How do I determine the average number of successes, variance, and standard deviation, for a given set of die rolls. Assuming that I roll 10 dice, and am going to call a 5 or 6 a 'success' on any single die, how can I determine the average number of successes, the variance, and the standard deviation for the associated bell curve?
If I use $N$ as the number of dice used, and $P$ as the probability of getting a success on a single die roll, I believe the average is simply $NP$, or in this case, $10×\frac26=3.33$ successes on average.  
What I'd like to know now, is what is my range of possible successes at 1 and 2 standard deviations.  Beyond that, I'd like to understand how to perform this calculation assuming I change either $N$ or $P$.
Thanks in advance!
 A: The distribution is binomial with p=1/3.  Use the standard form to get the first and second moment and therefore the variance and standard deviation.
A: As mentioned by callculus, normal approximation would not be a good method in this case. Instead I will use the binomial distribution.
The expected value can be obtained in two ways:
$$\begin{align*}
E(X)
&=\sum_{k=0}^ n {n\choose{k}} k\cdot p^k\left(1-p\right)^{n-k}\\\\
&=\sum_{k=0}^{10} {10\choose{k}} k\cdot \left(\frac{1}{3}\right)^k\left(1-\left(\frac{1}{3}\right)\right)^{10-k} \\\\
&= \frac{10}{3}\\\\
&\approx 3.333
\end{align*}$$
You could also just do $np$ to get $$E(X)=10\cdot\frac{1}{3}=\frac{10}{3}\approx 3.333$$
The variance can be obtained in two ways:
$$\begin{align*}
Var(X)
&=E(X^2)-E(X)^2 \\\\
&=\sum_{k=0}^{10} {n\choose{k}} \cdot k^2 \cdot p^k\left(1-p\right)^{n-k} - \left(\sum_{n=0}^{10}{n\choose{k}} \cdot k \cdot p^k\left(1-p\right)^{n-k}\right)^2 \\\\
&=\sum_{k=0}^{10} {10\choose{k}}\cdot k^2 \cdot \left(\frac{1}{3}\right)^k\left(1-\left(\frac{1}{3}\right)\right)^{10-k} - \left(\sum_{n=0}^{10} {10\choose{k}}\cdot k \cdot \left(\frac{1}{3}\right)^k\left(1-\left(\frac{1}{3}\right)\right)^{10-k}\right)^2 \\\\
&= \frac{40}{3} - \left({\frac{10}{3}}\right)^2\\\\
&= \frac{20}{9}\\\\
&\approx 2.222
\end{align*}$$
You could also just do $npq$ to get
$$Var(X)=10\cdot\frac{1}{3}\cdot\frac{2}{3}=\frac{20}{9}\approx 2.222$$
The standard deviation is just 
$$S.D.=\sqrt{Var(X)}=\sqrt{\frac{20}{9}}\approx 1.490711985$$
