Topic 14 (continued): More about Probability Histograms
|Learning objectives: Learn more about what probability histograms are. Develop an intuitive feel for a major result of the Central Limits Theorem: "As the number of draws from a box grows large, the probability histogram for the sum of the draws comes to resemble the normal curve."|
Remember that we began the present discussion by seeking license to work problems like this:
A charitable organization knows from long experience that responses to "begging letters" have an average of $2.90 and an SD of $10. If the organization sends out 10,000 letters to a random sample of potential donors, then about what are the chances that this appeal will bring in at least $30,000?
We reasoned that the entire experiment (send out 10,000 letters ; sum the donations ) could, conceptually, be repeated an arbitrarily large number of timesand that the massive list of sums thereby generated would tell us about the long-term performance of the sampling exercise. If we had access to such a collection of data, preferably in histogram form, then we could easily answer the above question: wed simply add up all the area to the right of $30,000.
Obviously it is not practicable to generate such a histogram empirically. However, I have stated that the histograms average and SD must be, respectively, the EV(sum) and SE(sum). Therefore, if we could somehow show that the histogram looked like the normal curve, we could perform a normal approximation. This is where high-level statistical theory comes in handy. Over the course of a century or two, a number of mathematicians collectively proved the Central Limits Theorem (CLT). This intellectual tour de force says, among other things, that as the number of draws from a box grows large, the probability histogram for the sum (or average or percent) of the draws comes to resemble the normal curve.
In our last class we began to define the probability histogram. We did this two ways. First, we suggested that hypothetical histograms of the type described aboverepeat the experiment an arbitrarily large number of times; summarize the sums by means of a histogramare roughly what we mean by "the probability histogram." Second, we attempted a definition by examples. Probably you remember how we drew the probability histogram for the simple experiment, "take the sum of one roll of a fair die." We listed possible outcomes (1, 2, 3, 4, 5, 6) along the horizontal axis. Then, above those outcomes, we drew rectangles whose areas were equal to the outcomes probabilities. In the "one roll" example, we made all rectangles of equal size because if the die be fair, then all possible outcomes are equally likely. Here is a sketch of that probability histogram:
Note that every rectangle has the same area. That is what we want; areas represent probabilities, and if the die is fair, then every spot should have the same probability.
At this point, then, you should know what a probability histogram is, and you might be willing to accept on faith the assertion about approach to normalityand thats good, for we are definitely not going to prove the CLT. On the other hand, perhaps we can at least offer some gentle persuasion that will allow your faith to grow. To do so, let us return to the problem of dice sums. Clearly the above probability histogram, for the one-die roll, looks nothing like the normal curve. But next lets consider the sum of two rolls. Clearly, 2 is not very large; still, it is bigger than 1, and if the CLT is true, perhaps we can begin to glimpse, even with only two draws, some approach to normality. To draw the probability histogram for the two-roll experiment, first erect your axes and then list the possible outcomes along the horizontal one. Now figure how big each of the eleven rectangles should be. Probably you should do this by brute force. There are 36 different, equally likely, draw combinations (the first draw has six different possibilities; for each of these, the second draw has six different possibilities ). How many of these draw-combinations give you a "2"? (Answer: only one.) So how much area goes in the rectangle to be erected above the "2"? (Answer: 1/36.) Fill in the table, and be sure you can draw the probability histogram.
|Number of ways to get it||1|
|Probability of sum||1/36|
Heres the probability histogram, with the dice-roll combinations written in the rectangles:
I admit that the probability histogram for the two-dice toss doesnt look very much like the normal curve. However, (1) it looks more like the normal curve than the histogram for the one-die toss, and (2) two draws from a box isnt a very large number. (I leave it to you to draw the histogram for the three-die toss or the four-die tossif you are entirely lacking in more useful employment.)
Now we can move on to another, easier experiment. Consider tossing a fair coin and counting "heads." Lets start with one toss, then do two tosses, then three tosses, then four tosses. That is, well increase the number of draws from one to two to three to four. If the CLT is true, then we should begin to observe some modest approach to normality. As a climax, you might want to complete the four-toss table:
|Number of "heads"||0||1||2||3||4|
|Combinations that give that number of "heads"||TTTT||TTTH
(You fill in the rest of them if you like .)
|Count of combinations||1||4||1|
Ill draw the probability histograms for the one-coin toss (one draw from the box) and the four-coin toss (four draws from the box). If you like, you can draw probability histograms for 2 and 3 tosses.
After you have sketched and examined the probability histograms, youll see why mathematicians suspected the truth of the CLT before it was proved. But 4 isnt really a very large number. On p. 316 your book draws probability histograms for 100, 400, and 900 draws from the "fair coin" box. Then p. 320 shows probability histograms for draw from a box for a very unfair coin. All this should help convince you that As the number of draws from a box grows large, the probability histogram for the sum (or average or percent) of the draws will come to resemble the normal curve.
Heres something else you should note. The list of numbers that gave rise to the Four-Coin-Toss probability histogram was [0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4]. You should figure the average and SD of that list the old way (like you learned back before the first test). You will find that the average is ____ and the SD is ____. Now you should also figure the EV(sum) and SE(sum). These are:
EV(sum) = ave(box)(# of draws) = (1/2)(4) = 2 (Is this the average of the probability histogram?)
SE(sum) = SD(box)[(# of draws)0.5] = (1)(2) = 2 (Is this the SD of the probability histogram?)
Now please do a bunch of the review exercises on pp. 327-329.