Independent solution

How to solve this Poisson Distribution question

Setup

Setup

For a Poisson count with mean λ, the complement of a zero count is the probability of one or more events.

Pr⁡(N≥1)=1−Pr⁡(N=0)=1−e−λ\Pr(N\ge1)=1-\Pr(N=0)=1-e^{-\lambda}

Model

Model

Apply the reported probability multiplier to the two means and isolate the second zero-count probability.

1−e−λ2=2(1−e−0.5)1-e^{-\lambda_2}=2(1-e^{-0.5})
e−λ2=2e−0.5−1e^{-\lambda_2}=2e^{-0.5}-1

Compute

Compute

Take the negative natural logarithm of the positive complement.

λ2=−ln⁡(2e−0.5−1)\lambda_2=-\ln(2e^{-0.5}-1)
λ2=1.5461752701\lambda_2=1.5461752701

Answer

Answer

The second Poisson mean rounds to 1.55.

1.55(D)\boxed{1.55\quad\text{(D)}}