Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Let's return to our Galton board for further exploration. In the previous section, we used the Law of Large Numbers to see that, for each
Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Let's return to our Galton board for further exploration. In the previous section, we used the Law of Large Numbers to see that, for each bin, the theoretical and experimental probabilities get closer and closer to one another as we put more and more marbles through the process. We're going to return to theoretical probabilities now and examine how the probabilities are distributed across all of the bins.
Recall that in our earlier examples the bins in the middle had higher probabilities than did the bins at the sides. This was because there are more paths that terminate in the middle bins than terminate in the side bins. Let's make a histogram that correlates to the bins and their probabilities.
First a simple 2 row machine:
Notice that the distribution is symmetric; the probabilities for both the far right and far left bins are equal. Let's look at the histogram for a Galton board with four rows.
Notice that the probabilities for each version of the system are distributed across all of the bins and that even though the individual probabilities change as we add more bins, they always sum to 1. This is in keeping with our intuition that any marble must indeed end up in exactly one of the bins. At this point, we're going to need to label the bins so that we can discuss results in more detail.
To do this, let's say that each marble, as it progresses through the system, gets 1 point for each movement to the right and 0 points for each movement to the left. Each bin can then be represented by the summative "score" of a marble that ends up there.
For example, a marble passing through a four-row system would have a maximum possible score of 4, corresponding to the right-most bin, and a minimum possible score of zero, corresponding to the left-most bin. The remaining bins would have scores as follows:
The average result, which would be the expected score for one average marble, can be found as before by multiplying the value of each bin by the probability of landing there and summing the results. This would give us.
Interestingly, we can also find this by looking at the number of rows and multiplying by the probability of going right. This would be 4 rows times = 2. Recall t