Probabilty Distribution (Poisson Distribution)

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SUMMARY

The discussion centers on the application of the Poisson Distribution to model the probability of rain in a town during October, where it rains on average 10.3 days. The probability of rain on any given day is calculated as 0.332, derived from the formula λ/31, where λ is the expected number of rainy days. The probability distribution for the number of rainy days, X, is expressed as Pr(X=x) = [(0.332^x)*(e^-0.332)] / (x!). This formulation confirms the use of the Poisson Distribution for independent events occurring over a fixed interval.

PREREQUISITES
  • Understanding of Poisson Distribution and its applications
  • Familiarity with the concept of expected value in probability
  • Basic knowledge of exponential functions and factorials
  • Ability to manipulate mathematical expressions and equations
NEXT STEPS
  • Study the properties and applications of the Poisson Distribution in real-world scenarios
  • Learn how to derive and interpret the expected value for different probability distributions
  • Explore the relationship between the Poisson Distribution and the Exponential Distribution
  • Practice solving problems involving Poisson processes and independent events
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Students studying probability theory, statisticians, and anyone interested in modeling independent events using the Poisson Distribution.

Daniel323
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Homework Statement


It rains on 10.3 days in the town in October on average. Let X denote the number of days in October on which it rains. Assume that rain falling on different days can be treated as independent events. (31 days in October).

Write down an expression for the probability that it rains on a given day in October, and hence state the probability distribution of X.

Homework Equations


I've identified the problem to be a Poission Distribution, with Pr(X=x) = [ (λ^x)*(e^-λ) ] / (x!)

The Attempt at a Solution


The probability of it raining on any given day in October = Expected no. of days it will rain / total no. of days = 10.3/31 = 0.332

Hence probability distribution of X would be Pr(x) = [ (0.332^x)*(e^-0.332) ] / (x!)

Sorry this is my first attempt at such a question and am not sure if I have done it right, or if I have even gone about it the right way.

If anyone knows how to tackle this problem, please let me know if I am right or wrong.

Thanks,
Daniel
 
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