Normal Distribution and a Discrete Amount From what I understand about normal distribution is that you make a discrete number continuous by adding .5 which every way the question asks for.
What if you were to have a discrete number with a set amount you cannot have half of. 
Would you keep it discrete?
Ex) say there were 30 prizes, would you calculate the z-score as 30 or 30.5?
If a game was attempted 100 times and the chance of winning is 30%, would 30 prizes be enough?
mean= np = 100*.2=20
standard deviation sqrt(100*0.2*0.8) = 16
Here is my dilemma:
z(30) 
=(30-20)/16 
=10/16 
=0.63 
0.63=0.7357 
z(30.5)
=(30.5-20)/16
=10.5/16
=0.66
0.66=0.7454
SO, the chances of having enough prizes for 100 attempts is either 73.57 or 74.54 depending on that 0.5
 A: Suppose we have a binomially distributed random variable $X$, and we want to use the normal approximation. If "$n$" is quite big, say $1500$, the continuity correction makes no big difference, so it is of no great practical importance whether or not you use it. (But if an exercise wants you to use it, you have no choice.)
For middling $n$, such as $n=50$, the continuity correction often yields a more accurate approximation. Here is how I would apply it. 
Method: Let $k$ be an integer. Then $\Pr(X\le k)$ gets approximated by the probability that the normal with the same mean $np$ and same variance $np(1-p)$ as $X$ is $\le k+0.5$. This latter probability is the same as the probability that $Z\le \frac{k+0.5-np}{\sqrt{np(1-p)}}$.
A similar continuity correction is made when we approximate a Poisson with not too small $\lambda$ by a normal. But the main place where you will be using the continuity correction is in the normal approximation to the Binomial.
A Caution: We do not automatically "add $0.5$."  Here is an example. Use the normal approximation with continuity correction to approximate the probability that $X\lt k$, where $k$ is a given integer. 
At first sight this looks like the approximation problem described under Method. But there is a difference, here we have $\lt k$, not $\le k$. So here is what we do. The probability that $X\lt k$ is exactly the same as the probability that $X\le k-1$. Now this looks exactly like the kind of problem discussed in Method, we have a $\le$.
So the required probability is approximated by the probability that a certain normal is $\le k-1+0.5$, in other words by the probability the normal is $\le k-0.5$.
So when you are using the continuity correction with the binomial, I would advise making sure to convert the problem into one that involves $\le$.
As another example, suppose we want to approximate the probability that $X\ge k$. Be careful in applying the continuity correction. We want $1$ minus the  probability that $X\lt k)$. The $X\lt k$ problem has been just discussed. 
I hope this short description has been helpful.
Remark: For the places where continuity correction can be helpful, it is less important than it used to be. For using software we can calculate binomial probabilities exactly. 
About the prizes question, I can answer it properly only if I get the full exact statement of the problem. 
