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3-171. Cabs pass your workplace according to a Poisson process with a mean of five cabs per hour. Suppose that you exit the workplace at 6:00 P.M. Determine the following: (a) Probability that you wait more than 10 minutes for a cab. (b) Probability that you wait fewer than 20 minutes for a cab. (c) Mean number of cabs per hour so that the probability that you wait more than 10 minutes is \(0.1\).

Short Answer

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(a) 43.46% chance to wait more than 10 minutes. (b) 86.47% chance to wait fewer than 20 minutes. (c) Need 13.8 cabs/hour for 0.1 probability of waiting over 10 minutes.

Step by step solution

01

Understanding the Poisson Process

The problem states that cabs pass at an average rate of 5 per hour, which defines our rate \( \lambda = 5 \) cabs/hour in the Poisson process.
02

Convert Waiting Time to Hours

To find the probability of waiting more than or fewer than a specific number of minutes, we first convert minutes to hours. 10 minutes is \( \frac{10}{60} = \frac{1}{6} \) hours, and 20 minutes is \( \frac{20}{60} = \frac{1}{3} \) hours.
03

Define Poisson Probability Formula

The probability that we will wait for time \( t \) without seeing a cab (zero occurrences) can be given by the cumulative distribution function of the Poisson process as: \[ P(T > t) = e^{-\lambda t} \] where \( t \) is the time period in hours and \( \lambda \) is the mean rate of occurrences.
04

Calculate Probability for Part (a)

For a wait time of more than 10 minutes, we use \( t = \frac{1}{6} \): \[ P(T > \frac{1}{6}) = e^{-5 \times \frac{1}{6}} = e^{-\frac{5}{6}} \approx 0.4346 \]. Thus, there is about a 43.46% chance that you will wait more than 10 minutes.
05

Calculate Probability for Part (b)

For a wait time of fewer than 20 minutes, we compute \[ P(T < \frac{1}{3}) = 1 - e^{-5 \times \frac{1}{3}} = 1 - e^{-\frac{5}{3}} \approx 0.8647 \]. Hence, there is an 86.47% probability that you wait fewer than 20 minutes.
06

Solve for Mean Rate in Part (c)

For part (c), we are given that \( P(T > \frac{1}{6}) = 0.1 \) and need to find the mean rate \( \lambda \). Using \[ 0.1 = e^{-\lambda \times \frac{1}{6}} \], we take the natural log: \[ -\frac{\lambda}{6} = \ln(0.1) \]. Solving for \( \lambda \), we get \[ \lambda = -6 \cdot \ln(0.1) \approx 13.8 \]. Hence, 13.8 cabs per hour is needed so the waiting chance is 0.1.

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Key Concepts

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

Waiting Time Calculations
In the Poisson process, waiting time is an essential calculation. It helps to determine the time before an event occurs, like waiting for the next cab. Imagine this as an invisible clock turning until the event – the cab’s arrival – happens. When dealing with waiting times, we often need to convert time into hours if not already provided in that unit.
For example, 10 minutes is equal to \( \frac{10}{60} = \frac{1}{6} \) of an hour. It's crucial to make this conversion because the Poisson process is typically defined in terms of an hourly rate or occurrences. This ensures consistency in calculations across different time spans. Once you have the waiting time in the right unit, you can apply it to various mathematical models.
The beauty of these calculations lies in how they predict the probability of waiting a certain period before an event occurs or doesn't occur, which is particularly useful in managing expectations in situations like catching a cab during rush hour.
Probability in Poisson Processes
Probability plays a key role in interpreting results within a Poisson process. At its core, the Poisson process models the likelihood of a given number of events happening in a fixed interval of time or space. With cabs passing by your workplace at a known average rate (like 5 cabs per hour), you can determine various probabilities.
For instance, in waiting time scenarios, we might be interested in the probability of not seeing any cabs for a certain waiting time, such as more than 10 minutes. This is calculated using the formula \( P(T > t) = e^{-\lambda t} \), where \( \lambda \) is the average rate of occurrence (5 cabs per hour in our case), and \( t \) is the time you are concerned about.
This formula essentially tells us how likely it is that no cabs are seen during the time \( t \). Understanding these probabilities helps make informed decisions, like whether to wait for a cab or consider alternate transport options.
Cumulative Distribution Function
A significant tool for analyzing Poisson processes is the cumulative distribution function (CDF). The CDF helps in understanding the likelihood of a random variable falling within a certain range. In the context of a Poisson process, it accounts for the total probability that events occur up to a specific time.
In our cab example, if we want to determine the probability of waiting fewer than 20 minutes, we use the complementary cumulative distribution function. The CDF is given by the expression: \( P(T < t) = 1 - e^{-\lambda t} \). In this instance, \( \lambda = 5 \) and \( t = \frac{1}{3} \) hours, translating into 20 minutes. This function allows us to calculate that probability effectively, reflecting a very high chance (about 86.47%) of catching a cab within the desired time.
Using the CDF, we can swiftly assess how often specific wait times occur, granting us deeper insights into the dynamics of such unpredictable events, hence aiding in time management and planning.

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