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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.


  
Figure 6.1: Probability model : spinner
\begin{figure}
\begin{center}
\epsfig{file=fig25.ps, height=5in, width=5in, angle=-90}\end{center}\end{figure}

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
\begin{figure}
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\epsfig{file=fig26.ps, height=5in, width=5in, angle=-90}\end{center}\end{figure}

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 : $\bar{x}_2$
\begin{figure}
\begin{center}
\epsfig{file=fig27.ps, height=5in, width=5in, angle=-90}\end{center}\end{figure}

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 : $\bar{x}_3$
\begin{figure}
\begin{center}
\epsfig{file=fig28.ps, height=5in, width=5in, angle=-90}\end{center}\end{figure}

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
\begin{figure}
\begin{center}
\epsfig{file=fig29.ps, height=5in, width=5in, angle=-90}\end{center}\end{figure}

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 : $\bar{x}_2$
\begin{figure}
\begin{center}
\epsfig{file=fig30.ps, height=5in, width=5in, angle=-90}\end{center}\end{figure}

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.


next up previous contents index
Next: Central Limit Theorem Up: Central Limit Theorem Previous: Central Limit Theorem

2001-01-01