Poisson vs Binomial distribution.

In summary, the conversation discusses the use of the Poisson distribution as an approximation for the Binomial distribution when dealing with large sample sizes and small success probabilities. This is due to the difficulty in computing binomial coefficients and the potential for computer overflow. The approximation is valid if the chance of 0 or n successes is negligible and if the sample size is sufficiently large.
  • #1
Jcampuzano2
5
0
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?
 
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  • #2
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.
 
  • #3
Jcampuzano2 said:
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?

Partly it is holdover from the old days when computation was expensive. The teaching of statistics hasn't changed much in the past 50 years, as far as I could tell. The binomial is still tricky to compute because the factorials in the intermediate results can be very large and you have to be careful not to get computer overflow.

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.
 

1. What are the main differences between Poisson and Binomial distributions?

The Poisson distribution is used to model the number of occurrences of a rare event in a fixed interval of time or space. It assumes that the events are independent and occur at a constant rate. On the other hand, the Binomial distribution is used to model the number of successes in a fixed number of trials, where each trial has only two possible outcomes (success or failure) and the probability of success is constant.

2. When should I use a Poisson distribution instead of a Binomial distribution?

You should use a Poisson distribution when the number of trials is very large (approaching infinity) and the probability of success is very small (approaching zero). This is because in this scenario, the Binomial distribution closely approximates the Poisson distribution. Additionally, the Poisson distribution is more appropriate when the outcome of the event is discrete (e.g. number of emails received) rather than continuous (e.g. height).

3. Can a Poisson distribution be used to model a Binomial distribution?

Yes, a Poisson distribution can be used to approximate a Binomial distribution when the number of trials is very large and the probability of success is very small. This is because as the number of trials increases, the Binomial distribution approaches the shape of a Poisson distribution.

4. What are the key assumptions of a Poisson distribution?

The key assumptions of a Poisson distribution are that the events are independent, they occur at a constant rate, and the probability of occurrence is very small for each individual event. Additionally, the events should be random and the time or space interval should be fixed.

5. How do I calculate probabilities with Poisson and Binomial distributions?

The formula for calculating probabilities with a Poisson distribution is: P(X = x) = (e^-λ * λ^x) / x!, where λ is the average number of occurrences and x is the specific number of occurrences. For a Binomial distribution, the formula is: P(X = x) = (nCx) * p^x * (1-p)^(n-x), where n is the number of trials, x is the specific number of successes, and p is the probability of success on each trial.

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