Why do we apply the sample mean version of the CLT for a problem involving a sample size of 1? I am having problems understanding the following question and answer.
It seems to me that the sample size is n = 1 and the population size is N=500.
If I read it this way then we do not have a large enough sample to use CLT.
The answer uses n=500 but this is also N, so doesn't that imply the mean is certain to be 750 dollars?
Question

A small micro-loan bank has 500 loan customers. If the total annual
loan repayments made by an individual is a random variable with mean
750 dollars and standard deviation 900 dollars. Approximate the probability that the
average total annual repayments made across all customers is greater
than 755 dollars.

Answer

Apply the (sample mean version of the) CLT using
$(750,900^2/500)=(750,1620)$ .
The z-value for this is  $\frac{755−750}{\sqrt{1620}}=0.1242$ .
Looking this up in the Z-table the closest value is 0.12 which
corresponds to a probability of  (<0.12)=0.5478 .
So the probability that the average total annual payments are greater
than $755 is approximately:
(>755)=(>0.12)=1−(<0.12)=1−0.5478=0.4522 .

 A: This is an example of a poorly worded question.  If one were to interpret it strictly as written, one has a sample size of $n=500$ and a population size of $N=500$, so yes, the population mean is certain to be \$750 (and so the probability that this mean is greater than \$755 is known to be zero).  If you were to give this answer to the question, that would be correct in my view.  Nevertheless, in view of the answer given, it appears that the writer of the question intended to treat the sample of $n=500$ customers as a random sample of a "large" population ($N=\infty$) and the resulting calculations are consistent with that.
For these types of questions, it is worth noting that the confidence interval formula for a population mean can be written in a way that allows a finite or infinite population $n \leqslant N \leqslant \infty$.  The general formula for the confidence interval for the population mean (see e.g., O'Neill 2014, pp. 285-286) is:
$$\text{CI}_N(1-\alpha) = \Bigg[ \bar{x}_n \pm \frac{t_{\alpha/2,DF_n}}{n} \cdot \sqrt{\frac{N-n}{N}} \cdot s_n \Bigg],$$
where $DF_n = n-1$ for a mesokurtic distribution (e.g., the normal distribution).  You can easily confirm that this interval reduces to a single point given by the sample mean in the special case where $n=N$ and reduces to the standard form used for a "large" population when $N=\infty$.

How to re-word the question: In your bounty request you have asked how the question could be better worded to properly express the query reflected in the posted answer.  To re-word the query, it would be important to be clear that $n=500$ and $N=\infty$ in this problem.  (For the latter we usually just refer to the population as "large" --- see this related answer for an explanation.)  It is also desirable to specify that the observed customers are a random sample of the population.  Something like this would be an appropriate wording:

Question: A micro-loan bank has a large number of loan customers.  Analysts at the bank take a random sample of 500 of their loan customers and examine the total annual loan repayments made by each of the sampled customers --- they find a mean of \$750 and a standard deviation of \$900 from these values.  Use this data to approximate the probability that the average total annual repayments made across all customers at the bank is greater than \$755.

A: The first Comment of @COOLSerdash is correct. The wording of the problem is somewhat confusing.
Moreover, the choice of numbers leads to a z-value that needs to be rounded for use of a printed table, thus we get a noticeable rounding error in the posted answer.
You have $\bar X =\bar X_{500} \sim\mathsf{Norm}(\mu=750, \,\sigma=900/\sqrt{500}),$ and you seek $P(\bar X > 755) = 1-P(\bar X \le 755) = 0.4505682,$ exactly. (Using R:)
1 - pnorm(755, 750, 900/sqrt(500))
[1] 0.4505682

If you were to standardize, then $z = 0.124226.$
z = (755-750)/(900/sqrt(500)); z
[1] 0.124226

Then the exact answer is $P(Z > z) = 1 - P(Z \le z) =0.4505682,$ exactly (same as above). So there is no essential error from standardizing.
1 - pnorm(z)
[1] 0.4505682


However, using printed tables without interpolation, you have to round $z$ to two places in order to enter a table that rounds probabilities to four places.
As in the posted 'answer' you
would get $0.4522,$ which results from rounding twice.
round(1 - pnorm(round(z,2)), 4)
[1] 0.4522

There may be little practical difference between 0.4506 (correct to four places) and 0.4522. But it can be frustrating to use badly designed
homework software that requires results "correct to four places," if you
give the correct value to four places and the software emulates imprecise
use of printed tables.
