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 .


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.
