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 1/3Better 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 *S*_{2} 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 *S*_{2} = 4 and
.
What is the probability model for *S*_{2}? 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 *S*_{2}/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 1/9See 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 *S*_{3} 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 *S*_{3} = 7 and
.
What is the probability model for *S*_{3}? Just put the set of third branches on the above tree diagram. How about ?
Well that's just *S*_{3}/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 1/27See 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 1/2See 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 *S*_{2} be the sum of the numbers spun and let
be the average of the numbers spun. What is the probability model for *S*_{2}? Just use a tree diagram (i.e., two spins, top branch is 1 and 1, etc.: SKETCH IT!). How about ? Well that's just *S*_{2}/2. The distribution of
is given by:

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

- 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 isRange 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.