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Poisson distribution is defined as: $$P(X=i)=\frac{\lambda^i}{i!}e^{-\lambda}$$ How do you find out upper limit for this distribution ?

Lets say that $\lambda=0.7$. My question is that is there any other way for solving $k$ from this equation than using sum like this: $$P(X≈1)=\sum_{i=0}^{k}\frac{0.7^i}{i!}e^{-0.7}$$ For example with $k=11$ i would consider this close enough to one. $$\sum_{i=0}^{11}\frac{0.7^i}{i!}e^{-0.7}≈1$$

Tuki
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  • You would need to select some negligible probability tolerance. – Ian Sep 27 '17 at 20:32
  • Then you can use the Lagrange remainder to show $P(X \geq k) \leq \lambda^k/k!$ and set that bound to be below your tolerance. – Ian Sep 27 '17 at 20:40

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