Rolling dice Probability that Sum

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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

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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.

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Let's $x_1,\ldots x_{20}\in\{1,2,3,4,5,6\}$. For a exact solution $$ P\left(30\leq x_1+\ldots+x_{20}\leq 40 \right)=\frac{1}{20^6}\sum_{30\leq N\leq 40}\quad\sum_{x_1+\ldots+x_{20}\leq N} \frac{N!}{x_1!\cdot \ldots \cdot x_{10}!} $$