# Rolling a die, probability of the sum

I need help with a particular problem in my statistics textbook.

Imagine rolling a normal die with six sides $100$ times. I need to find the probability that the sum of the values rolled in those $100$ times is less than $300$.

So for the sum, we want to look at the random variable $Y=X_1+X_2+...+X_{100}$. Every $X_i$ should have the same mean and variance, let's say the mean is $\mu$ and the variance is $\sigma^2$. Then the mean of $Y$ would be $100\mu$ and the variance is $100\sigma^2$.

We want to find $P(Y<300)$. Using a continuity correction, we want to look at $P(Y<300.5)$. Then $P(Y<300.5)=P(Z<\frac{300.5-100\mu}{\sqrt{100 \sigma^2}})$. I could find the answer from here.

However, I am having a hard time figuring out what $\mu$ and $\sigma^2$ should equal. I know that $\mu=np$ and $\sigma^2=np(1-p)$ for a normal approximation. But I don't know what $n$ and $p$ would be for the $X_1, X_2, ..., X_{100}$ or if that is the correct way to go about this problem.

Any help is appreciated, thank you.

Finding $\mu$ and $\sigma$ requires you to make use of the properties of the expectation and variance, namely:

$$\mu = \mathbb{E}[Y] = \mathbb{E}[X_1 + \dots + X_{100}] = \mathbb{E}[X_1] + \dots + \mathbb{E}[X_{100}] = 100 \frac{1+2+3+4+5+6}{6} = 350 \\ \sigma^2 = \mathrm{Var}[Y] = \mathrm{Var}[X_1 + \dots + X_{100}] = \mathrm{Var}[X_1] + \dots + \mathrm{Var}[X_{100}] = 100\left(\frac{91}{6} - 3.5^2\right) = 2741.6666...$$

The variance of $X$ comes from first principles, i.e. $\mathbb{E}[X^2] - (\mathbb{E}[X])^2$

Ask yourself this: where did the formulas $\mu=np$ and $\sigma^2=np(1-p)$ come from?

If you got them in the usual way, they are formulas for the mean and variance of a binomial distribution, and they come from the fact that the binomial distribution is the sum of $n$ independent Bernoulli variables, each of which is $1$ with probability $p.$ Each of those Bernoulli variables has mean $p$ and variance $p(1-p),$ and there are $n$ of them ...

But dice are not Bernoulli variables; they don't have mean $p$ and variance $p(1-p),$ and your formula doesn't apply. But you can easily find the mean and variance of one six-sided die -- just use the definitions of mean and variance for a discrete variable, write out the possible values of the die and their probabilities, plug these into the definitions and add up the values -- so you should be able to find the mean and variance of the sum of the $100$ dice.

The mean roll of one die is $\frac 16(1+2+3+4+5+6)=3.5$. By the linearity of expectation, the mean of the sum of $100$ rolls is ???. You should compute the variance of the roll of one die. The variance of the sum is the sum of the variances of the individual rolls.