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The exponential distribution is frequently applied to the waiting times between successes in a Poisson process. If the number of calls received per hour by a telephone answering service is a Poisson random variable with parameter \(X=6,\) we know that the time, in hours, between successive calls has an exponential distribution with parameter \(\beta=1 / 6\). What is the probability of waiting more than 15 minutes between any two successive calls?

Short Answer

Expert verified
The probability of waiting more than 15 minutes between any two successive calls is approximately 0.223.

Step by step solution

01

Understanding the problem

The problem is to find the probability of waiting more than 0.25 hours between two consecutive calls. The interval between the two calls is an exponential random variable. The exponential distribution has a probability density function given by \[f(t ; \beta)=\frac{1}{\beta} e^{-t / \beta}\], where \(t\) is the time, and \(\beta=1/6\) is the distribution parameter.
02

Applying the formula for exponential distribution

In order to find the probability of waiting more than 0.25 hours, we need to calculate the integral of the distribution function from 0.25 to infinity. The cumulative distribution function is given by \[F(t ; \beta)=1-e^{-t / \beta}\], from which we can deduce that the probability of waiting more than a certain amount of time is given by \(P(T > t) = 1 - F(t ; \beta)\). Now applying it for \(t = 0.25\), \[P(T > 0.25) = 1 - F(0.25 ; 1/6) = 1 - (1 - e^{-0.25 / (1/6)})\]
03

Calculating the probability

Substitute the values and solve the equation: \[P(T > 0.25) = e^{-0.25 / (1/6)} = e^{-1.5}\]. Using the exponential table or a calculator, you can find that \(e^{-1.5} \approx 0.223\). This is the probability that the wait time between two successive calls is greater than 0.25 hours.

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Key Concepts

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

Poisson Process
A Poisson process is a statistical model that represents events occurring randomly over a continuous interval of time or space. This model is particularly useful when dealing with the occurrence of rare or random events that don't have a predictable pattern such as the arrival of calls at a telephone service center, radioactive decay, or traffic flow.

The main feature of a Poisson process is that the events are independent, meaning the occurrence of one event does not affect the probability of another event happening. Additionally, the process has a constant average rate, denoted by the parameter \(\beta\), which represents the rate at which the events occur.

For example, in a scenario where a call center receives 6 calls per hour, the time between each call can be modeled by an exponential distribution as part of a Poisson process with a rate \(\beta = 1/6\) per hour. This means on average, there is one call every 10 minutes, since \(1/6\) of an hour is approximately 10 minutes.
Probability Density Function
The probability density function (PDF) is a function that describes the likelihood of a random variable taking on a given value. The PDF for continuous variables, such as the time between events in a Poisson process, depicts how the probabilities are distributed over the range of possible values.

For the exponential distribution, which is continuously defined, the PDF is given by the formula: \[f(t ; \beta)=\frac{1}{\beta} e^{-t / \beta}\] where \(t\) represents the time and \(\beta\) is the distribution parameter. The graph of a PDF for an exponential distribution will show a curve that starts off high at \(t=0\) and rapidly decreases as \(t\) increases.

In our call center example, we use the PDF of the exponential distribution to determine the likelihood of waiting a specific amount of time between calls. With \(\beta = 1/6\), the probability of wait times are more likely to be short, reflecting the frequent arrival of calls.
Cumulative Distribution Function
The cumulative distribution function (CDF), unlike the PDF, provides the probability that a random variable \(T\) is less than or equal to a given value \(t\). Essentially, it gives us the area under the PDF curve from the lowest possible value of \(T\) up to \(t\). For the exponential distribution, the CDF is mathematically defined as: \[F(t ; \beta)=1-e^{-t / \beta}\]

The CDF is a non-decreasing function that starts at 0, when the given value \(t\) is 0, and asymptotically approaches 1 as \(t\) increases to infinity. This fits with the concept that as you wait longer, the probability of having received a call approaches certainty.

In practical terms, when we want to calculate the probability of the time between calls being greater than a certain value (such as 15 minutes in our textbook exercise), we use the complement of the CDF, which is \(1 - F(t ; \beta)\). Doing so gives us the probability that the time until the next event is longer than \(t\), which is what our original exercise question aims to find.

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