When are real limits used for calculating z score?

Suppose $X sim mathsf{Binom}(n=75,, p=0.2)$ and you want to find $P(X le 12).$ You have several choices.

Binomial formula: Use the formula for the binomial PDF (or PMF) to evaluate each of 13 individual terms. A somewhat tedious task. $$P(X le 12) = P(X = 0) + P(X = 1) + cdots + P(X = 12) = sum_{k=0}^{12} {75 choose k}(.2)^k(.8)^{75-k}.$$

Tables: Some books have tables for binomial probabilities. Your textbook may have a few such tables in an appendix for various (smallish) values of $n$ and a few values of $p.$ Typically, $n = 75$ and $p=0.2$ will not be found there. Before the computer era, there were entire books of binomial probabilities, but those have mainly disappeared now.

Software: Maybe you have a statistical calculator that will do such computations. But nowadays there is good software that will do the job easily. One type of software that will handle this problem is R. In R, the function pbinom is a binomial CDF function. The CDF consists of probabilities of the form $P(X le k).$ The R code below shows that, for $X sim mathsf{Binom}(n=75,, p=0.2),$ we have $P(X le 12) = 0.2397,$ correct to four places, with more places of accuracy available if needed.

pbinom(12, 75, .2)
[1] 0.2396826

Normal approximation: For $n$ sufficiently large and $p$ not too near $0$ or $1,$ the distribution $mathsf{Binom}(n,p)$ can be approximated by using a normal distribution with matching mean and variance: $mathsf{Norm}(mu, sigma),$ where $mu = np$ and $sigma = sqrt{np(1-p)}.$

An often-useful rule of thumb is that the normal approximation gives about two-place accuracy if both $np > 5$ and $n(1-p) > 5.$ Both are satisfied for $n=75$ and $p = 0.2.$ So we find $mu = np = 15$ and $sigma = sqrt{np(1-p)} = 3.4641.$

Because the binomial distribution is discrete and the normal distribution is continuous, we have to be a little careful in order to get best results from the normal approximation. For $mathsf{Binom}(75,,0.2)$, we have $$P(X le 12) = P(X < 12.5) = P(X < 13).$$ But for $mathsf{Norm}(mu = 15,, sigma=0.6928),$ these are three different probabilities. Briefly put, the 'continuity correction' uses the second of the three because it (usually) gives the best approximation.

$$P(X le 12.5) = Pleft(frac{X-mu}{sigma} le frac{12.5-15}{3.461}right) approx P(Z le -0.7217 ) \ approx P(Z le -0.72) = 0.24,$$ where $Z sim mathsf{Binom}(0,1).$ The second approximation is necessary if you are using printed tables of the standard normal distribution because (without interpolation) these tables provide z-values to only two places.

pnorm(-0.7217);  pnorm(-0.72)
[1] 0.2352395
[1] 0.2357625

In the figure below, the exact binomial probability $P(X le 12)$ is the sum of the heights of the vertical black bars to the left of the broken red line. The normal approximation is the area under the blue normal density curve to the left of the red line.

Notice that, according to the normal curve, the probability $P(X = 12)$ is represented by the area under the curve and above the interval $(11.5, 12.5].$ By using the normal approximation, we have included this entire probability--instead of just part of it.