Using Central Limit Theorem to determine number of simulations required I am trying to determine how many simulations are adequate to estimate number of claims in the next twelve months using a homogenous poisson process.
What I did is simulate 1000 times, find the mean and standard deviation of those simulations, and used CLT to find N (5% confidence interval), what I found is a really small number (<1), this can't be possible. How come CLT doesn't work in this case? is it because 1000 is too small?
I simulated 1000 times and I have a vector with 1000 elements. I calculated the average of these elements and the standard deviation, and I used the formula:
(0.05*Average)/(standard deviation/sqrt(n))=1.96 for a 5% CI with 95% probability. 
edit: I followed the formula in the photo, when I calculated the standard deviation, it is very large (around 80).
 A: Let's say you've made several experiments and got your means as $X_1,X_2,...X_N$. The essence of CLT is approximating sample mean as normal, i.e. $\bar{X}=\frac{1}{N}\sum_{i=1}^N{X_i}$ is assumed to be normal. And, rest assured that $1000$ is not a small number in this case. So, your estimated sample mean is approximately normally distributed. 
If you want to have a $95 \%$ CI for your estimate, you can just add and subtract $2\sigma_{\bar{X}}$ to your observed mean. Here, if you don't know the true deviation of $X_i$ (if you know it,  $\sigma_{\bar{X}}=\frac{\sigma_{X_i}}{\sqrt{N}}$, you can also estimate it using your sample set. 
Your question, "what should $N$ be for 95% CI?", is not complete. For every $N$, you can have a 95 % CI. Of course, for small $N$, it'll be large since you're more uncertain and vice versa. Here, you can adjust $N$ to limit the width of your CI, i.e. what should N be such that my CI has width = $w$? For this, we first calculate the width (for 95 % CI), and that is $4\sigma_{\bar{X}}=4\frac{\sigma}{\sqrt{N}}$, and you'll have $N\approx \left(\frac{4\sigma}{w}\right)^2$
