Chapter 12: Problem 82
Suppose \(X\) is Poisson distributed with parameter \(\lambda=1.2 .\) Find the probability that \(X\) is at most 3 .
Short Answer
Expert verified
The probability is approximately 0.9661.
Step by step solution
01
Understanding the problem
We are given a Poisson distribution with the parameter \( \lambda = 1.2 \). We need to find the probability that the random variable \( X \), which follows this distribution, is at most 3. This is written mathematically as \( P(X \leq 3) \).
02
Applying the Poisson Probability Mass Function
The probability mass function for a Poisson distribution is given by: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where \( \lambda = 1.2 \) and \( k \) is the number of occurrences. We need to calculate this probability for \( k = 0, 1, 2, \text{ and } 3 \) and then sum them up to find \( P(X \leq 3) \).
03
Calculating for k=0
Substitute \( k = 0 \) into the Poisson probability mass function: \[ P(X = 0) = \frac{e^{-1.2} \times 1.2^0}{0!} = e^{-1.2} \] Using a calculator, we find \( P(X = 0) \approx 0.3012 \).
04
Calculating for k=1
Substitute \( k = 1 \) into the function: \[ P(X = 1) = \frac{e^{-1.2} \times 1.2^1}{1!} = 1.2e^{-1.2} \] Using a calculator, \( P(X = 1) \approx 0.3614 \).
05
Calculating for k=2
Substitute \( k = 2 \) into the function: \[ P(X = 2) = \frac{e^{-1.2} \times 1.2^2}{2!} = \frac{1.44 e^{-1.2}}{2} \] Using a calculator, \( P(X = 2) \approx 0.2168 \).
06
Calculating for k=3
Substitute \( k = 3 \) into the function: \[ P(X = 3) = \frac{e^{-1.2} \times 1.2^3}{3!} = \frac{1.728 e^{-1.2}}{6} \] Using a calculator, \( P(X = 3) \approx 0.0867 \).
07
Summing the Probabilities
Now, sum the probabilities calculated for \( k = 0, 1, 2, \text{ and } 3 \): \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \approx 0.3012 + 0.3614 + 0.2168 + 0.0867 \approx 0.9661 \]
08
Conclusion
The probability that \( X \) is at most 3 is approximately 0.9661. This means there is a 96.61% chance that \( X \) will be 3 or less.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Mass Function
The Probability Mass Function (PMF) is a fundamental concept when dealing with discrete probability distributions, such as the Poisson distribution. For a Poisson-distributed random variable, the PMF provides the probability of observing a specific number of events, say \(k\), in a fixed interval of time or space. In mathematical terms, it is expressed as: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where:
- \(e\) is the base of the natural logarithm, approximately equal to 2.71828.
- \(\lambda\) is a positive real number, known as the rate parameter.
- \(k!\) (k factorial) is the product of all positive integers up to \(k\).
Parameter Lambda
In the context of a Poisson distribution, the parameter \(\lambda\) is a pivotal term because it characterizes the distribution. It represents the average number of times an event occurs within a given timeframe or area. Understanding what \(\lambda\) stands for can help you interpret results much more effectively:
- When \(\lambda\) is small, events occur less frequently.
- When \(\lambda\) is larger, events occur more often.
Summation of Probabilities
To find the probability of a Poisson-distributed variable being at most a certain value, such as 3 in this instance, involves summing multiple probabilities. Individual PMF values for each \(k\) are combined in such a way to yield cumulative probabilities:
- Begin by calculating probabilities for \(k = 0, 1, 2, \) and \(3\).
- Then, sum these probabilities: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]