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
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.
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.
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.
Let's examine why the other options are not the correct distributions for the inter-arrival time in a continuous-time Poisson process:
Therefore, the inter-arrival time of a customer in this scenario follows the Exponential distribution.
| 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) |
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.
The length of time X, needed by an examinee of competition to complete a 1-hour exam, is a random variable with
PDF \(f(x)=\dfrac{6}{5}(x^2+x);0 \le x \le 1.\) , The value of F(0.5) is:
If X follows a binomial distribution with n = 6 and \(p=\dfrac{1}{4}\) then the skewness of X is:
If the customers arrive in a shop in Poisson fashion with parameter λ, the fourth raw moment \(\mu_4^{'}\) for the inter-arrival time is:
A discrete random variable X has the probability functions as:
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
f(x) | K | 2k | 3k | 5k | 5k | 4k | 3k | 2k | k |
What percentage of scores falls within three standard deviations from the mean for the normal variate?