Suppose that a proportion $p$ of a given population verifies a given characteristic.
Let $X$ denote the random variable that equals the number of times this characteristic is observed in a sample of size $n$ (so $X$ follows a binomial distribution of parameters $(n,p)$), and let $F=\frac{X}{n}$ denote the corresponding frequency.
Sampling theory tells us that under some conditions, $F$ approximately follows a normal distribution with parameters $(p,\sqrt{p(1-p)/n})$, and this can be used to compute the probability $P(F \le f)$.
My question is, instead of using this approximation, why can't we use the fact that the distribution of $X$ is known, and that $ P(F \le f) = P(\frac{X}{n} \le f) = P(X \le f \cdot n) $ ? Wouldn't this give the exact probability ? Is there a reason why the normal distribution is preferred ?