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Problem with Papoulis Exercise, Probability question

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claudiocamera

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papoulis 0.49 0.52

Solving exercises in Papoulis’ book, I came across a doubt regarding an exercise and its answer provided.

The problem is:

A fair coin is tossed n times. Find n such that the probability that the number of heads is between 0.49n and 0.52n is at least 0.9.

The answer is : G(0.04√n) - G(0.02√n) >1.9 , hence n> 4556.

Usually the numbers given is this kind of problems are simetrical in relation to np in order that we have an equation in the form {2G(f(n)) -1 > probability required} . These kind of problems are easily solved looking at normal table. In the way it was present I can’t figure out how to solve it, since in my opinion we have two icognits for only one equation. How to find it from the table since I have two different values G(0.04√n) and G(0.02√n) ?

Any help in claryfing it ?
 

Quibbler

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solved exercises papoulis

The probability is symetrical about the 50:50 point so take 0.5 of the probability between .48-.52 plus 0.5 of the probability between .49-.51. I would have done it another way as the probability of any head/tail combination is

(H+T)!/H!*T!

and the total number of possible combinations is 2^N

the answer I got was 4200 (it needs to be a multiple of 100 because you can't get a fraction of a head).
 

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