Problem 145 - Applied Statistics and Probability for Engineers

Question

The time between arrivals of customers at an automatic teller machine is an exponential random variable with a mean of five minutes. (a) What is the probability that more than three customers arrive in 10 minutes? (b) What is the probability that the time until the fifth customer arrives is less than 15 minutes?

Step-by-Step Solution

Answer

(a) 0.1429, (b) 0.5981.

Step 1: Understand the Problem for Part (a)

The time between customer arrivals is an exponential random variable with a mean of 5 minutes. We need the probability that more than 3 customers arrive in 10 minutes.

Step 2: Convert the Mean to Rate for Part (a)

The rate \( \lambda \) for an exponential distribution is the reciprocal of the mean: \( \lambda = \frac{1}{5} \) per minute.

Step 3: Identify the Associated Poisson Process for Part (a)

The number of arrivals in a given time period follows a Poisson process. We can use \( \lambda = \frac{1}{5} \) to find the rate for 10 minutes: \( \lambda' = \frac{1}{5} \times 10 = 2 \).

Step 4: Use the Poisson Formula for Part (a)

The Poisson probability mass function is \( P(X=k) = \frac{e^{-\lambda'} \lambda'^k}{k!} \). We want \( P(X > 3) \), so we calculate \( 1 - (P(X=0) + P(X=1) + P(X=2) + P(X=3)) \).

Step 5: Calculate the Probabilities for Part (a)

Compute each: \[ P(X=0) = \frac{e^{-2} \, 2^0}{0!} = 0.1353 \] \[ P(X=1) = \frac{e^{-2} \, 2^1}{1!} = 0.2707 \] \[ P(X=2) = \frac{e^{-2} \, 2^2}{2!} = 0.2707 \] \[ P(X=3) = \frac{e^{-2} \, 2^3}{3!} = 0.1804 \]

Step 6: Compute Probability for Part (a)

\( P(X > 3) = 1 - (0.1353 + 0.2707 + 0.2707 + 0.1804) = 0.1429 \)

Step 7: Understand the Problem for Part (b)

The time until the fifth customer arrives is considered. This follows a gamma distribution or can use the fact that it's the sum of 5 exponential random variables.

Step 8: Utilize Gamma Distribution for Part (b)

The sum of 5 exponential variables (with mean 5 minutes) is a gamma distribution with shape parameter \( k = 5 \) and rate parameter \( \lambda = \frac{1}{5} \).

Step 9: Calculate Probability Using Gamma Distribution for Part (b)

We need \( P(T < 15) \) where \( T \) is the time for the fifth arrival. The required probability can be found using \( P(T < 15) = \text{Gamma}(5, \frac{15}{5}) \), which is the cumulative distribution function.

Step 10: Find Cumulative Probability for Part (b)

Using tables or software to calculate: \( P(T < 15) = F_{Gamma}(3) = 0.5981 \).

Key Concepts

Poisson Process

A Poisson process is a mathematical model used to represent events occurring randomly over a fixed period. In the context of customer arrivals at an ATM, the Poisson process helps us understand and predict the number of arrivals within a given time frame. This is incredibly important in fields such as queue management and operations research.

The key aspect of a Poisson process is the concept of a 'rate', often denoted by the symbol \( \lambda \). This rate represents the average number of events (in this case, customer arrivals) expected to occur per unit of time.

• Constant mean rate: The events occur with a constant mean rate, independent of the time since the last event.
• Independence: The number of events occurring in disjoint intervals are independent.
• State transitions: The time between successive events follows an exponential distribution.

Understanding the Poisson process is vital for calculating the probability of a certain number of events occurring within a specified time frame, as illustrated by how we determined the likelihood of more than three customers arriving in ten minutes.

Exponential Distribution

The exponential distribution describes the time between events in a Poisson process. It gives the probability of the time until the next event, like the time between customer arrivals at an ATM.

In this distribution, the mean and rate are closely tied. The mean is the expected time between events, and the rate \( \lambda \) is the reciprocal of this mean.

• Memoryless property: The exponential distribution is memoryless. This means the probability of an event occurring in the future is independent of past events.
• Parameter: The single parameter \( \lambda \) captures the entire distribution with the probability density function given by \( f(x) = \lambda e^{-\lambda x} \) for \( x \geq 0 \).

By converting a mean of 5 minutes into a rate of \( \frac{1}{5} \) per minute, we applied this principle during the original exercise to determine the interval probabilities.

Gamma Distribution

The gamma distribution appears when considering multiple events in a Poisson process, particularly when dealing with the time until the 'k-th' event occurs. It's an extension of the exponential distribution, used when we need to sum several exponential random variables.

For example, in our exercise, we were interested in finding the time until the fifth customer arrival, which is less than 15 minutes. This required us to use the gamma distribution as it's suited to handle the summary of such stochastic events.

• Shape and scale parameters: The shape parameter \( k \) corresponds to the number of events (in this case, 5 customers), and the scale is given by \( \theta = \frac{1}{\lambda} \), which is the mean of the exponential variables.
• Probability calculation: The cumulative distribution function (CDF) of the gamma distribution allows us to calculate the probability of the time interval until a particular number of events has occurred.

This distribution was the main tool to solve part (b) of our problem, allowing us to use the CDF and conclude on the necessary probability using either tables or statistical software.

Probability Calculations

Probability calculations are essential in determining the likelihood of events within a Poisson process. They allow us to make informed decisions based on statistical estimates.

In the exercise provided, calculating the probability involved using both the Poisson and gamma distributions depending on the question. Here's how these work:

• Poisson probability mass function: This function is used when determining the probability of observing exactly \( k \) events in a fixed interval, formulated as \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \).
• Gamma cumulative distribution function: This function is used to determine the likelihood that the time until a certain number of events occurs is less than a given value. It’s calculated in practice with tabs or software, reflecting complex integrations.

By following these standard procedures, we were able to solve the questions effectively, highlighting the power and relevance of proficient probability calculations in applied statistics.