Approximating probability P (X<0.56) for n=10 using central limit theorem

Orbit

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Hi all!

I'm confused about one of the exercises and I haven't been able to find any clues on how to solve it from the book.

There are X random numbers, independent from eachother and they all have the uniform distribution on interval 0,1
We look at the sample mean.

Approximate the probability P (X<0.56) for n=10
Use the central limit theorem.

What I got so far:
It is an uniform distribution so all the options have the same chance. As the exercise wants to know what the probability is for X<0.56, I assume that we will be taking steps of 0.01 from 0 to 1. Which leaves us with chance p of 0.01

This is where it gets confusing, if I look in the book it says you can use the normal approximation for the binomial distribution.
However, it says nothing about an uniform distribution and how to use it for CLT.
If I would ignore that and continue as if I had a binomial distribution, I get stuck at the next step which is approximating μ and σ as n*p is not larger than 5.

Can someone point me in the right direction?
 
Hi all!

I'm confused about one of the exercises and I haven't been able to find any clues on how to solve it from the book.

There are X random numbers, independent from eachother and they all have the uniform distribution on interval 0,1
We look at the sample mean.

Approximate the probability P (X<0.56) for n=10
Use the central limit theorem.

What I got so far:
It is an uniform distribution so all the options have the same chance. As the exercise wants to know what the probability is for X<0.56, I assume that we will be taking steps of 0.01 from 0 to 1. Which leaves us with chance p of 0.01

This is where it gets confusing, if I look in the book it says you can use the normal approximation for the binomial distribution.
However, it says nothing about an uniform distribution and how to use it for CLT.
If I would ignore that and continue as if I had a binomial distribution, I get stuck at the next step which is approximating μ and σ as n*p is not larger than 5.

Can someone point me in the right direction?

Rather than argue with the hint, why not think about what the Central Limit Theorem does an give the question more thought?

The Central Limit Theorem doesn't care much about the starting distribution.

p(X <0.56)? It's uniform. Why would it be anything but 0.56?

Should we be considering p((X1+X2+...X10)/10 < 0.56)?
 
Rather than argue with the hint, why not think about what the Central Limit Theorem does an give the question more thought?

The Central Limit Theorem doesn't care much about the starting distribution.

p(X <0.56)? It's uniform. Why would it be anything but 0.56?

Should we be considering p((X1+X2+...X10)/10 < 0.56)?

Thank you for helping me!

I think I understand it better now. I was a bit confused about the uniform distribution part but I think this site explains it clearly (http://www.statisticalengineering.com/central_limit_theorem.htm)
If you throw a dice, you have 1/6 chance of hitting 1 or 2 or 3 ..., but if you throw it 10 times, the chance of getting the sample mean of 1 is very small because there is only one possibility of getting that.

So I think this should be the formula:
p((X1+X2+...X10)/10 < 0.56)
(X1+X2+...X10)/10 is the sample mean, which you can also write as X with a bar above it, I will write it as a bar below it for computer reasons for now.

p(X < 0.56)

This results in
p(Z < 0.56-mu/(sigma/sqrt n)

Now for mu, this is the mean. I'd say that this is 0.5 as this is the usual estimator.
As for sigma, S is the usual estimator or the sqrtsample variance.
usual.png

This is the formula I should be using. I know n=10 but I'm not sure on how to calculate the X and X

(Xn-(X1+X2+...X10)/10)2/9
I don't have any results for X1 to X10, how can I calculate this? Is it possible to pick X1=0 and X10=1 and for X2 to X9 values in between?
 
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