Probability of a certain sum of values from a set of dice rolls A six-sided die is rolled 100 times. Using the normal approximation, find the probability that the face showing six turns up between 15 and 20 times. Find the probability that the sum of the face values of the 100 trials is less than 300.
For the first part of the question, I did the following:
$P(15 \le X \le 20) = \sum_{15 \le i \le 20} C(100,i)(\frac{1}{6})^i(\frac{5}{6})^{100-i}$
Where X is the number of sixes rolled. My answer was about 0.56.
I have no idea how to do the second part. I know I have to do something like
$P(Y<300|N=100)$
Where Y is the sum and N is the number of times rolled. But I don't know the probability of the sum so I'm stuck.
 A: Due to the CLT, a sum of i.i.d. random variables is distributed:
$$
\sum_{i=1}^nX_i \sim N\left(\mu =n\cdot\mu_{X_i},\sigma^2 = n\cdot\sigma^2_{X_i}\right)
$$
The mean of a single dice roll ($X_i$) is 3.5 and the variance is 35/12.
That should help you find the answer.
A: In the comments to Glen's answer you seem to have used a normal approximation pnorm(300, 350, sqrt(3500/12)) to get 0.001707396.  This is not a bad answer, though you can do better.
If you used the continuity correction the continuity correction pnorm(299.5, 350, sqrt(3500/12)) you would get 0.001553355.  I suspect this is what was being asked for.
It is in fact possible to calculate this more precisely. The following R code does so (yes, I know it has for loops).
sides <-  6   
throws <- 100 

## p[j,i] is probability of exactly (j+sides) after (i+1) throws 
p <- matrix(rep(0, sides*(throws+1)^2 ), ncol=throws+1 )
p[sides,1] <- 1 # probability 1 of score of 0 after 0 throws  

for (i in 2:(throws+1) ){
  for (j in (sides+1):(sides*(throws+1)) ){
     p[j,i] <-  sum(p[(j-sides):(j-1), i-1]) / sides
                                          } 
                            }
sum( p[0:(299+sides), throws+1] ) 

This gives the result 0.001505810.  
The normal approximation with continuity correction is within 0.00005, which looks good, though the relative error is about 3%, which looks slightly less impressive; this often happens using the normal approximation in the tail of the distribution.  
