# Binomial cdf (Or something completely different? look inside)

• Jul 30th 2008, 05:06 AM
Lore
Binomial cdf (Or something completely different? look inside)
Our good friend Jones wants to explore South America's wild jungle. You see, Jones wants to see at least 10 500 people of the native inhabitants in South America. He wants to go from tribe to tribe, and we assume that he is an excellent counter or that he got some kind of device that makes him able to count the amount of inhabitants in the tribe in no time.

The problem for Jones, however, is that he's a tad unsure about how many tribes he has to visit to achieve his goal on seeing 10 500 people. There are a 60% probability that a tribe has a population on between 5000 and 2700, and a 40% probability that the tribe has a population on between 3000 and 1620.

The tribes have the same probability for each population. That is, the chance for the 60% probability-tribes to have a population of 5000, 2700 or any number in between is the same. This also counts for the 40% probability tribes.

a) Find the probability that Jones only needs to visit 3 tribes.

b) Find the probability that Jones needs to visit 4, 5, 6 and 7 tribes.

For finding the probabilities, I assume that normal distribution is in the picture somewhere. If the probability was 100% that the population was e.g. between 5000 and 2700, you could just multiply the difference between max and min with the amount of tribes he met and divide by 6 (to get σ). Then put the mean value of the max and min multiplied by the amount of tribes as μ into the cdf, which will give a "rough" estimate of the answer you're looking for (by putting in 1- P(X<10499.5)). You're most likely going to use this when you're doing 0.6^n, 0.4^n, and all 0.6^(n/2)*0.4^(n/2) (as long n>3), correct?

However, this is not the case for the majority of the solutions, and that's why I'm here (Hi)

Also, I sincerely apologize if this is in the wrong section, as I am not familiar with the American school system nor what they learn.