91Ó°ÊÓ

How many cars show up? For Monday through Thursday when there isn't a holiday, the average number of vehicles that visit a particular retailer between \(2 \mathrm{pm}\) and \(3 \mathrm{pm}\) each afternoon is \(6.5,\) and the number of cars that show up on any given day follows a Poisson distribution. (a) What is the probability that exactly 5 cars will show up next Monday? (b) What is the probability that \(0,1,\) or 2 cars will show up next Monday between \(2 \mathrm{pm}\) and \(3 \mathrm{pm} ?\) (c) There is an average of 11.7 people who visit during those same hours from vehicles. Is it likely that the number of people visiting by car during this hour is also Poisson? Explain.

Step by step solution

01

Understand the Poisson distribution

The Poisson distribution is useful for modeling the number of times an event occurs in a fixed interval of time or space. The average rate of occurrence (mean) is needed, which in this case is given as 6.5 cars per hour.

02

Calculate Probability for Part (a)

For Part (a), we want to find the probability that exactly 5 cars show up. The Poisson probability formula is \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \) where \( \lambda \) is the average rate (6.5) and \( k \) is the number of events (5).Substitute the values:\[ P(X=5) = \frac{e^{-6.5} \times 6.5^5}{5!} \] Calculate the actual numerical value using a calculator.

03

Calculate Probability for Part (b)

For Part (b), we need to calculate the probability that 0, 1, or 2 cars show up. Calculate each separately and sum the probabilities.- \( P(X=0) = \frac{e^{-6.5} \times 6.5^0}{0!} \)- \( P(X=1) = \frac{e^{-6.5} \times 6.5^1}{1!} \)- \( P(X=2) = \frac{e^{-6.5} \times 6.5^2}{2!} \)Add these probabilities together to get the total probability.

04

Analyze Part (c)

Consider whether the same Poisson process applies to people. While the number 11.7 is also an average rate, the assumption that the arrivals are independent events and the rate is constant over the interval must hold true for the Poisson distribution. If there's a direct proportionality to the number of cars (e.g., a fixed number of people per car), then a Poisson distribution may not accurately model it. Thus, we determine it's less likely to be Poisson-distributed due to variability in people per car.

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Key Concepts

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

Probability Calculation

To understand how to calculate probabilities using the Poisson distribution, we rely on the formula

• \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \)

This formula helps us determine the probability of observing exactly \( k \) events happening in a fixed time interval, given an average rate \( \lambda \) of occurrence.
For example, when calculating the probability of exactly 5 cars arriving, use an average rate \( \lambda = 6.5 \) and \( k = 5 \).
Here's how you calculate it:

• Substitute \( \lambda = 6.5 \) and \( k = 5 \) into the formula.
• Use a calculator to determine the result of \( P(X=5) = \frac{e^{-6.5} \times 6.5^5}{5!} \).

This process provides the exact chance of having 5 cars show up, crucial for situations requiring precise forecasting.

Average Rate

In Poisson distribution problems, understanding the average rate is pivotal.
The average rate, or \( \lambda \), represents the expected number of occurrences within a fixed span of time or space.
For the problem in hand, the retailer observes an average of 6.5 cars arriving per hour.

• This rate is stable and calculated over similar conditions (such as time and day, e.g., Monday through Thursday).

Knowing the mean value aids in predictive modeling and planning, allowing businesses or researchers to gauge regular patterns.
Average rates may differ when analyzing different elements, such as the number of people, and require analyzing other conditions and variables that might affect these rates.

Independent Events

For a Poisson distribution, one crucial requirement is that the events are independent.
This means that the arrival of one car is not affected by the arrival or absence of another.

• Events take place independently of one another in both time and influence.

The independence of arrivals ensures that the average rate remains consistent over time.
If one event could affect another (for instance, due to traffic congestion that varies at different times), it violates one basic assumption of the Poisson model.
Considerations of independence help affirm the distribution's appropriateness but identifying breaches could lead to a more suitable alternative model.

Distribution Assumptions

Adopting a Poisson distribution comes with specific assumptions:

• Events are independent.
• The average rate (\( \lambda \)) holds constant over the given interval.
• No two events occur simultaneously.

Analyzing the problem, if these conditions aren't met, the Poisson model may not fit.
For example, in evaluating whether human visitors during this time period also fit a Poisson distribution, consider:

• Is there a consistent number of individuals-per-car factor?
• Do multiple arrivals or certain events skew average rates?

If these adjustments occur, the variability must be considered and may suggest another distribution model better represents the data, like binomial or normal distributions, based on other characteristics, such as spread and event causation.