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?


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.


Apply the (sample mean version of the) CLT using $𝑁(750,900^2/500)=𝑁(750,1620)$ .

The z-value for this is $\frac{755βˆ’750}{\sqrt{1620}}=0.1242$ .

Looking this up in the Z-table the closest value is 0.12 which corresponds to a probability of 𝑃(𝑍<0.12)=0.5478 .

So the probability that the average total annual payments are greater than $755 is approximately:

𝑃(π‘π‘Žπ‘¦π‘šπ‘’π‘›π‘‘π‘ >755)=𝑃(𝑍>0.12)=1βˆ’π‘ƒ(𝑍<0.12)=1βˆ’0.5478=0.4522 .

  • 1
    $\begingroup$ The question boils down to asking what's the (approximate) distribution of $\bar{X}$, the average payment made by $n=500$ customers. So the sample size is $500$. $\endgroup$ Commented Sep 11, 2021 at 18:00
  • $\begingroup$ Isn't it at all relevant that the sample size is the whole population? $\endgroup$
    – Kirsten
    Commented Sep 11, 2021 at 18:08
  • 3
    $\begingroup$ That's a good question. David Spiegelhalter calls this a metaphorical population where what we observe has been drawn from some imaginary space of possibilities. At the end of the year, we know with certainty how much loan payments the customers made. But we can imagine counterfactual years in which different individuals might have paid a different amount. Another point of view is that what will happen in the next year is due to chance (according to the question). $\endgroup$ Commented Sep 11, 2021 at 18:21
  • $\begingroup$ Why do you say the question is about the next year? $\endgroup$
    – Kirsten
    Commented Sep 11, 2021 at 18:51
  • $\begingroup$ Sorry, I meant any yet unobserved year. $\endgroup$ Commented Sep 11, 2021 at 18:55

2 Answers 2


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.

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    $\begingroup$ +1 for the answer, but the re-wording is unlikely IMHO to reflect what the questioner was thinking. Those 500 loan customers weren't a random sample ("the bank has 500 customers"), but they were being viewed as part of a stationary process and the respondent was being asked to view them as if they were a random sample of that process. Although this might sound like nitpicking, we have fielded many questions revolving around confusing a sample with a population or process, especially in such situations, indicating this is important to clarify. See a comment by @BruceET. $\endgroup$
    – whuber
    Commented Sep 14, 2022 at 17:38
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    $\begingroup$ Okay, but there is no bank; it is made up. There is no bank and no customers. So I'm basing the rewording on altering the situation to conform to their stipulated answer in the simplest way possible. $\endgroup$
    – Ben
    Commented Sep 14, 2022 at 21:18

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

enter image description here

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.

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    $\begingroup$ Thank you. Is the fact that we are using the whole population irrelevant? I am so confused because the question is using sampling and the whole population at the same time. $\endgroup$
    – Kirsten
    Commented Sep 11, 2021 at 22:13
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    $\begingroup$ You seem to believe there is one fixed population of 500. As bank clients come and go, the number of current clients will fluctuate over time. So the only way I know to make sense of the problem is to say we're interested in a year when there are 500 clients, to say the dist'n of their payments is NORM(750, 900), and use that to get eh dist'n of the "sample' mean of 500 clients. (And I imagine you may just have studied the distributions of sample means.) // It would be clearer to say we pick 500 clients from a huge bank chain, give the $\mu$ and $\sigma.$ and and ask about $\bar X_{500}.$ $\endgroup$
    – BruceET
    Commented Sep 11, 2021 at 22:32

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