A Poisson distribution is such that the probability is the same for .pdf
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In a Poisson distribution, the probabilities for x = 1 and x = 2 are equal when the mean u is set to 1. The calculations show that the probabilities for p(0) and p(1) result in both being e^(-1), approximately 0.3679.
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![A Poisson distribution is such that the probability is the same for X = 1 and X = 2. What is this
probability?
Solution
Note that for a Poisson distribution,
P(x) = u^x e^(-u)/x!
Thus,
P(0) = e(-u)
P(1) = u (e^-u)
Thus,
u(e^-u) = e^(-u)
e^(-u) (1-u) = 0
Thus,
u = 1
Thus,
P(0) = P(1) = e^(-1) = 0.367879441 [ANSWER]](https://image.slidesharecdn.com/apoissondistributionissuchthattheprobabilityisthesamefor-230703085520-425f48d1/85/A-Poisson-distribution-is-such-that-the-probability-is-the-same-for-pdf-1-320.jpg)
A Poisson distribution is such that the probability is the same for .pdf
- 1. A Poisson distribution is such that the probability is the same for X = 1 and X = 2. What is this probability? Solution Note that for a Poisson distribution, P(x) = u^x e^(-u)/x! Thus, P(0) = e(-u) P(1) = u (e^-u) Thus, u(e^-u) = e^(-u) e^(-u) (1-u) = 0 Thus, u = 1 Thus, P(0) = P(1) = e^(-1) = 0.367879441 [ANSWER]