I want to determine how many instances of a random variable I need, so that the probability that the sum of them is bigger than 8000 is over 95%.
My statistics lectures have been a few years ago, so I cannot find the right approach. I think it's related to the sampling size determination, however I could not apply the formulas to this problem.
Let $X$ have the distribution:
$P(X=1)=0.53925,$
$P(X=2)=0.03572,$
$P(X=3)=0.00039,$
$P(X=4)=0.05596,$
$P(X=5)=0.00004,$
$P(X=6)=0.00002,$
$P(X=7)=0.00008,$
$P(X=8)=0.36853.$
I want to determine the smallest n that fullfills:
$P((\sum_{i=0}^n X)\ge 8000) \ge 0.95.$
I am happy about any kind of help.
You are expected to use the normal approximation. For the sum of random variables, the mean of the sum is the sum of the means. The variance of the sum is the variance of the means. Use that variance to compute a standard deviation.