# If 10 dice are rolled, what is the probability that their sum is within an interval?

Let $X$ be a random variable representing the roll of one die.

Then,

$$E(X) = \dfrac{1+2+3+4+5+6}{6} = 3.5$$

$$E(X^2) = \dfrac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \dfrac{91}{6}$$

$$Var(X) = E(X^2)-E(X)^2 = \dfrac{91}{6} - \dfrac{49}{4} = \dfrac{70}{24}$$

$$\sigma = \sqrt{Var(X)} = \sqrt{\dfrac{70}{24}} \approx 1.71$$

Let $Y$ be a random variable representing sum of the rolls of $N$ dice and $S$ representing the set of possible outcomes.

How can I find $P\{ A \leq Y \leq B\}$ for some $A,B \in S, A < B$?

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One could calculate the probability that the sum is $k$ explicitly, and add up over our interval. Not so pleasant to do by hand, but a task easy to write a program for.
For a possibly crude approximation, note that $Y$ has mean $35$ and variance $10\dfrac{70}{24}$. Although $10$ is not at all a large number, one can approximate $Y$ by a normal with the same mean and variance. It may be useful to cross one's fingers.
You can look at my comment on this question. I don't have the spreadsheet on this computer, but it isn't hard to reproduce. Then sum for your favorite $A$ and $B$