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In our course notes, it says: "The CLT states that if random samples of size $n$ are repeatedly drawn from any population with mean $\mu$ and variance $\sigma^2$, then when $n$ is large the distribution of the sample means will be approximately normal."

I bolded "repeatedly" because that's the part I'm confused about. In the examples and the homework problems, it seems that even when we have $n$ supposedly large enough ($>30$), we're still only talking about a single sample of size $n$.

For example, one of the problems states: "Suppose a population has mean $\mu=5$ and standard deviation $\sigma = 2$, then supposed a random sample of size 38 is selected. What is the probability that the sample mean is between $4$ and $6$?"

Then we go on to use the CLT to solve the problem, supposedly because $n > 30$ allows us to use the CLT. But from that previous paragraph in italic I quoted, it seems that the CLT only works when many many samples of size n are drawn from a population, even when $n > 30$.

Here we're only randomly selecting a single sample, yet it seems to be good enough to use the CLT. Why is that?

EDIT: When I say "all CLT problems" I mean all the ones I ran into, including the famous swan problem, in which we randomly select $n>30$ swans once.

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    $\begingroup$ You ask about the meaning of probability (explained as a long-run phenomenon) in a context where just a single result (such as the sample mean) is available. Here is a perfect analogy. When we say a coin is "fair" we mean that in a sufficiently long run of repeated independent flips, the proportion of heads will be close to $1/2$. If I ask you about the chance that one flip of this fair coin will be heads, if you are rational you will answer "1/2". But that's only a statement about one flip, not a long run of flips! Why was your answer of "1/2" good enough? $\endgroup$
    – whuber ♦
    Commented Jul 11, 2016 at 18:19
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    $\begingroup$ Possible duplicate of Why does the central limit theorem work with a single sample? $\endgroup$
    – Greenparker
    Commented Jul 11, 2016 at 18:20
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    $\begingroup$ @whuber, you have a knack for understanding the source of people's confusions :) and you're right, that's exactly the same thing; for some reason I couldn't see it. I think stuff is thrown at us so quickly (summer course) that I'm getting confused with all the terminology. $\endgroup$
    – jeremy radcliff
    Commented Jul 11, 2016 at 18:56
  • $\begingroup$ @Greenparker, you're right, it's a very similar question. Thanks for the link. $\endgroup$
    – jeremy radcliff
    Commented Jul 11, 2016 at 18:58
  • $\begingroup$ You're confused about the meaning of the word "sample." The trials $X_1,\ldots,X_n$ contitute one sample of size $n.$ $\qquad$ $\endgroup$
    – Michael Hardy
    Commented May 27, 2019 at 17:41

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The CLT is a statement about fluctuations of averages. Specifically, a single sample of size $n$ will give you an expression:

$$\mu_n:=\frac{X_1+\cdots+X_n}{n},$$

which gives you an estimate of the true mean (average). Here $\mu_n$ is a random variable, with some complicated distribution. The CLT states that as $n$ gets larger, $\mu_n$ will be closer and closer to the true mean $\mu$, with fluctuations that look like a normal distribution, centered on the mean with variance $\sigma^2/n$. Said another way, $\mu_n$ is approximately distributed like $N(\mu,\sigma^2/n)$. In other words, if you were to repeat your sampling a bunch of times, and plot a histogram of the $\mu_n$'s that you drew each time, you'd get something that looks like a normal distribution with the above parameters.

Thus it makes perfect sense to ask questions like: what is the probability that $\mu_n>2$? In this case we'd use the CLT to approximate the distribution of $\mu_n$ as a normal distribution. This means that you could numerically verify the CLT by repeatedly drawing samples of size $n$ and computing $\mu_n$, thereby plotting a histogram and calculating probabilities from it.

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  • $\begingroup$ @AlexR. I think that makes sense, thank you. Does it matter whether we know that the population is normally distributed if $n$ is large enough? In the "swan problem", we are told "the weight of an swan is normally distributed", then told we get a random sample of $30+$ swans, and we go on to use the CLT. but if our sample of size $n$ is already $>30$, does it even matter what the original population's distribution looks like? For ex if we weren't told the swan weights are N distributed, would it matter? $\endgroup$
    – jeremy radcliff
    Commented Jul 11, 2016 at 18:53
  • $\begingroup$ @jeremyradcliff to my knowledge, the CLT can still be used when the Swan weights are not normally distributed (that's kind of the nice thing about the CLT). The CLT is really about the distribution of the means of the swan weights. Depending on the distribution of the swan weights (so not the distribution of the mean weights) you'll probably need a larger or smaller sample size n to state that n is large enough. Also note that the n > 30 is just a simple rule of thumb, it could for example be that you need at least 100, or a 1000. $\endgroup$
    – Amonet
    Commented Feb 13, 2019 at 9:30

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