Central Limit Theorem Homework: Find Expected Deviation

[PullMeOut]

Homework Statement

consider an experiment with 2 possible outcomes, 1 and 0, with a priori probabilities p and
1-p. we would like to find out the average (expected) deviation after N trials, of the relative frequency of the "1"s, N1/N
Use the central limit theorem to find expected deviation.

Homework Equations

N1=[tex]\sum[/tex][tex]^{N}_{i=1}[/tex] ni , where ni is the outcome of the ith trial

The Attempt at a Solution

I know expected deviation of N1 is the square root of its variance.
and variance is:
[tex]\sigma[/tex][tex]^{2}[/tex]=<x[tex]^{2}[/tex]> - <x>[tex]^{2}[/tex]

<x>=[tex]\sum[/tex][tex]^{N}_{i=1}[/tex]pi xi

but i have to use central limit theorem
[tex]\lim_{N \rightarrow inf } \left(N1/N)[/tex]

and I'm lost , the question looks so easy but i have no idea what to do?

 

[Jhero]

Hello, thank you for sharing your question. The central limit theorem states that the sum of a large number of independent and identically distributed random variables will be approximately normally distributed. In this case, we can consider N1/N as the sum of N independent and identically distributed random variables, each representing the relative frequency of "1"s in a single trial.

To find the expected deviation, we can use the formula for the standard deviation of a normal distribution:

\sigma = \sqrt{N} \cdot \sigma_{1/N}

where \sigma_{1/N} is the standard deviation of the individual random variables, which can be calculated as:

\sigma_{1/N} = \sqrt{p(1-p)}

Therefore, the expected deviation of N1/N after N trials would be:

\sigma = \sqrt{N} \cdot \sqrt{p(1-p)} = \sqrt{Np(1-p)}

I hope this helps. Please let me know if you have any further questions.