**Here is the probability model.** Consider the fair spinner with the numbers 1, 2, and 3 on it.
Let *X* denote the number spun. Then the probability model for *X* is

Range 1 2 3 Probabilities 1/3 1/3 1/3

In practice we won't know the probability model for *X*. In this case, we won't know if the spinner is fair or not.
But in practice we can take a sample from the probability model. Based on the sample, perhaps we can say something about the probability model.
Now it is very important that the sample is a **random sample**.

A sample is a **random sample** if:

- 1.
- The items in the sample are drawn independently of one another.
- 2.
- Conditions do not change as the sample is drawn.

In this case, the spins of the spinner are independent of one another and we are not changing the chances of a 1, 2 or 3 from spin to spin.

**Here is a Sample drawn from the probability model.** Suppose we decide on a sample size of 100. The following is a sample of the probability model; i.e.,
I spun a fair spinner 100 times:
So here are the results of **my** random sample:

2 2 1 1 2 2 1 2 1 1 1 1 1 1 1 1 2 3 1 3 1 3 1 1 2 2 3 3 2 1 2 3 2 3 3 1 2 1 2 1 3 2 1 3 3 1 2 2 2 1 3 3 1 1 1 1 3 1 1 1 1 2 1 3 2 2 1 2 2 2 3 2 3 2 3 3 3 3 1 1 1 1 2 2 1 1 3 2 3 2 1 1 3 2 1 3 3 1 1 2

We of course tally up the sample and get the sample distribution:

Range 1 2 3 Frequency 43 31 26 Relative Frequency .43 .31 .26

This sample distribution is an **estimate** of the probability model for *X*. That is, .43 is our estimate of *p(1)*, .31 is our estimate of *p(2)*, and .26 is our estimate of *p(3)*. The histogram of the sample

Each dot represents 3 points . : : . : : . : : : : : : : : : : : : -------+---------+---------+---------+---------+------- 1 2 3is our estimate of the graph of the probability model

1/3 1/3 1/3 : : : : : : : : : : : : : : : -------+---------+---------+---------+---------+------- 1 2 3

What's that you're thinking? It seems a little off if the spinner is *fair*?
Be careful, you are starting to think statistically. You may be even thinking of a formal test statistic to see if such a sample could be generated by the fair spinner probability model. Hey, we'll get there soon!

Suppose you compute the sample mean of sample distribution. You get the value . Now since the histogram is an estimate of the probability model, what is 1.83 an estimate of?

It's not hard to see. There are two ways to calculate ,
here. One way is to add up the 100 numbers and divide by the sample size 100. However, in adding up these numbers you have added 3 to itself 26 times. Hence a much easier way is to use the tallying and add as follows:

By the last line it is easy to see what is estimating. Again, .43 is our estimate of

That's the Greek letter

This estimate is the result of one sample! If I spin the spinner another 100 times, I am going to get a different estimate of .
If you spin it 100 times you are going to get a different estimate, too. In fact, if everyone in class spins the spinner 100 times we are going to get
different estimates, (there may be a few ties because the probability model is discrete).

So the important thing to determine is: "How much does
miss
by?" That's the way to think. Keep it up!

In the spinner example , if the spinner is *fair* then
*p*(1) = *p*(2) = *p*(3) = 1/3 and .
So
is an estimate of ,
in this case. We missed by .17.

In practice, we will not know the population mean. But, hopefully, we will have a random sample. We will calculate .
We will estimate with a degree of confidence, "How much does
miss
by?"

In general, the probability model mean, ,
is called a **parameter** of the probability model. For a discrete random variable *X*, to determine the mean, as in the spinner problem, we simply cross multiply the range values by the associate probabilities and total it up; that is,

- 1.
- Let S denote the sum of two numbers spun on a fair spinner with the numbers 1, 2, and 3 on it. The range of S is 2, 3, 4, 5, 6. Determine the probability
model
**mean**of S. - 2.
- Repeat the last problem if the spinner is spun 3 times.
- 3.
- Let X denote the number of aces in a 2 card hand drawn without replacement from a
*well shuffled*standard deck 0f 52 cards. Then the range of X is {0, 1, 2}. Determine the probability model**mean**of X. - 4.
- Repeat the last problem under sampling with replacement.
- 5.
- Let X denote the number of hearts in a 2 card hand drawn without replacement from a
*well shuffled*standard deck 0f 52 cards. Then the range of X is {0, 1, 2}. Determine the probability model**mean**of X. - 6.
- In the urn problem (with the balls well mixed ) discussed above, let Z denote the number of red balls in the sample (without replacement) of size 3. Determine the probability model
**mean**of Z.