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Chapter 4: Problem 161

When a bus service reduces fares, a particular trip from New York City to Albany, New York, is very popular. A small bus can carry four passengers. The time between calls for tickets is exponentially distributed with a mean of \(30 \mathrm{~min}-\) utes. Assume that each call orders one ticket. What is the probability that the bus is filled in less than three hours from the time of the fare reduction?

Short Answer

Expert verified
The probability that the bus is filled in less than 3 hours is approximately 0.848.

Step by step solution

01

Understanding the Problem

We need to calculate the probability that the bus, which can hold 4 passengers, is filled in less than 3 hours. The calls arrive following an exponential distribution with a mean of 30 minutes.
02

Convert Time to Consistent Units

Convert the 3-hour time limit into minutes to match the mean time between calls. Since 1 hour is 60 minutes, 3 hours equals 180 minutes.
03

Determine Mean Rate (Lambda)

For an exponential distribution with mean 30 minutes, the rate \(\lambda\) is the reciprocal, \(\lambda = 1/30\) per minute.
04

Use the Poisson Process Approximation

The exponential distribution of time between events relates to the Poisson distribution for the number of events. We want to find the probability that 4 calls occur in less than 180 minutes. Here, mean number of events \(\lambda t = 180/30 = 6\).
05

Calculate the Probability with the Poisson Distribution

The Poisson probability formula is: \[ P(X=k) = \frac{e^{-\lambda t} (\lambda t)^k}{k!} \]where \( k = 4 \), \( \lambda t = 6 \). Calculate \[ P(X \geq 4) = 1 - P(X < 4) = 1 - \sum_{k=0}^{3} \frac{e^{-6} 6^k}{k!} \]
06

Compute Cumulative Probability for Events 0 to 3

Calculate individual probabilities and their sum:\[ P(X=0) = \frac{e^{-6} 6^0}{0!} \]\[ P(X=1) = \frac{e^{-6} 6^1}{1!} \]\[ P(X=2) = \frac{e^{-6} 6^2}{2!} \]\[ P(X=3) = \frac{e^{-6} 6^3}{3!} \]Sum is \( P(X<4) \).
07

Final Calculation

Subtract the cumulative probability of fewer than 4 calls from 1 to find the probability that at least 4 calls (so the bus is filled) occur in less than 3 hours:\[ P(X \geq 4) = 1 - P(X<4) \].
08

Solve the Equation

Calculate using the sums from Step 6:\[ P(X=0) + P(X=1) + P(X=2) + P(X=3) \]. Then:\[ P(X \geq 4) = 1 - (P(X=0) + P(X=1) + P(X=2) + P(X=3)) \].

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Key Concepts

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

Exponential Distribution
The exponential distribution is a popular statistical model for describing the time between independent events that occur continuously and with a constant average rate. Imagine you're waiting for calls to buy bus tickets. These calls would arrive at random intervals.

  • This distribution is memoryless, meaning the probability of an event occurring in the future is independent of any past events.
  • The exponential distribution is characterized by a single parameter \( \lambda \) (or 1 divided by the mean time between events), representing the average rate at which events occur.
For example, if the mean time between calls is 30 minutes, \( \lambda \) is \( \frac{1}{30} \) per minute, since this is the reciprocal of 30 minutes. This property is crucial in scenarios like the bus ticket sales, where one is interested in the frequency of calls over time.
Mean Rate (Lambda)
The mean rate, denoted as \( \lambda \), is a central concept in both exponential and Poisson distributions. This rate provides us with an understanding of how often an event is expected to occur over a specific period.

  • For exponential distributions, \( \lambda \) represents the average number of occurrences per unit time. It's calculated as the reciprocal of the mean time between events.
  • In the problem scenario, the mean time between calls is 30 minutes, which makes \( \lambda = \frac{1}{30} \) per minute.
This value of \( \lambda \) helps us extend the concept to a Poisson process, where it acts as the expected number of occurrences over a longer time period by multiplying it with the total time considered (here, 3 hours or 180 minutes). Thus, for 180 minutes, the expected number of calls is 6.
Cumulative Probability
Cumulative probability refers to the probability that a random variable is less than or equal to a certain value. In our bus ticket example, this helps us determine the likelihood of at least a specific number of calls being made within a time frame.

  • We utilize the Poisson probability formula to calculate the cumulative probability for events from 0 to a certain number.
  • This involves summing up probabilities of individual events (from 0 to 3 in this case), calculated using the Poisson formula: \( P(X=k) = \frac{e^{-\lambda t} \cdot (\lambda t)^k}{k!} \).
The cumulative probability for fewer than 4 calls is found by summing \( P(X=0) \), \( P(X=1) \), \( P(X=2) \), and \( P(X=3) \). Subtracting this sum from 1 gives us the probability that 4 or more calls are received in the given time period, meaning the bus will be full.

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