A six sided die is rolled 100 times. Using the normal approximation, find the probability that the sum of the face values of the 100 trials is less than 300.
This question is the second part to a question asked before: Central Limit Theorem (Normal Approximation to Binomial)
I have no issues with the first part. My question here is how could I find $\mu$ and $\sigma^2$ ? In part 1, I had the ability to use the fact that the random variable behaved like a Binomial R.V. That would not work here. I am doing this by hand and was hoping to get more of the theoretical reason behind finding a proper $\mu$ and $\sigma^2$.
Let $X_i$ be iid with uniform distribution, i.e. \begin{equation} P(X_i = k) = \frac{1}{6} \end{equation} for all $k = 1 \ldots 6$ and $i = 1 \ldots 100$.
Consider \begin{equation} Y = \sum X_i \end{equation} and \begin{equation} S_{100} = \frac{Y}{100} \end{equation} Assuming $n = 100$ is large enough, then by the Central Limit Theorem \begin{equation} \sqrt{100}(S_{100} - \mu ) \rightarrow \mathcal{N}(0,\sigma^2) \end{equation} where $(\mu,\sigma)$ are mean and std. deviation of $X_i$. The mean is \begin{equation} \mu = E(X) = \frac{1}{6}(1 + 2 + 3+ 4+ 5 +6) = \frac{21}{6} = 3.5 \end{equation} The variance is \begin{equation} \sigma^2 = Var(X) = E(X^2) - \mu^2 \end{equation} \begin{equation} E(X^2) = \frac{1}{6}(1^2 + 2^2 + 3^2+ 4^2+ 5^2 +6^2) = \frac{91}{6} \end{equation} So \begin{equation} \sigma^2 = \frac{91}{6} - \frac{21^2}{36} \simeq 2.916 \end{equation} This means that \begin{equation} 10(S_{100} - 3.5 ) \rightarrow \mathcal{N}(0,2.916) \end{equation} or \begin{equation} S_{100} - 3.5 \rightarrow \mathcal{N}(0,\frac{2.916}{100}) \end{equation} or \begin{equation} S_{100} \rightarrow \mathcal{N}(3.5,0.02916) \end{equation} So you are now interested in finding \begin{equation} P(S_{100} < \frac{300}{100}) \end{equation} or \begin{equation} P(S_{100} < 3) \end{equation} Standardize \begin{equation} P(S_{100} < 3) = P(\frac{S_{100} - 3.5}{0.02916} < \frac{3-3.5}{0.02916}) \end{equation} Let $Z = \frac{S_{100} - 3.5}{0.02916} \sim \mathcal{N}(0,1)$ is the standard normal \begin{equation} P(S_{100} < 300) = P(Z < \frac{3-3.5}{0.02916}) \end{equation}