- #1
ChrisVer
Gold Member
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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?
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?