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.

  

Figure 6.1: Probability model : spinner

Note from the picture that the mean is 2; i.e., $\mu = 2$. Show that the standard deviation, $\sigma$, is $\sqrt{2/3}$. 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 $\bar{X}_2$  be the average of the numbers spun. For example, if the numbers 1 and 3 are spun then S₂ = 4 and $\bar{X}_2 = 2$. 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.

  

Figure 6.2: Probability model : S₂

How about $\bar{X}_2$? 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 $\bar{X}_2$ 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.

  

Figure: Probability model : $\bar{x}_2$

Note that the range of $\bar{X}_2$ is from 1 to 3, which is the same range  as X. The probability distribution  of $\bar{X}_2$ is quite different than that of X. The distribution of X is flat and uniform. On the other hand, the distribution of $\bar{X}_2$ is unimodal and mound shaped.

Note from the picture that the mean is 2; i.e., $\mu = 2$. You can show that $\sigma = \frac{\sqrt{2/3}}{\sqrt{2}}$.

Suppose we spin it three times. Let S₃ be the sum of the numbers spun and let $\bar{X}_3$ be the average of the numbers spun. For example, if the numbers 1, 3 and 3 are spun then S₃ = 7 and $\bar{X}_3 = 7/3$. What is the probability model for S₃? Just put the set of third branches on the above tree diagram. How about $\bar{X}_3$? 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 $\bar{X}_3$ 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.

  

Figure: Probability model : $\bar{x}_3$

Note from the picture that the mean is 2; i.e., $\mu = 2$. You can show that $\sigma = \frac{\sqrt{2/3}}{\sqrt{3}}$.

Ready to jump? What's happening here? As the number of spins n increases, the mass is moving towards the center. It seems that $\bar{X}_n$ is more "likely" to be in the middle. The mean of $\bar{X}_n$ 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 $\bar{X}_n$ 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.

  

Figure 6.5: Probability model : Spinner with no 2

Note from the picture that the mean is 2; i.e., $\mu = 2$. You can show that $\sigma =\sqrt{5}$.

Now suppose we spin it twice. Let S₂ be the sum of the numbers spun and let $\bar{X}_2$ 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 $\bar{X}_2$? Well that's just S₂/2. The distribution of $\bar{X}_2$ is given by:

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

Better yet, see Figure 6.6.

  

Figure: Probability model : $\bar{x}_2$

Even for this probability model, the mass for $\bar{X}_2$ is moving to the middle. The mean of the distribution of $\bar{X}_2$ and its standard deviation is $\sqrt{4.5}$.

Exercise 7.1.1  

1.

Suppose the population has the distribution

      Range   0    1
      Prob.   .3  .7

Let $\bar{X}_3$ be the sample average of a sample of size 3 from this population. Using a tree diagram, show that the distribution of $\bar{X}_3$ 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 $\bar{X}_3$ have the same value .7.

3.

Show that the variance of the population in #1 is .21.

4.

Show that the variance of $\bar{X}_3$ in #1 is .07.