I am studying for the final tomorrow and have been using a quiz as reference.
A fair die is rolled 720 times. Let the random variable Y be the number of 6's obtained. Approximate the probabilities.
Since n is such a high number we can treat it like a normal distribution using the Central Limit Theorem. In my professor's answer he changes the Y to X and add/sub a .5 depending on which "tail" we are trying to find.
Ex. P(Y<98) -> P(X<98.5) P(Y>115) -> P(X > 114.5)
What exactly is happening here?
If $X$ is the number of 6's seen in $n = 720$ rolls of a fair die, then $X \sim \mathsf{Binom}(n = 720, p = 1/6).$ Thus, $\mu = E(X) = np = 120,$ $\sigma^2 = Var(X) = np(1-p) = 100,$ and $\sigma = SD(X) = 10.$
It seems your professor may be trying to find $P(X \le 98) = P(X < 99) = P(X < 98.5),$ where all three probabilities are the same for the discrete random variable $X$, which takes only integer values. In making a normal approximation, it is best to use $P(X < 98.5).$ This is called the 'continuity correction' because it uses the value most compatible with the continuous normal distribution. (See the reference suggested by @spaceisdarkgreen.)
Then $$P(X < 98.5) \approx P\left(Z = \frac{X-\mu}{\sigma} < \frac{98.5 - 120}{10} = -2.15\right) = 0.0158,$$ where $Z$ is standard normal; you can use printed normal CDF tables (or software) to find this approximate probability.
The exact binomial probability from R statistical software is 0.0141, which is not well approximated by the normal distribution. Ordinarily, one can expect two or three of places of accuracy from a normal approximation when $n$ is so large, but results can be poor when $p$ is far from $1/2,$ as here. Nowadays when software is available to find exact binomial probabilities, it is not a good idea to use normal approximations in practical situations.
The figure below shows the binomial distribution (black bars) and the approximating normal density function.