# central limit therom help needed

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https://i.imgur.com/jUvEXm3.jpg

so for part b I said we assumed that central limit theorem is used since sample is large. but the markschme also adds no need to assume normal distribution. doesnt the central limit therom also assume normal distribution hence doesn't this statement contradict it self?

so for part b I said we assumed that central limit theorem is used since sample is large. but the markschme also adds no need to assume normal distribution. doesnt the central limit therom also assume normal distribution hence doesn't this statement contradict it self?

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#3

https://en.wikipedia.org/wiki/Bean_machine

where at each level, the ball bounces left or right according to a binomial distribution, but the distribution at the bottom is approximately normal.

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#4

(Original post by

doesnt the central limit therom also assume normal distribution?

**assassinbunny123**)doesnt the central limit therom also assume normal distribution?

There are various forms of the central limit theorem, but a straightforward one states that if you draw samples of size n from any probability distribution with finite mean and variance, then the means of those samples will be normally distributed.

If you want an example where the central limit theorem does not hold, try the Cauchy Distribution - a distribution where the mean and variance do not exist!

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#5

(Original post by

https://i.imgur.com/jUvEXm3.jpg

so for part b I said we assumed that central limit theorem is used since sample is large. but the markschme also adds no need to assume normal distribution. doesnt the central limit therom also assume normal distribution hence doesn't this statement contradict it self?

**assassinbunny123**)https://i.imgur.com/jUvEXm3.jpg

so for part b I said we assumed that central limit theorem is used since sample is large. but the markschme also adds no need to assume normal distribution. doesnt the central limit therom also assume normal distribution hence doesn't this statement contradict it self?

I recommend watching a video from the youtube channel DONG, its quite cool.

[link]https://youtu.be/UCmPmkHqHXk[/link]

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#6

(Original post by

No, the "best" example is the bean machine

https://en.wikipedia.org/wiki/Bean_machine

where at each level, the ball bounces left or right according to a binomial distribution, but the distribution at the bottom is approximately normal.

**mqb2766**)No, the "best" example is the bean machine

https://en.wikipedia.org/wiki/Bean_machine

where at each level, the ball bounces left or right according to a binomial distribution, but the distribution at the bottom is approximately normal.

At each level, the ball bounces left or right according to a

**Bernoulli distribution**, and since the sum of n independent Bernoulli random variables is a binomial distribution, the distribution at the bottom is Binomial - which can then (after a fair amount of algebra!) be shown to converge (in a sense that needs to be made rigourous) to a normal distribution.

So these little machines demonstrate the special case of the central limit theorem for Bernoulli random variables.

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(Original post by

Have a look at galton boards, its a neat way of demonstrating the central limit theorem.

I recommend watching a video from the youtube channel DONG, its quite cool.

[link]https://youtu.be/UCmPmkHqHXk[/link]

**Shaanv**)Have a look at galton boards, its a neat way of demonstrating the central limit theorem.

I recommend watching a video from the youtube channel DONG, its quite cool.

[link]https://youtu.be/UCmPmkHqHXk[/link]

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#8

so whatever kind of distribution the parent population has, you will always find that the distribution of sample means is normal.

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#9

My stats lecturer basically said if the number of trials is greater than or equal to 30, the distribution (whatever it was originally, binomial, bernoulli etc.) can be approximated by a normal distribution now.

I'm careful with the wording as perhaps that is different to "assuming" the variable is normally distributed. I don't think weight loss is "naturally" normally distributed, but we can approximate it as, with a large enough sample size

I'm careful with the wording as perhaps that is different to "assuming" the variable is normally distributed. I don't think weight loss is "naturally" normally distributed, but we can approximate it as, with a large enough sample size

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(Original post by

No, and this is an important point, as it explains why you get to use the normal distribution so much in probability and statistics!

There are various forms of the central limit theorem, but a straightforward one states that if you draw samples of size n from any probability distribution with finite mean and variance, then the means of those samples will be normally distributed.

If you want an example where the central limit theorem does not hold, try the Cauchy Distribution - a distribution where the mean and variance do not exist!

**Gregorius**)No, and this is an important point, as it explains why you get to use the normal distribution so much in probability and statistics!

There are various forms of the central limit theorem, but a straightforward one states that if you draw samples of size n from any probability distribution with finite mean and variance, then the means of those samples will be normally distributed.

If you want an example where the central limit theorem does not hold, try the Cauchy Distribution - a distribution where the mean and variance do not exist!

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#11

(Original post by

but you just said that central limit therom says that the distribution of mean and variance are normally distributed. here the answer says that central limit therom can be assumed but normal distribution cannot,how is this possible?

**assassinbunny123**)but you just said that central limit therom says that the distribution of mean and variance are normally distributed. here the answer says that central limit therom can be assumed but normal distribution cannot,how is this possible?

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(Original post by

Gregorius is trying to help you. the population distribution does not have to be normal; the distribution of sample means is normal.

**the bear**)Gregorius is trying to help you. the population distribution does not have to be normal; the distribution of sample means is normal.

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#13

**assassinbunny123**)

but you just said that central limit therom says that the distribution of mean and variance are normally distributed. here the answer says that central limit therom can be assumed but normal distribution cannot,how is this possible?

Suppose that you start with a probability distribution that has a finite mean (mu) and variance (sigma-squared). If you draw repeated samples of size n from that distribution, then the distribution of the

**means**of those samples will be approximately normal with mean mu and variance sigma-squared divided by n. That's the central limit theorem.

Let's turn to the question you posted. Here you have a single random sample of size 80 from one underlying population and a single sample of size 65 from another underlying population and you're asked whether these two samples give sufficient evidence to show that the means of the underlying populations are different.

To do this, you use the

**means**of each sample - the central limit theorem implies that

**these**can be taken as having come from normal distributions.

So, you don't need to assume normality of the distributions of weight losses, as the central limit theorem tells you that the means of these distributions will be (approximately) normally distributed.

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