In binomial distribution number of successes (usually denoted as $x$) must be between between $0$ and $n$, inclusive ($n$ is the number of trials). So for example there can be a problem which asks for probability that something happens 3 times at most, such as at most 3 heads if flipping a coin 10 times. In this case the probability would just be $P(0)+P(1)+P(2)+P(3)$.
However, when using normal distribution as an approximation to binomial distribution, $x$ can theoretically take negative values as well. I have seen textbooks where a question asks for probability of, for example, 3 successes at most and in the solution the probability is denoted as $P(x \leq3.5)$ (using continuity correction). However, I have been thinking shouldn't $x$ still be bounded, so the correct way to write it would be $P(-0.5 \leq x \leq3.5)$. I have calculated probabilities for some problems in both ways and in some cases differences were relatively large.
Question: When using normal approximation of binomial distribution is it correct to use -0.5 as the lower boundary for calculations? Or does an assumption of unboundedness produce a better approximation?
Hopefully my question is clear enough.
In cases with a small number of trials, like flipping a coin $10$ times, there will be some difference between the binomial probability and the normal probability. For these cases you should use the binomial one since, for instance, $P(\text{getting at most }3 \text{ heads})$ can be calculated by hand.
In cases with a bigger number of trials, like flipping a coin $100$ times, there will be smaller difference between the two probabilities since the binomial distribution, compared with the smaller case, has narrower "gaps" and hence tends to be continuous. Therefore, whether to set a lower boundary to restrict the result does not make a great difference.