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# Some Probability Examples

Why are you taking this class? There are several reasons but the main reason is the Central Limit Theorem . It's the reason why statistics works. In this chapter we consider this theorem. In the rest of book, we will make use of it. We begin with a few probability examples which illustrate it.

Consider our favorite example: a fair spinner  with the numbers 1, 2 and 3 on it. Let X be the number spun. Then the probability model for X is:

Range    1    2   3
Prob.   1/3  1/3  1/3

Better yet, a picture of it is found in Figure 6.1.

Note from the picture that the mean is 2; i.e., . Show that the standard deviation, , is . If you cannot show it, ask your instructor in class to show you.

Now suppose we spin it twice. Let S2 be the sum of the numbers spun and let   be the average of the numbers spun. For example, if the numbers 1 and 3 are spun then S2 = 4 and . What is the probability model for S2? Just use a tree diagram (i.e., two spins, top branch is 1 and 1, etc.: SKETCH IT!). My sketch is given in Figure 6.2.

How about ? Well that's just S2/2. So the probabilities don't change, just the range values. If you have done your tree diagram  correctly, the probability model for is given by:

Range    1    3/2   2   5/2    3
Prob.   1/9   2/9  3/9  2/9   1/9

See Figure 6.3.

Note that the range of is from 1 to 3, which is the same range  as X. The probability distribution  of is quite different than that of X. The distribution of X is flat and uniform. On the other hand, the distribution of is unimodal and mound shaped.

Note from the picture that the mean is 2; i.e., . You can show that .

Suppose we spin it three times. Let S3 be the sum of the numbers spun and let be the average of the numbers spun. For example, if the numbers 1, 3 and 3 are spun then S3 = 7 and . What is the probability model for S3? Just put the set of third branches on the above tree diagram. How about ? Well that's just S3/3. So the probabilities don't change, just the range values. If you have done your tree diagram correctly, the probability model for is given by:

Range    1     4/3    5/3     6/3   7/3   8/3     3
Prob.   1/27   3/27   6/27    7/27  6/27  3/27  1/27

See Figure 6.4.

Note from the picture that the mean is 2; i.e., . You can show that .

Ready to jump? What's happening here? As the number of spins n increases, the mass is moving towards the center. It seems that is more "likely" to be in the middle. The mean of is 2, the same mean as in the original model. The variance is, however, getting smaller. Again the mass is moving towards the center, the distribution of is becoming less dispersed.

Lets try one more example.

Suppose 2 can never be spun. That is the distribution of X, the number spun, is

Range    1    2   3
Prob.   1/2   0  1/2

See Figure 6.5.

Note from the picture that the mean is 2; i.e., . You can show that .

Now suppose we spin it twice. Let S2 be the sum of the numbers spun and let be the average of the numbers spun. What is the probability model for S2? Just use a tree diagram (i.e., two spins, top branch is 1 and 1, etc.: SKETCH IT!). How about ? Well that's just S2/2. The distribution of is given by:

Range    1     2     3
Prob.   1/4   2/4   1/4

Better yet, see Figure 6.6.

Even for this probability model, the mass for is moving to the middle. The mean of the distribution of and its standard deviation is .

Exercise 7.1.1
1.
Suppose the population has the distribution
      Range   0    1
Prob.   .3  .7

Let be the sample average of a sample of size 3 from this population. Using a tree diagram, show that the distribution of is
      Range  0       1/3   2/3   1
Probs  0.027 0.189 0.441 0.343

2.
Show that the means of the population and the distribution of have the same value .7.
3.
Show that the variance of the population in #1 is .21.
4.
Show that the variance of in #1 is .07.

Next: Central Limit Theorem Up: Central Limit Theorem Previous: Central Limit Theorem

2001-01-01