Problem 41 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

HIGHWAY ACCIDENTS A report models the number of automobile accidents on a particular highway as a random variable with a Poisson distribution. Suppose it is found that on average, there is an accident every 10 hours. a. Find the probability that there are no accidents on this highway during a randomly selected 24-hour period. b. Find the probability that there is at least one accident on this highway during a randomly selected 12 -hour period. c. Find the probability that there are no accidents on this highway during a randomly selected hour.

Step-by-Step Solution

Answer

a) 0.0907, b) 0.6988, c) 0.9048.

Step 1: Define the Poisson parameter

The Poisson distribution is characterized by the parameter \( \lambda \), which represents the average rate of occurrence within a given time period. Given that there is one accident every 10 hours, \( \lambda \) is \( \frac{1}{10} = 0.1 \) accidents per hour.

Step 2: Determine \( \lambda \) for the given periods

For part (a), the time period is 24 hours, so \( \lambda = 24 \times 0.1 = 2.4 \) accidents in 24 hours. For part (b), the time period is 12 hours, so \( \lambda = 12 \times 0.1 = 1.2 \) accidents in 12 hours. For part (c), the time period is 1 hour, so \( \lambda = 1 \times 0.1 = 0.1 \) accidents in 1 hour.

Step 3: Apply the Poisson formula for part (a)

The Poisson probability of having \( k \) accidents in a period with parameter \( \lambda \) is given by \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]. For zero accidents in 24 hours (k=0), \[ P(X = 0) = \frac{2.4^0 e^{-2.4}}{0!} = e^{-2.4} \]. Calculate this to find the probability.

Step 4: Calculate the numerical value for part (a)

Using the formula, \( P(X = 0) = e^{-2.4} \). Estimate the value of \( e^{-2.4} \) using a calculator: \( e^{-2.4} \approx 0.0907 \). Thus, the probability is approximately 0.0907.

Step 5: Apply and calculate for part (b)

The probability of at least one accident in a 12-hour period is \[ P(X \geq 1) = 1 - P(X = 0) \]. Similarly, \[ P(X = 0) = \frac{1.2^0 e^{-1.2}}{0!} = e^{-1.2} \]. Calculate \( e^{-1.2} \) to find \( P(X \geq 1) = 1 - e^{-1.2} \approx 1 - 0.3012 = 0.6988 \).

Step 6: Apply and calculate for part (c)

The probability of no accident in a 1-hour period is \[ P(X = 0) = \frac{0.1^0 e^{-0.1}}{0!} = e^{-0.1} \]. Calculate \( e^{-0.1} \): \( e^{-0.1} \approx 0.9048 \).

Key Concepts

Poisson Distribution

A Poisson distribution is a probability distribution used to model the number of events occurring within a fixed interval of time or space when these events happen with a known constant mean rate and independently of the time since the last event. This characteristic is measured by the parameter lambda (\( \lambda \)).

Lambda (\( \lambda \)) is both the mean and variance of the distribution. For example, in the highway accidents problem, if there's an average of one accident every 10 hours, then \( \lambda \) is \frac{1}{10} = 0.1 accidents per hour. This tells us how many accidents we expect per hour on average.

The Poisson formula is used to find the probability of observing \(k\) events (accidents in this case) over the time period, given by:
< br>\[ P(X = k) \= \frac{\lambda^k e^{-\lambda}}{k!} \] where:

\item \( P(X = k) \) is the probability of \( k \) events occurring in a fixed interval.
\item \( \lambda \) (lambda) is the average rate of events (accidents per hour).
\item \( k \) is the number of events (accidents) we are calculating the probability for.
\item \( e \) is the base of the natural logarithm (approximately equal to 2.71828).

Probability

Probability measures the likelihood that an event will occur. It ranges from 0 (impossible event) to 1 (certain event). In Poisson distribution problems, we calculate the probability of a specific number of events happening within a certain time frame.

For example, in part (a) of the exercise, the probability of zero accidents in 24 hours (\( k = 0 \)) can be calculated with the formula:

\[ P(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} \= e^{-2.4} \]

This is because \( \lambda \) for 24 hours is calculated as \( 24 \times 0.1 \) which is 2.4. Thus, \( P(X = 0) \) simplifies to \( e^{-2.4} \), yielding approximately 0.0907.

Part (b) asks for the probability of at least one accident in 12 hours. We first calculate the probability of zero accidents and subtract it from 1:

\[ P(X \geq 1) = 1 - P(X = 0) \]

Given \( \lambda \) for 12 hours is 1.2, we find:

\[ P(X = 0) = e^{-1.2} \approx 0.3012 \]

Therefore, the probability of at least one accident is:
\[ P(X \geq 1) \approx 1 - 0.3012 = 0.6988 \]

In part (c), we find the probability of no accidents in 1 hour with \( \lambda = 0.1 \). Thus:

\[ P(X = 0) = e^{-0.1} \approx 0.9048 \]

Understanding probability enables solving such events accurately using Poisson distribution.

Random Variable

A random variable is a variable whose possible values are outcomes of a random phenomenon. It's crucial in calculating probabilities in Poisson distribution problems.

In our highway accident scenario, the random variable \( X \) represents the number of accidents occurring in a specified time period.

Three key points about random variables in the context of Poisson distribution include:

\item **Discrete Nature**: Poisson random variables are discrete; they can take on integer values like 0, 1, 2, etc.
\item **Mean and Variance**: For a Poisson random variable, the mean (expected value) and variance are both equal to \( \lambda \).

For instance, in part (a), the random variable is defined over a 24-hour period where \( \lambda = 2.4 \). This means on average, we expect 2.4 accidents in 24 hours.
\item **Independence of Events**: The occurrence of one event (accident) does not affect another. Each hour's accident occurrence is independent of others in Poisson processes.

Understanding that \( X \) is random helps in grasping why and how we calculate these probabilities; we rely on the average rate (\( \lambda \)) and the key event outcomes (\( k \)).