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Toss coin probabilities (Bino vs Gauss approx)

  1. May 24, 2015 #1


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    Hi, I calculated the probability to this scenario:
    getting between 3 and 6 Heads after tossing a coin for n=10 trials...

    The binomial probability for this is:
    [itex]P(3 \le k \le 6) = \sum_{k=3}^6 Bi(k;p,n)= \sum_{k=3}^6 \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} =\sum_{k=3}^6 \frac{10!}{k!(10-k)!} (0.5)^{10} = 0.7734375 [/itex]

    Whereas for the Gaussian approximation to the binomial:
    [itex] G(x) = \frac{1}{\sqrt{2 \pi n p (1-p) }} \exp\Big( - \frac{ (x-np)^2}{2np(1-p)} \Big)[/itex]

    I get:
    [itex]P(3 \le x \le 6) = \int_3^6 dx~ G(x) = ... = 0.633505 [/itex]

    My question is why are the two probabilities so off from each other? Is it because [itex]n[/itex] cannot be considered as "large" ? I think when extracting the Gauss approximation, the additional higher orders go as [itex]\frac{1}{n}= 0.1[/itex] but that's for the gaussian distribution ##G## and not the integral of it.
    Any feedback?
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  3. May 24, 2015 #2

    Simon Bridge

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  4. May 24, 2015 #3


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    So the difference is mainly due to the fact that [itex]np= n(1-p)=5[/itex] is not large enoguh? So Gauss can go a bit off in its predictions (here ~10%)...

    The strange thing is that this is the only rule the approximation break, whereas the other two rules give:
    [itex] \Big| \frac{1}{\sqrt{n}} \Big( \sqrt{\frac{q}{p}} -\sqrt{\frac{p}{q}} \Big) \Big| = 0 <0.3 [/itex]
    [itex] np \pm 3 \sqrt{npq} \in [0.26 , 9.74 ] [/itex] which is in [0,n=10]...

    For the calculations I try to build my own source codes just for a practice :smile: (or check with the formulas in wolframalpha), and so far my numerical methods work fine.
  5. May 30, 2015 #4
    You are forgetting the continuity correction: $$ P(3 \le x \le 6) \approx \int_{2.5}^{6.5} G(x)~dx \approx 0.7717 $$
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