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Automobiles arrive at the Elkhart exit of the Indiana Toll Road at the rate of two per minute. The distribution of arrivals approximates a Poisson distribution. a. What is the probability that no automobiles arrive in a particular minute? b. What is the probability that at least one automobile arrives during a particular minute?

Short Answer

Expert verified
a. Probability is 0.1353; b. Probability is 0.8647.

Step by step solution

01

Understand the Problem

Given a Poisson distribution with an average rate of arrival \( \lambda = 2 \) automobiles per minute. We need to find probabilities based on this distribution.
02

Use Poisson Formula for No Arrivals

To find the probability of no automobiles arriving in a particular minute, apply the Poisson probability formula: \( P(X=k) = \frac{e^{-\lambda} \lambda^{k}}{k!} \) where \( k = 0 \) and \( \lambda = 2 \).
03

Calculate Probability for No Arrivals

Substitute the values into the Poisson formula: \( P(X=0) = \frac{e^{-2} \times 2^{0}}{0!} = \frac{e^{-2}}{1} \). Use a calculator to find \( e^{-2} \approx 0.1353 \).
04

Determine Probability for At Least One Arrival

The probability of at least one arrival can be found by subtracting the probability of no arrivals from 1: \( P(X \geq 1) = 1 - P(X=0) \).
05

Calculate Probability for At Least One Arrival

Using the calculated value for no arrivals, \( P(X \geq 1) = 1 - 0.1353 = 0.8647 \).

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Key Concepts

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

Probability Calculations
The Poisson distribution is a helpful tool for computing the probability of a given number of events, like automobile arrivals, in a specific period of time. This distribution is particularly suited for situations where the events occur independently and at a constant average rate. Probability calculations can be accomplished using the Poisson probability formula.
The formula is \( P(X=k) = \frac{e^{-\lambda} \lambda^{k}}{k!} \), where \( \lambda \) represents the average rate of occurrences (here, 2 automobiles per minute) and \( k \) is the number of occurrences you are interested in. When \( k = 0 \), you are looking at the probability of no automobile arrivals.
  • Plug in \( \lambda = 2 \) and \( k = 0 \) to find the probability of no arrivals.
This straightforward calculation is important, especially in scenarios involving predictions over time, such as traffic flow or entry rates in a queue.
Arrival Rate
The arrival rate, denoted as \( \lambda \), is a fundamental part of the Poisson distribution. It defines the expected number of events, such as cars arriving, in a given period. In our scenario, the arrival rate is 2 cars per minute.
This constant rate allows us to apply the Poisson distribution effectively. When using a Poisson model, the assumption is that the number of cars arriving in different intervals of one minute follows the same distribution pattern.
  • The arrival rate helps determine the probability of a range of outcomes, such as zero cars arriving, or more than one.
  • It is a crucial parameter that models the randomness of events in time-bound processes.
Understanding \( \lambda \) helps in planning and managing resources effectively; whether it be optimizing toll booth staffing or managing flow at busy exits.
Mathematical Formulae
Mathematical formulae are central to calculating and understanding probabilities in the Poisson distribution model. They not only allow us to compute exact probabilities but also to express them in a language that can be universally understood.
The primary formula for the Poisson distribution is \( P(X=k) = \frac{e^{-\lambda} \lambda^{k}}{k!} \).
  • \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
  • \( k! \) (k factorial) means multiplying all positive integers up to \( k \).
For instance, when calculating the probability of no arrivals: substitute \( k = 0 \), which simplifies the multiplication due to \( 0! = 1 \). The formula highlights the elegance of mathematics in making complex predictions, like estimating traffic, simple and manageable.

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Most popular questions from this chapter

Recent crime reports indicate that 3.1 motor vehicle thefts occur each minute in the United States. Assume that the distribution of thefts per minute can be approximated by the Poisson probability distribution. a. Calculate the probability exactly four thefts occur in a minute. b. What is the probability there are no thefts in a minute? c. What is the probability there is at least one theft in a minute?

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 70 percent of the cases. Suppose the 15 cases reported today are representative of all complaints. a. How many of the problems would you expect to be resolved today? What is the standard deviation? b. What is the probability 10 of the problems can be resolved today? c. What is the probability 10 or 11 of the problems can be resolved today? d. What is the probability more than 10 of the problems can be resolved today?

Industry standards suggest that 10 percent of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 12 Nissans yesterday. a. What is the probability that none of these vehicles requires warranty service? b. What is the probability exactly one of these vehicles requires warranty service? c. Determine the probability that exactly two of these vehicles require warranty service. d. Compute the mean and standard deviation of this probability distribution.

In a Poisson distribution \(\mu=4\). a. What is the probability that \(x=2 ?\) b. What is the probability that \(x \leq 2 ?\) c. What is the probability that \(x>2 ?\)

New Process, Inc., a large mail-order supplier of women's fashions, advertises same-day service on every order. Recently the movement of orders has not gone as planned, and there were a large number of complaints. Bud Owens, director of customer service, has completely redone the method of order handling. The goal is to have fewer than five unfilled orders on hand at the end of 95 percent of the working days. Frequent checks of the unfilled orders at the end of the day reveal that the distribution of the unfilled orders follows a Poisson distribution with a mean of two orders. a. Has New Process, Inc., lived up to its internal goal? Cite evidence b. Draw a histogram representing the Poisson probability distribution of unfilled orders.

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