Vtyjkyuk

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.
beginequation
P(X_i = k) = frac16
endequation
for all $k = 1 ldots 6$ and $i = 1 ldots 100$.
Consider
beginequation
Y = sum X_i
endequation
and
beginequation
S_100 = fracY100
endequation
Assuming $n = 100$ is large enough, then by the Central Limit Theorem
beginequation
sqrt100(S_100 - mu ) rightarrow mathcalN(0,sigma^2)
endequation
where $(mu,sigma)$ are mean and std. deviation of $X_i$. The mean is
beginequation
mu = E(X) = frac16(1 + 2 + 3+ 4+ 5 +6) = frac216 = 3.5
endequation
The variance is
beginequation
sigma^2 = Var(X) = E(X^2) - mu^2
endequation
beginequation
E(X^2) = frac16(1^2 + 2^2 + 3^2+ 4^2+ 5^2 +6^2) = frac916
endequation
So
beginequation
sigma^2 = frac916 - frac21^236 simeq 2.916
endequation
This means that
beginequation
10(S_100 - 3.5 ) rightarrow mathcalN(0,2.916)
endequation
or
beginequation
S_100 - 3.5 rightarrow mathcalN(0,frac2.916100)
endequation
or
beginequation
S_100 rightarrow mathcalN(3.5,0.02916)
endequation
So you are now interested in finding
beginequation
P(S_100 < frac300100)
endequation
or
beginequation
P(S_100 < 3)
endequation
Standardize
beginequation
P(S_100 < 3) = P(fracS_100 - 3.50.02916 < frac3-3.50.02916)
endequation
Let $Z = fracS_100 - 3.50.02916 sim mathcalN(0,1)$ is the standard normal
beginequation
P(S_100 < 300) = P(Z < frac3-3.50.02916)
endequation