I was presented two formulas, if random variables ${X_1,...,X_n}$ form a random sample of a distribution of mean $\mu$ and standard deviation $\sigma$ and $n \to \infty$: \begin{align} &P\left(\frac{\overline{X_n}-\mu}{\sigma/\sqrt n}< a\right) = \Phi(a) \:\:\:\: (1) \\ &P\left(\frac{\overline{X_n}-n\mu}{\sigma\sqrt n}< a\right) = \Phi(a) \:\: (2) \end{align} where $\Phi$ is the cdf of a standard normal distribution.
But I not sure when to use which.
For example. Each minute a machine produces $4$ meters of rope with standard deviation of $0.4$ meters. Assuming the amount produced in differet minutes are i.i.d, what is the probability that the machine will produce at least $250$ meters.
I'm supposed to use $(2)$ but why?
Another example: $16$ digits are chosen at random from the set $\{0,...,9\}$. What is the probability that their average will lie between $4$ and $6$?
I'm supposed to use $(1)$ but why?
Your second formula is not correct. It's missing an $n$, it should be
$$P\left(\frac{\sum_{i=1}^n X_n -n\mu}{\sigma\sqrt{n}} < a \right) = P\left(\frac{n\overline{X_n} -n\mu}{\sigma\sqrt{n}} < a \right) = Φ(\alpha).$$
Then both of the formulas say the exact same thing, except the second one has both the numerator and the denominator multiplied by $n$. You can absolutely use either in both cases, the reason a specific one is recommended is that formula $(1)$ has the average in the numerator, whereas $(2)$ is using the entire sum of your observations, making it easier to understand what you're plugging into the formula.
In your first problem, if you wanted to use $(1)$ you would need to change the wording from "Will produce at least $250$ meters" to "Will produce at least $250/n$ meters per minute on average", which means the same thing, but adds unnecessary work because you have to do the division.