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Poisson vs Binomial distribution. |
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| Feb19-13, 01:22 PM | #1 |
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Poisson vs Binomial distribution.
Hello PF
This might be a fairly simple question to most of you, but I was given this problem (don't worry, I already solved it just wondering about something) Suppose the probability of suffering a side effect of a certain flu vaccine is 0.005. If 1000 persons are inoculate, find the approximate probability that (a) at most 1 person suffers, (b) 4,5, or 6 persons suffer. I already solved it, but this problem is in the chapter on the Poisson distribution. Unfortunately my teacher didn't cover this distribution in detail, but when I first looked at the problem it look like a typical Binomial distribution problem? I later figured out I was supposed to approximate with the Poisson distribution. Why would we use an approximation for the Binomial when we could just apply it, and under what circumstances am I allowed to make this approximation in the first place? |
| Feb19-13, 03:11 PM | #2 |
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The problem with the binomial distribution is that it is very hard to calculate.
So the second question would be [tex]\sum_{k=4}^6 \binom{1000}{k} (0.005)^k0.995^{1000-k}[/tex] This is the correct answer. But computing those binomial coefficients is not very fun. However, we can show that if we are working with binomial(n,pn) distributions and if [itex]np_n\rightarrow \lambda[/itex] for some [itex]\lambda[/itex], then [tex]\binom{n}{k} p^k (1-p)^{n-k} \rightarrow e^{-\lambda} \frac{\lambda^k}{k!}[/tex] So, if n is very large and p is very small, then the Binomial(n,p) distribution is very close to the Poisson(np) distribution. So, in our case, p=0.005 is small and n=1000 is large. The product is medium: 5. So we can approximate the answer by [tex]\sum_{k=1}^6 e^{-5} \frac{5^k}{k!}[/tex] And we are also rid of that pesky binomial coefficient. This approximation is also theoretically interesting. The sum of two (independent) Poisson distributions is always a Poisson distribution, for example. But the sum of two (independent) binomial distributions is not binomial. |
| Feb20-13, 01:47 AM | #3 |
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You can use the Poisson approximation when n is large (greater than 50 is probably enough) and when the chance of 0 successes or n successes is negligible. It depends on how much accuracy you need, so there can be no hard and fast rule. |
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