Rolling dice Probability that Sum If 10 pairs of fair dice are rolled, approximate the probability that the sum of the
values obtained (which ranges from 20 to 120) is between 30 and 40 inclusive.
I dont know where to start with this one. I have been looking all over the web for example, but nothing i find is applicable for finding the sum of numbers.
any advice would be great
 A: We are tossing $20$ dice. Let $X_i$ be the number showing on the $i$-th die. We are interested in the random variable $X_1+\cdots+X_{20}$.
The $X_i$ are independent, mean $\frac{7}{2}$, variance $\frac{35}{12}$ (please verify).
So $Y$ has mean $\mu=20\cdot\dfrac{7}{2}$, variance $\sigma^2=20\cdot \dfrac{35}{12}$.
We cross our fingers and use the normal approximation. This is because $Y$ is the sum of a not terribly small number of independent identically distributed respectable random variables $X_i$. 
So we want the probability that if $W$ is normal with mean $\mu$ and variance $\sigma^2$, then $30\le W\le 40$.
The rest depends to some degree on whether you are expected to use the continuity correction. 
Without the continuity correction, we want the probability that a standard normal $Z$ satisfies $\frac{30-\mu}{\sigma}\le Z\le \frac{40-\mu}{\sigma}$.
With continuity correction, replace $40$ by $40.5$, and $35$ by $34.5$.
If you have any difficulty finishing, please leave a comment. 
A: Let's $x_1,\ldots x_{20}\in\{1,2,3,4,5,6\}$. For a exact solution 
$$
P\left(30\leq x_1+\ldots+x_{20}\leq 40  \right)=\frac{1}{20^6}\sum_{30\leq N\leq 40}\quad\sum_{x_1+\ldots+x_{20}\leq N} \frac{N!}{x_1!\cdot \ldots \cdot x_{10}!}
$$
