Math Problem Statement

p1=p0 in Poisson distribution what is the mean

Solution

In a Poisson distribution, the probability mass function is given by:

P(X=k)=λkeλk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}

where:

  • λ\lambda is the mean and variance of the distribution,
  • kk is the number of occurrences (non-negative integers),
  • ee is the base of the natural logarithm.

If P1=P0P_1 = P_0, we have:

λ1eλ1!=λ0eλ0!\frac{\lambda^1 e^{-\lambda}}{1!} = \frac{\lambda^0 e^{-\lambda}}{0!}

Simplifying the factorial terms: λeλ=eλ\lambda e^{-\lambda} = e^{-\lambda}

Divide both sides by eλe^{-\lambda} (since eλ0e^{-\lambda} \neq 0): λ=1\lambda = 1

Thus, the mean (λ\lambda) of the Poisson distribution in this case is 1.


Would you like further clarification or details? Here are 5 related questions to explore:

  1. What is the variance of a Poisson distribution, and how does it relate to the mean?
  2. How is a Poisson distribution used in real-life scenarios?
  3. What happens to the shape of the Poisson distribution as λ\lambda increases?
  4. How can you calculate probabilities for specific events using the Poisson formula?
  5. What is the relationship between the Poisson and exponential distributions?

Tip: In a Poisson distribution, the mean and variance are always equal. Keep this property in mind for quick checks!

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Math Problem Analysis

Mathematical Concepts

Probability
Poisson Distribution

Formulas

P(X = k) = (λ^k * e^(-λ)) / k!
λ = mean and variance in a Poisson distribution

Theorems

Properties of the Poisson Distribution

Suitable Grade Level

Grades 10-12