Binomial ExperimentsDate: 09/05/2003 at 19:21:26 From: Confused Subject: What is the Binomial Distribution? How can I explain the binomial distribution to my six grade math class? Date: 09/05/2003 at 20:50:36 From: Doctor Tom Subject: Re: What is the Binomial Distribution? Well, I'm not sure how far you can hope to get with sixth graders, but there are some things you can do. A binomial distribution arises any time an event with exactly two outcomes is repeated a bunch of times. The simplest example is the case where the two outcomes are equally likely -- flipping a coin, for example. For the class, what I'd do is pick a fixed number of flips, say 10. And rather than flip one coin ten times, it's easier if they flip 10 coins once and count the number of times "heads" appears. Have everybody in the class do this 10 times (or more) and record the number of "heads" for each attempt. There are 11 possible outcomes, from 0 heads to 10 heads. Obviously, it'll be unlikely to get any zeros or tens, but just collect the data and draw it as a bar chart and notice that the most likely outcome seems to be 5, and that it does seem to form a bell-like distribution. With more and more repeats of the experiement, you'll find that the proportions of times each outcome occurs matches better and better the theoretical results (which you can obtain from the tenth row of Pascal's triangle, where I count the top row as row zero). In other words, if you add the eleven numbers in that tenth row, they will add to 2^10 = 1024, and so as the number of experiments (where an experiment consists of flipping a coin ten times) with results 0 through 10 heads should occur with the following likelihoods: 1 10 45 120 210 252 210 120 45 10 1 ----, ----, ----, ----, ----, ----, ----, ----, ----, ----, ---- 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 In other words, about 1/4 (252/1024, to be precise) of the times that you do the experiment, there will be exactly 5 heads. Only 1 time in about a thousand will all ten be heads, and so on. You can compare these theoretical results with those generated by your experiments. That may be good for the whole lesson, but a binomial distribution does not require that the likelihoods of the results be the same. For example, suppose your experiment is to roll a die with six sides, and you consider that you "win" if you obtain a 5 or a 6, and that you "lose" if you obtain 1, 2, 3 or 4. If you run that experiment ten times (ten rolls of a die or 1 roll of ten dice), there are again eleven possible outcomes: zero through ten wins. But it stands to reason (since you usually lose) that it will be FAR more likely to get ten losses than ten wins. With some dice and time, you can do the same thing here as you did with the coins. You'll see that you also get a bell-shaped curve, but it'll be lopsided, with the bump nearer zero than 10. In fact, since you win 1/3 of the time, the bump will be near the 1/3 point -- near 3 wins in 10. The expected results can be calculated, but it's quite a bit messier to do so. You can use the same fraction numerators that we worked out with Pascal's triangle (1, 10, 45, 120, 210, 252, 210, 120, 45, 10, and 1), but they need to be multiplied by (1/3)^W(2/3)^L to get the probability of W wins and L losses, where W+L = 10. For example, the probability of zero wins is: 1 x (1/3)^0 x (2/3)^10 = .01734... The probability of ten wins is: 1 x (1/3)^10 x (2/3)^0 = .000016935... The probability of exactly 3 wins (most likely result) is: 120 x (1/3)^3 x (2/3)^7 = .26012... I'll bet this is beyond a sixth grade class, however. You might look at it not with 10 rolls, but rather with 2. Then there's a hope of seeing what's going on by drawing a tree for the possible outcomes: start at a root and go left (win) with 1/3 probability, and go right (lose) with 2/3 probability. From those nodes repeat for the probabilites after two events. You can then sum up the probabilities to obtain: 1/3 x 1/3 = 1/9 chance of two wins 1/3 x 2/3 + 2/3 x 1/3 = 4/9 chance of one win (could be first or second) 2/3 x 2/3 = 4/9 chance of zero wins Maybe you could extend this to three experiments and notice that there is: 1 way to get 3 losses 3 ways to get 2 losses and 1 win 3 ways to get 1 loss and 2 wins 1 way to get 3 wins. Point out that this looks like Pascal's triangle's third row and that the only difference is that as there are more and more wins, the likelihood gets smaller and smaller. Look in any introductory book on statistics for more material, but I'm pretty sure it will be beyond what you can use in a sixth grade class. Good luck! - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ |
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