I have a problem with the following task: If you roll one 6 sided die 1000 times, what is probability that sum of fallen numbers will be between 3000 and 4000?
So,
mean = 3.5
std. dev = 1.71
I tried using the following formula but then I encounter $\phi(9.26)$ and I don't know what to do after. I'd aprreciate any help with this problem, thanks :).
$\phi$($\frac{4000-3500} {\sqrt{1000}*1.71}$) - $\phi$($\frac{3000-3500} {\sqrt{1000}*1.71}$)
I assume that the rolls are independent. Let $X$ denote the sum of the rolled numbers after 1000 rolls. Then, indeed by the central limit theorem, the distribution of $(X-3500)/(1.71\sqrt{1000})$ will be approximately standard normal. All your calculations so far are correct. The last step missing is to evaluate the cdf at those points. First, the symmetry of the standard normal distribution implies $\phi(-x)=1-\phi(x)$ for $x\in\mathbb{R}$. Hence, $$\phi(\frac{4000-3500} { 1.71\sqrt{1000}})-\phi(\frac{4000-3500} {1.71\sqrt{1000}})= \phi(\frac{500} {1.71\sqrt{1000}})-\phi(\frac{-500} {1.71\sqrt{1000}}) \\ = 2\phi(\frac{500} {1.71\sqrt{1000}})-1 = 2\phi(9.25)-1.$$ Unfortunately the cdf the normal distribution cannot be written down using elementary functions. So, we have to rely on numerical approximations which are usually implemented in software packages (e.g. pnorm(x) in R) or written down as tables in statistics books. Note that 9.25 is more than 9 standard deviations (=1) away from the mean (=0). We have that $\phi(3)\approx 0.9986501$ or $\phi(5)\approx0.9999997$ are already quite close to 1. Remembering that the tails of the normal distribution get thinner and thinner the more we move away from the mean, $\phi(9.25)$ will be a number very, very close to 1. Hence, the probability is approximately $2\phi(9.25)-1\approx 1.$
Note that most software packages will evaluate $\phi(9.25)$ simply as 1 (because there are only a limited numbers of digits in floating point numbers available).