Problem 28 - Basic Statistics for Business & Economics

Question

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?

Step-by-Step Solution

Answer

a) The probability is approximately 0.1353. b) The probability is approximately 0.8647.

Step 1: Understand the Problem

We have a Poisson distribution for the number of automobiles arriving per minute, with an average rate \( \lambda = 2 \) cars per minute. The problem requires finding the probability for two different scenarios.

Step 2: Set Up the Poisson Probability Formula

The Poisson probability mass function is given by: \[ P(X=k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \] where \( \lambda = 2 \) and \( k \) is the number of successes (arrivals).

Step 3: Part a: Calculate the Probability for No Arrivals (k=0)

For no automobiles arriving in a particular minute, use \( k = 0 \):\[ P(X=0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2} \approx 0.1353 \]

Step 4: Part b: Calculate the Probability for At Least One Arrival

To find the probability of at least one automobile arriving, we use the complement rule:\[ P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-2} \approx 1 - 0.1353 = 0.8647 \]

Step 5: Verify the Solution

Verify that the calculations for both parts are consistent and accurate. The complements and individual probabilities are aligned with the properties of Poisson distribution.

Key Concepts

Probability

Probability is a key concept in statistics that measures how likely an event is to occur.
It's a value between 0 and 1, where 0 means the event won't happen and 1 means it will definitely happen.
In our context of automobiles arriving at a toll road exit, probability is used to predict the number of cars that might show up in a given minute.
Probability helps us understand randomness and uncertainty in different scenarios.
Here, the Poisson distribution allows us to calculate probabilities related to specific numbers of events (like car arrivals) within a fixed interval.
For this particular problem, we utilize probability to determine the chance of having zero car arrivals in one minute, or at least one car arrival during that time.

Probability Mass Function

The Probability Mass Function (PMF) is crucial when working with discrete random variables like in our Poisson distribution problem.
A PMF specifies the probability that a discrete random variable is exactly equal to some value, say \( k \).
For a Poisson distribution, which is appropriate when predicting the number of events (e.g., car arrivals) in fixed intervals, the PMF is given by the formula: \[ P(X=k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \] Here, \( \lambda \) represents the average rate of people or events happening, and \( e \) is the base of natural logarithms, approximately 2.718.
In our problem, with \( \lambda = 2 \) arrivals per minute, the PMF lets us calculate probabilities for specific arrival counts, such as zero cars (\( k = 0 \)).
Understanding how the PMF works helps in effectively interpreting and solving Poisson-related probabilistic questions.

Complement Rule

The complement rule is a useful trick in probability that helps calculate the probability of an event by using the probability of its opposite.
This rule states: \[ P(A') = 1 - P(A) \] where \( P(A') \) is the probability of the complement of event \( A \).
In simple terms, if you know the probability that something does not happen, you can find the probability that it does happen by subtracting from 1.
In our exercise, the complement rule was used to find the probability of at least one automobile arriving in a minute.
Initially, we found the probability of the opposite event (no cars arriving) as \( P(X=0) = e^{-2} \).
Then, using the complement rule, we determined the probability of one or more cars arriving: \[ P(X \geq 1) = 1 - P(X = 0) \approx 0.8647 \] This shows how effective the complement rule is, simplifying computations involving multiple scenarios of probabilistic outcomes.