Binomial DistributionDate: 12/13/2001 at 08:43:08 From: lisa Garrard Subject: Conditional probability 37% of people over age 80 will not be around within the next 10 years. In a random sample of 113 senior citizens, find the probability that in the next 10 years, at most 90 will die. Date: 12/13/2001 at 10:50:28 From: Doctor Mitteldorf Subject: Re: Conditional probability Lisa- Have you been working with binomial distributions? That's what this is about. Each person individually has a .37 chance of dying and a .63 chance of living. The probability that all 113 will die is (.37)^113 power, and that all 113 will live is (.63)^113. For in-between values (say n living and 113-n dying) there's a factor like this for each group: (.63)^n * (.37)^(113-n). There's also a "binomial factor" that comes from Pascal's triangle and says how many ways this can happen. For example, there are 113 people, so if n = 1, then there are 113 choices for how to make 1 person live and 112 die. Therefore the binomial factor for n = 1 is 113. For n = 2, it's the number of ways to choose pairs of people who will be the two who live: (113*112)/2 The general formula is 113! C(113,n) = -------------- n!(113-n)! When you're done putting all this together, you should add up the probabilities that the number of people who die will be 91 all the way through 113. My guess is that you'll have a pretty low number, since you really only expect an average of 42 out of 113 to die (37%). The chance that you have as many as 91 unlucky people is very small, and for 92 or more it's even smaller. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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