrival time between *single* consecutive events is specifically the exponential distribution (which is a Gamma distribution with $k=1$). The Gamma distribution is used for the waiting time until the $k$-th arrival.
- Exponential distribution: As explained above, the time between events in a continuous-time Poisson process is exponentially distributed.
Therefore, the inter-arrival time of a customer in this scenario follows the Exponential distribution.
Revision Table: Process and Time Distributions
| Process Type | Events | Distribution for Count of Events in Interval | Distribution for Time Between Events (Inter-arrival Time) |
|---|---|---|---|
| Poisson Process (Continuous Time) | Random events over time/space | Poisson Distribution | Exponential Distribution |
| Bernoulli Process (Discrete Trials) | Success/Failure in trials | Binomial Distribution (Count of successes) | Geometric Distribution (Trials until first success) |
Additional Information: Properties of Exponential Distribution
The Exponential distribution is characterized by a single parameter, the rate $\lambda$. In the context of a Poisson process with rate $\lambda$, the mean of the exponential distribution for inter-arrival time is $1/\lambda$. The variance is $(1/\lambda)^2$.
For this problem, the mean inter-arrival time is $1/10$ hours, which equals 6 minutes.
A unique property of the exponential distribution is the memoryless property. This means that the probability that an event has not occurred by time $t+s$, given that it has not occurred by time $t$, is the same as the probability that it has not occurred by time $s$. Mathematically, $\text{P(X > t + s | X > t) = P(X > s)}$. In simpler terms, knowing how long you have already waited does not change the probability distribution of how much longer you still have to wait.
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