Problem 21 - Applied Statistics and Probability for Engineers

Question

A popular clothing manufacturer receives Internet orders via two different routing systems. The time between orders for each routing system in a typical day is known to be exponentially distributed with a mean of 3.2 minutes. Both systems operate independently. (a) What is the probability that no orders will be received in a 5-minute period? In a 10 -minute period? (b) What is the probability that both systems receive two orders between 10 and 15 minutes after the site is officially open for business? (c) Why is the joint probability distribution not needed to answer the previous questions?

Step-by-Step Solution

Answer

(a) 0.0439 for 10 minutes; (b) Calculate \( P_{both} = (P(X = 2))^2 \); (c) Independence of systems allows separate calculations.

Step 1: Understanding the Exponential Distribution

The time between orders is described by an exponential distribution with a mean of 3.2 minutes. The probability density function of an exponential distribution is given by \( f(x;\lambda) = \lambda e^{-\lambda x} \) where \( \lambda \) is the rate parameter (1 divided by the mean). Therefore, \( \lambda = \frac{1}{3.2} \approx 0.3125 \).

Step 2: Calculating No Orders in 5 Minutes for One System

For no orders in 5 minutes, we need the cumulative distribution function (CDF) for \( x \), \( P(X > x) = e^{-\lambda x} \). With \( x = 5 \), for one system, \( P(X > 5) = e^{-0.3125 \times 5} = e^{-1.5625} \). Calculate this to find the probability.

Step 3: Probability Calculation for 5 Minutes

Using the exponential CDF: \( P(X > 5) = e^{-1.5625} \approx 0.2096 \). This is the probability for one system to receive no orders in 5 minutes. For both systems independently, multiply the probabilities: \( P_{both} = (0.2096)^2 \).

Step 4: Calculating No Orders in 10 Minutes for One System

Repeat the CDF calculation for 10 minutes, \( P(X > 10) = e^{-0.3125 \times 10} = e^{-3.125} \). Calculate this to determine the probability for one system.

Step 5: Probability Calculation for 10 Minutes

Using the exponential CDF: \( P(X > 10) = e^{-3.125} \approx 0.0439 \). For both systems: \( P_{both} = (0.0439)^2 \).

Step 6: Probability of Two Orders Between 10 and 15 Minutes

The number of orders follows a Poisson distribution. For a time interval of 5 minutes, \( \lambda_5 = \frac{5}{3.2} \approx 1.5625 \). We want \( P(X = 2) = \frac{\lambda_5^2 e^{-\lambda_5}}{2!} \). Calculate this for both systems.

Step 7: Joint Probability for Both Systems

For two orders in both systems, multiply the probabilities: \( P_{both} = (P(X = 2))^2 \). Each system is independent, so their combined probability is the product of their probabilities.

Step 8: Joint Probability Distribution Explanation

Exponential and Poisson processes assume independence between events in different systems, which allows separate probability calculations for each system. Hence, we can handle each system independently without needing a combined distribution.

Key Concepts

Poisson Distribution

The Poisson Distribution is essential when analyzing the number of events occurring within a fixed interval of time or space, provided the events happen independently and at a constant average rate. In this exercise, such a scenario manifests as the number of customer orders received by two independent routing systems over particular time intervals.

To determine probabilities using the Poisson distribution, we utilize the formula: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] where:

• \( \lambda \) is the average rate of occurrence (mean), adjusted for the interval of interest.
• \( k \) is the number of events (e.g., orders) we want the probability for.

In the exercise, you're calculating the probability of two orders being placed in a 5-minute window. Knowing that the mean time between orders is 3.2 minutes allows you to utilize these insights.

Probability Calculation

Probability calculation involves determining the likelihood of one or more outcomes happening. Here, we primarily use two types of calculations:

• Exponential Cumulative Distribution Function (CDF): Used to determine the probability of receiving no orders in a given time by applying \( P(X > x) = e^{-\lambda x} \). This is crucial for understanding scenarios where no events (orders) occur within the specified time period.
• Poisson Probability: Used when calculating the probability of receiving a specific number of orders in a time interval, such as calculating the chance of both systems receiving exactly two orders between 10 to 15 minutes.

To calculate for both systems independently, multiply individual probabilities. This multiplication takes advantage of independent event properties and is vital for complex probability scenarios.

Independent Events

Independent events occur when the outcome of one event does not influence the outcome of another. In our problem, the orders received by the two routing systems are independent. This means the probability of events occurring in one system is not affected by what happens in the other system.

When dealing with independent events, probabilities can be computed separately for each system and then combined by multiplication. If each system operates without being influenced by the other, it's valid to assume this independence for probability calculations.

This concept is especially important when considering multiple systems or processes as it simplifies probability calculations, allowing each process to be handled individually.

Exponential Distribution Function

The Exponential Distribution is characterized by a constant hazard rate and is typically used to model times to events, such as the time until a new order arrives in our scenario. Its probability density function (PDF) is expressed as: \[ f(x; \lambda) = \lambda e^{-\lambda x} \] where:

• \( \lambda \) is the rate parameter, calculated as the reciprocal of the mean (\( \lambda = \frac{1}{3.2} \) in our exercise).
• \( x \) is the time variable.

With the exponential distribution, the cumulative distribution function (CDF) \( P(X > x) = e^{-\lambda x} \) helps in finding probabilities for scenarios where no events occur for a set time period.

This distribution is particularly useful when the "memoryless" property is at play—where the probability of an event occurring in the next interval is independent of how much time has already passed. Therefore, whether calculating for a 5-minute interval or any other period, this principle makes the exponential distribution a powerful tool.