I ran into this expression while reading through a chapter a book:
$$\sum_{i=0}^{n}\frac{e^{-1500}1500^i}{i!} \ge 0.95$$
And they solve for $n$ and got $n = 1564$ from the expression, but there's no detail on how to get $n$. Can someone explain how to get that number for $n$?
The book is Discrete Event System Simulation by Jerry Banks. This is in the 5th Edition, Chapter 6 Example 6.15, if anyone wants to look this up.
On
From a purely algebraic point of view $$\sum_{i=0}^{n}\frac{e^{-1500}1500^i}{i!} =\frac{\Gamma (n+1,1500)}{\Gamma (n+1)}$$ So, you could consider that we look for the zero of function $$f(x)=\frac{\Gamma (x+1,1500)}{\Gamma (x+1)}-0.95$$ which is not very well conditioned. Better will be to search for the zero of $$g(x)=\log \left(\frac{\Gamma (x+1,1500)}{\Gamma (x+1)}\right)-\log(0.95)$$ Remembering that $\frac{\Gamma (m,m)}{\Gamma (m)} < \frac 12$, let us use Newton method with $x_0=1500$. Since $g(x_0) <0$ and $g''(x_0) <0$, by Darboux theorem, we shall not face any overshoot of the solution which is $x=1563.485020$.
Just for the fun of it, ask Wolfram Alpha to plot $g(x)$ for $1500 \leq x \leq 1600$.
The probability distribution is the Poisson distribution with $\lambda=1500$. For large $\lambda$ the distribution is approximately normal with mean $\lambda$ and variance $\lambda$, so standard deviation $\sigma=\sqrt \lambda$. For a one-sided normal distribution you have $5\%$ of the area above mean + $1.648 \sigma$, which here is $1500+1.648 \sqrt {1500} \approx 1564$