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Consider a Poisson distribution with \(\mu=3\). a. Write the appropriate Poisson probability function. b. Compute \(f(2)\). c. Compute \(f(1)\). d. Compute \(P(x \geq 2)\).

Short Answer

Expert verified
a. Use \(P(x; \mu)=\frac{e^{-\mu} \mu^x}{x!}\). b. \(f(2) \approx 0.112\). c. \(f(1) \approx 0.672\). d. \(P(x \geq 2) \approx 0.104\).

Step by step solution

01

Poisson Probability Function

The probability function for a Poisson distribution is written as: \[ P(x; \mu) = \frac{e^{-\mu} \mu^x}{x!} \] where \(x\) is the number of occurrences, \(\mu\) is the average rate of occurrence, and \(e\) is the base of the natural logarithm (approximately 2.71828).
02

Compute f(2)

To find the probability of exactly 2 occurrences, substitute \(x = 2\) and \(\mu = 3\) into the formula: \[ f(2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{e^{-3} \cdot 9}{2} \approx \frac{0.224 \times 9}{2} = 1.008 \times 0.5 = 0.112 \] Hence, \(f(2) \approx 0.112\).
03

Compute f(1)

For \(f(1)\), substitute \(x = 1\) and \(\mu = 3\) into the formula: \[ f(1) = \frac{e^{-3} \cdot 3^1}{1!} = \frac{e^{-3} \cdot 3}{1} = 0.224 \times 3 \approx 0.672 \] Therefore, \(f(1) \approx 0.672\).
04

Compute P(x ≥ 2)

To find \(P(x \geq 2)\), use the complement rule: \[ P(x \geq 2) = 1 - P(x < 2) = 1 - (f(0) + f(1)) \] First, compute \(f(0):\) \[ f(0) = \frac{e^{-3} \cdot 3^0}{0!} = e^{-3} \approx 0.224 \] Then, use \(f(1) \, (\approx 0.672)\) from Step 3. So, \[ P(x \geq 2) = 1 - (0.224 + 0.672) = 1 - 0.896 = 0.104 \] Thus, \(P(x \geq 2) \approx 0.104\).

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Key Concepts

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

Probability Function
The Poisson probability function is fundamental for understanding random events with a known average rate. In simple terms, it helps determine how likely it is for a specific number of events to occur over a designated period. The mathematical expression for the Poisson probability function is given by:\[P(x; \mu) = \frac{e^{-\mu} \mu^x}{x!}\]where:
  • \(P(x; \mu)\) represents the probability of observing \(x\) events.
  • \(e\) is approximately 2.71828, known as Euler's number, the base of the natural logarithm.
  • \(\mu\) stands for the average rate of occurrence.
  • \(x!\) is the factorial of \(x\), representing the product of all positive integers up to \(x\).
This probability function is particularly useful when dealing with events that occur independently and are equally likely across a specific time frame. For example, predicting how many cars will pass through a toll bridge each minute can be modeled using this.Petting this theory into practice, when you want to know, say, the probability of exactly 2 events occurring when the average rate is 3, substitute these values into the Poisson formula.
Complement Rule
The complement rule in probability is a simple but handy trick. It's often easier to calculate the probability of something not happening, and then subtract that from 1. In the context of Poisson distribution problems, this becomes clear when you're asked to find probabilities like \(P(x \geq 2)\), which is quite tedious directly.To use the complement rule for \(P(x \geq 2)\):\[P(x \geq 2) = 1 - P(x < 2)\]This implies calculating \(P(x < 2)\), which includes \(P(0)\) and \(P(1)\). Break things down as:
  • Calculate \(P(0)\) using the probability function. With a mean \(\mu = 3\), the result here is \(\approx 0.224\).
  • Use your previous computation from the probability function for \(P(1)\), which was found as \(\approx 0.672\).
  • Plug these values in: \(P(x < 2) = P(0) + P(1) = 0.224 + 0.672 = 0.896\)
Now subtract from 1: \(P(x \geq 2) = 1 - 0.896 = 0.104\). This method smartly flips the complexity by dealing with smaller probabilities and then using a simple subtraction.
Rate of Occurrence
Understanding the rate of occurrence is crucial when dealing with Poisson distributions. The parameter \(\mu\), commonly used in the Poisson formula, represents this rate of occurrence—specifically, it's the average number of events happening in a given time interval or space.For example, if \(\mu = 3\), this means, on average, three events are expected to occur over the chosen time period. It's essential to have accurate data for this average rate to precisely predict probabilities of different numbers of occurrences.The rate of occurrence is directly linked to the probability of each possible outcome:
  • A higher \(\mu\) increases the likelihood of observing more events within the time frame.
  • Meanwhile, a lower \(\mu\) suggests events will be less frequent.
Each specific scenario where Poisson distribution is applicable should consider this rate. Examples include predicting website hits per hour or the frequency of rare genetic mutations in a sample population. Consequently, knowing and applying the correct rate of occurrence allows for reliable probabilistic modeling and insightful predictions.

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