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Central Limit Theorem

I think I misunderstand this. I think it means that if you take enough of the same sized random samples from a population then the means of the samples are distributed normally.

But supposing that the sample mean happens to be nowhere near the population mean then how does using a sample mean help.

This question probably makes no sense but that's because I just don't get what's going on in CLT.
Reply 1
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Smaug123
15
Original post by maggiehodgson
I think I misunderstand this. I think it means that if you take enough of the same sized random samples from a population then the means of the samples are distributed normally.

But supposing that the sample mean happens to be nowhere near the population mean then how does using a sample mean help.

This question probably makes no sense but that's because I just don't get what's going on in CLT.

The point is that as you take more and more samples, it gets less and less likely that they're all skewed away from the mean. Some will be skewed in the opposite direction to others, and they'll all average out. The CLT says "whatever you sample, in whatever distribution, if you take enough samples then the mean follows the normal distribution" - that is, if m is the population mean and s the standard deviation of the population, P(a < (Sn-n m)/(sqrt(n) s) < b) tends to phi(b)-phi(a). The square root on the denominator is there because of a reason that I can't convey without sketching, but at A-level (IIRC), that's not important.
Thank you.

So in the questions where you are told x(bar) is something then that x(bar) in the mean of ALL the sample means not just the mean of one of the samples where it could be quite iffy?
Reply 3
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Smaug123
15
Original post by maggiehodgson
Thank you.

So in the questions where you are told x(bar) is something then that x(bar) in the mean of ALL the sample means not just the mean of one of the samples where it could be quite iffy?

It depends on context - the sample means will (nearly) follow the normal distribution, so you end up with a sample from the normal distribution. I can't tell which xbar is without context - if there's more than one of it, then it's the sample means, while if there's just one, then it's the mean of the sample means.
Sitting here banging my head on the desk. I'm just not going to get CLT I think. I will mull over some questions bearing in mind what you've told me and hope that a practical or 12 will help this concept drop into place.

Thanks very much.
Reply 5
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Smaug123
15
Original post by maggiehodgson
Sitting here banging my head on the desk. I'm just not going to get CLT I think. I will mull over some questions bearing in mind what you've told me and hope that a practical or 12 will help this concept drop into place.

Thanks very much.

I can tell you why the CLT is true, if you like (it's not a fully rigorous proof, it relies on MGFs which aren't necessarily defined for all variables) - although if you've not already done a fair bit of probability, it probably will take a while to understand (or even a whale, as I first typed). For me at A-level it just took lots of questions.

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