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Magazine Chapter 3: Problem 1
Suppose that \(X\) has a Poisson distribution. Determine the following probabilities when the mean of \(X\) is 4 and repeat for a mean of 0.4 : a. \(P(X=0) \quad\) b. \(P(X \leq 2)\) c. \(P(X=4)\) d. \(P(X=8)\)
Short Answer
Expert verified
With mean 4: (a) 0.0183, (b) 0.2381, (c) 0.1954, (d) 0.0031. With mean 0.4: (a) 0.6703, (b) 0.9856, (c) 0.0003, (d) 0.
Step by step solution
01
Understanding the Poisson Distribution Formula
The Poisson probability mass function (PMF) is given by the formula: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] where \( \lambda \) is the average rate (mean) of occurrences in the interval, \( e \) is approximately 2.71828, and \( k \) is the exact number of events we want to find the probability for.
02
Calculating \( P(X=0) \) with Mean 4
For \( P(X=0) \) when the mean \( \lambda = 4 \), substitute into the Poisson PMF: \[ P(X = 0) = \frac{4^0 \cdot e^{-4}}{0!} = e^{-4} \approx 0.0183 \]
03
Calculating \( P(X \leq 2) \) with Mean 4
For \( P(X \leq 2) \), sum the probabilities from \( X = 0 \) to \( X = 2 \): \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \] Compute each probability: \[ P(X = 0) = \frac{4^0 \cdot e^{-4}}{0!} = e^{-4} \] \[ P(X = 1) = \frac{4^1 \cdot e^{-4}}{1!} = 4e^{-4} \] \[ P(X = 2) = \frac{4^2 \cdot e^{-4}}{2!} = 8e^{-4} \] Sum these probabilities: \[ P(X \leq 2) = e^{-4} + 4e^{-4} + 8e^{-4} = 13e^{-4} \approx 0.2381 \]
04
Calculating \( P(X=4) \) with Mean 4
For \( P(X=4) \) when the mean \( \lambda = 4 \), use the PMF: \[ P(X = 4) = \frac{4^4 \cdot e^{-4}}{4!} = \frac{256e^{-4}}{24} = \frac{256}{24}e^{-4} \approx 0.1954 \]
05
Calculating \( P(X=8) \) with Mean 4
For \( P(X=8) \) when the mean \( \lambda = 4 \), use the PMF: \[ P(X = 8) = \frac{4^8 \cdot e^{-4}}{8!} = \frac{65536e^{-4}}{40320} \approx 0.0031 \]
06
Calculating \( P(X=0) \) with Mean 0.4
For \( P(X=0) \) when the mean \( \lambda = 0.4 \), substitute into the PMF: \[ P(X = 0) = \frac{0.4^0 \cdot e^{-0.4}}{0!} = e^{-0.4} \approx 0.6703 \]
07
Calculating \( P(X \leq 2) \) with Mean 0.4
For \( P(X \leq 2) \), sum probabilities from \( X = 0 \) to \( X = 2 \): \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \] Compute probabilities: \[ P(X = 0) = e^{-0.4} \] \[ P(X = 1) = \frac{0.4^1 \cdot e^{-0.4}}{1!} = 0.4e^{-0.4} \] \[ P(X = 2) = \frac{0.4^2 \cdot e^{-0.4}}{2!} = 0.08e^{-0.4} \] Sum: \[ P(X \leq 2) = (1 + 0.4 + 0.08)e^{-0.4} = 1.48e^{-0.4} \approx 0.9856 \]
08
Calculating \( P(X=4) \) with Mean 0.4
For \( P(X=4) \) when the mean \( \lambda = 0.4 \), use the PMF: \[ P(X = 4) = \frac{0.4^4 \cdot e^{-0.4}}{4!} = \frac{0.0256e^{-0.4}}{24} \approx 0.0003 \]
09
Calculating \( P(X=8) \) with Mean 0.4
For \( P(X=8) \) when the mean \( \lambda = 0.4 \), the probability is calculated as: \[ P(X = 8) = \frac{0.4^8 \cdot e^{-0.4}}{8!} \approx 0 \] This is effectively zero due to the small value of \( 0.4^8 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Mass Function
A Poisson distribution describes the probability of a given number of events happening in a fixed interval of time or space. The probability mass function (PMF) for the Poisson distribution helps us calculate the likelihood of a specific number of occurrences. The formula is given by:
This function is crucial because it allows us to calculate probabilities for different values. If you're working with a Poisson distribution, knowing how to apply this formula is key to solving probability problems effectively.
- \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \)
This function is crucial because it allows us to calculate probabilities for different values. If you're working with a Poisson distribution, knowing how to apply this formula is key to solving probability problems effectively.
Poisson Probability Calculation
To perform a Poisson probability calculation, we use the PMF to find the probability of certain numbers of events occurring.
For example, calculating \( P(X = 0) \) involves substituting into the formula:
Moreover, we can sum probabilities for a range of values, like calculating \( P(X \leq 2) \). We compute:
For example, calculating \( P(X = 0) \) involves substituting into the formula:
- \( P(X = 0) = \frac{\lambda^0 \, e^{-\lambda}}{0!} = e^{-\lambda} \)
Moreover, we can sum probabilities for a range of values, like calculating \( P(X \leq 2) \). We compute:
- \( P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \)
Mean of Poisson Distribution
The mean of a Poisson distribution, represented by \( \lambda \), is both a measure of central tendency and the rate parameter of the distribution. \( \lambda \) indicates the average number of events expected to occur in a specified interval.
Understanding this mean is important because it impacts all probability calculations. A higher \( \lambda \) means more expected occurrences, affecting our entire probability landscape.
For instance, if \( \lambda = 4 \), we expect around four events in our interval. This value guides our expectations and computations using the PMF.
Understanding this mean is important because it impacts all probability calculations. A higher \( \lambda \) means more expected occurrences, affecting our entire probability landscape.
For instance, if \( \lambda = 4 \), we expect around four events in our interval. This value guides our expectations and computations using the PMF.
- In practical terms, different \( \lambda \) values allow for modeling various scenarios, from rare occurrences (small \( \lambda \)) to frequent ones (large \( \lambda \)).
