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Suppose that the average number of cars abandoned weekly on a certain highway is 2.2. Approximate the probability that there will be (a) no abandoned cars in the next week; (b) at least 2 abandoned cars in the next week.

Short Answer

Expert verified
The approximate probabilities are: (a) no abandoned cars in the next week: \(0.1108\) or \(11.08\%\) and (b) at least 2 abandoned cars in the next week: \(0.6454\) or \(64.54\%\).

Step by step solution

01

Understand Poisson Distribution

The Poisson distribution is a probability distribution which represents the number of times an event (in this case, abandoned cars) occurs over a specific time period (a week). The key formula for the Poisson distribution probability mass function (PMF) is as follows: \(P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}\), where \(X\) is the number of events (abandoned cars), \(k\) is the desired number of events, \(\lambda\) is the average rate of events (2.2 abandoned cars/week), and \(e\) is the Euler's number - approximately 2.71828.
02

Calculate the probability of no abandoned cars in the next week

For part (a), the student needs to find the probability that there will be no abandoned cars in the next week. This means finding the probability when \(k = 0\). Use the Poisson distribution formula with \(\lambda = 2.2\) and \(k = 0\): \(P(X = 0) = \frac{e^{-2.2}(2.2)^0}{0!}\)
03

Simplify the expression for part (a)

Now, simplify the expression for \(P(X = 0)\): \(P(X = 0) = \frac{e^{-2.2}(1)}{1} = e^{-2.2} \approx 0.1108\) So, the probability of no abandoned cars in the next week is approximately 0.1108 or 11.08%.
04

Calculate the probability of at least 2 abandoned cars in the next week

For part (b), the student needs to find the probability that there will be at least 2 abandoned cars in the next week. This means finding the probability for \(k = 2, 3, 4, ...\). However, it's easier to calculate the complementary probability for \(k = 0\) and \(k = 1\) and then subtract the sum of those probabilities from 1. Use the Poisson distribution formula with \(\lambda = 2.2\) and \(k = 1\): \(P(X = 1) = \frac{e^{-2.2}(2.2)^1}{1!}\)
05

Simplify the expression for part (b)

Now, simplify the expression for \(P(X = 1)\): \(P(X = 1) = \frac{e^{-2.2}(2.2)}{1} = 2.2e^{-2.2} \approx 0.2438\) So, the probability of 1 abandoned car in the next week is approximately 0.2438 or 24.38%.
06

Find the probability of at least 2 abandoned cars

Finally, to find the approximate probability of at least 2 abandoned cars in the next week, subtract the probabilities of 0 and 1 abandoned cars from 1: \(P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 1 - 0.1108 - 0.2438 \approx 0.6454\) The probability of at least 2 abandoned cars in the next week is approximately 0.6454 or 64.54%.

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Key Concepts

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

Probability Mass Function
The concept of the Probability Mass Function (PMF) is vital when dealing with discrete probability distributions such as the Poisson distribution. The PMF helps us find the probability of a discrete random variable exactly taking a specific value. In the case of the Poisson distribution, which is suitable for counting the number of events occurring within a fixed interval of time or space (like our car abandonment problem), the PMF is expressed as:\[P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}\]Here's what each component means:
  • \(X\) is the random variable, representing the number of abandoned cars.
  • \(k\) is the specific number of abandoned cars we are calculating the probability for.
  • \(\lambda\) is the average number of events (in this case, a weekly average of 2.2 cars).
  • \(e\) is approximately 2.71828, a mathematical constant and the base of the natural logarithm.
  • \(k!\) represents the factorial of \(k\).
By plugging our values into this formula, we can calculate probabilities for different numbers of abandoned cars in a week.
Expected Value
The Expected Value, in the context of the Poisson distribution, is relatively straightforward. It provides us with a measure of the center of the distribution, representing the average number of occurrences over a specified interval. For a Poisson distribution, the expected value is essentially the parameter \(\lambda\), which is also the mean of the distribution.The expected value of a Poisson distribution is given by:\[E(X) = \lambda = 2.2\]This means that on average, 2.2 cars are abandoned every week on the highway. When dealing with repeated processes like this, understanding the expected value helps us set our expectations on likely outcomes and plan accordingly. It's a powerful tool in decision making. For instance, city planners might use such statistics to deploy resources efficiently.
Complementary Probability
The concept of Complementary Probability revolves around the idea that the sum of probabilities of all possible outcomes in a sample space equals 1. In practice, it can be useful to calculate the probability of the complementary event, especially when it is simpler than directly finding the probability of the event in question.For example, in the exercise given, calculating the probability of at least 2 abandoned cars (\(k \geq 2\)) by directly summing the infinite possibilities isn’t practical. Instead, we calculate
  • the probability of having 0 abandoned cars
  • the probability of having 1 abandoned car
and then subtract these from 1 to find the probability of 2 or more abandoned cars.Mathematically, it is expressed as:\[P(X \geq 2) = 1 - P(X = 0) - P(X = 1)\]Using complementary probability simplifies our calculations while ensuring accuracy. It's an efficient way to find probabilities for events with many scenarios, like in large population studies or long-term predictions.

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