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Question

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:

The correct answer is

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:

  1. 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.
  2. 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.
  3. 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.
  4. 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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Important Questions from Random Variables Basics

  1. 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:

  2. If X follows a binomial distribution with n = 6 and \(p=\dfrac{1}{4}\) then the skewness of X is:

  3. 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:

  4. 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


    The value of E(X) is:
  5. What percentage of scores falls within three standard deviations from the mean for the normal variate?

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