Main Lesson: The Mean and Variance of a Discrete Random Variable

A. Introduction: Review of Probability Distributions

Ask learners to toss three coins ten times and record the number of heads on each toss. Divide the class into groups of five learners. Tell each learner to get the average of the number of heads obtained for the first five tosses, and the average of all the ten tosses. Then, in groups of three to five, get the average of the

averages that the groups got. If possible, ask learners to get the average for the entire class.

Next, ask learners what the possible number of heads was. They should say 0, 1, 2, 3. Next, ask them what the range of the average of the first five tosses was. Ask what the highest and lowest values are in the class. Record the number on the board. Do this also for the average of the ten tosses, the average of each group, and, finally, record the average of the class. Ask learners what they notice about the average of the number of heads. Also, ask them what they notice about the range of values as the number of tosses is increased. They should have noticed

fluctuations in the averages, but the averages approach 1.5, and the range of values from the averages gets narrower with more data (i.e. more students giving

information).

B. Main Lesson: The Mean and Variance of a Discrete Random Variable

Recall the lesson three for this chapter (on Probability Distributions of Discrete Random Variables). List down the distribution of the number of heads in tosses of three fair coins (or three independent tosses of one fair coin). If possible, ask learners to complete the first two columns of the table. Then, add another column for the product of the entries of the first and second columns, (X) P(X). After

completing the table, ask them to get the total of the row. Ask them to fill out the entries. Leave a fourth row which will be filled out later.

X=number of heads P(X) (X)P(X)

Ask them what multiplying X and P(X) resembled. They should have seen that it resembled getting a “weighted average” of data, with the probabilities serving as weights. Ask them about how the sum compares to the value that they got from the activity. They should notice that the average they got gets closer and closer to 1.5 as the number of values you average increases. This is related to the concept of the Mean of a random variable

Now, formally define the mean of a discrete random variable.

Definition

Given a discrete random variable X, the mean, denoted by μ, is the sum of the products formed from multiplying the possible values of X with their corresponding probabilities. It is also called the expected value of X, and given a symbol . More formally:

Help learners recall that empirical probabilities tend toward theoretical probabilities and, in consequence, the mean is also a long-run average. This can be observed from the results of the activity. As the number of trials of a statistical experiment increases, the empirical average also gets closer and closer to the value of the theoretical average. Inform learners that this is why we can interpret the mean as a long-run average.

Remind learners that if they have three coins tossed (and if all coins are fair), then they would have eight possible outcomes—HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. And if they would repeat tossing these coins 8000 times, they would expect 1000 tosses of each of the outcomes and, thus, the expected frequency of 3 heads would be 1000 tosses; of 2 heads, would be 3000 tosses; of 1 head, would be 3000 tosses; and no heads, would be 1000 tosses. If we average these, we would have

Tell learners that although the Mean is called an “expected value,” this should not be interpreted as the actual expected result when they do an experiment. In the example for tossing three coins, the mean is 1.5. Point out that when you toss three dice, you cannot get 1.5 heads out of it. This indicates that the Mean is not

necessarily a possible value of the random variable. So learners cannot simply say that the Mean is what they expect to be the number of heads when they toss three coins. Rather, it is to be interpreted as a long-run average. Mention also to them that the Mean is the value that we expect the long-run average to approach and it is not the value of the random variable X that we expect to observe.

Next, have learners recall that the average of a given set of data is a measure of central tendency. Inform them that the expected value—being an average—

measures the center of the distribution of the possible values of X.

Mention that the mean of a (discrete) random variable X can be given as a physical interpretation. Suppose we imagine that the x-axis is an infinite see-saw in each direction, and at each possible value of X, we place weights equal to the

corresponding probabilities. Then, the mean is at a point which will make the see-saw balanced. In other words, it is at the centre of gravity of the system.

It may be helpful to give other examples to help learners gain more insights.

C. Examples of Finding and Interpreting the Mean and Variance I. Example with biased dice

Recall the biased six-sided dice with a probability distribution for X, the number of spots on the upward face when the die is rolled, given as follows:

I 1 2 3 4 5 6

P(X=i)

The expected value of the distribution may be calculated as follows:

= + 2 ( ) + 3 ( ) + 4 ( ) + 5 ( ) + 6 ( ) =

If q = 0, then this reduces to a fair dice, for which, we would have a long-run average of 3.5 for the number of spots on the upward face.

Provide another example that is used in the real world, such as the following:

II. Practical Example Used in Insurance

An insurance company sells life insurance of Php500,000 for a premium or payment of Php10,400 per year. Actuarial tables show the probability of normal death in the year following the purchase of this policy is 0.1%. What is the expected “gain” for this life insurance policy?

Inform learners there are two simple events here. Either the customer will live this year or will die (a normal death). The probability of a normal death, as given by the problem, is 0.001, which will yield a negative gain to the insurance company in the amount of

-489,600= Php10,400 - Php500,000. The probability that the customer will live is 1-0.001=0.999. Thus, the insurance company’s expected gain X from this insurance policy in the year after the purchase has the following probability distribution:

Gains Outcome Probability

10,400 Live 0.999

-489,600 Normal Death 0.001 m = (10,400)(0.999) + (-489,600)(0.001) =9,900

Learners should take note that if the insurance company were to sell a very large number N of the Php500,000 insurance policies to many people, with the long-run average profit per insurance policy is hp,900, the company would be expected to make a total profit of N times Php9,900.

Next, ask learners whether a measure of central tendency is the only relevant

summary measure. They should remember that for a set of data, we also need other summary measures, such as measures of variability. Learners have already met the concepts of variance and standard deviation when summarizing data. Assist them to remember that the variance and standard deviation of a set of data are measures of spread. Tell learners that random variables also have a variance (and a standard deviation). The variance is derived by getting the expected value of (X-μ)² where μ is the Mean.

To illustrate, go back to the Table on flipping three coins and get the number X of heads in these three coins. Now, add the two columns with the following heading in bold (See below.), and fill the corresponding values.

X=number of heads P(X) (X)P(X) (X-μ)² (X-μ)²P(X)

0 1/8 0 (0-1.5)²=2.25 0.28125

1 3/8 3/8 (1-1.5)²=0.25 0.09375

2 3/8 6/8 (2-1.5)²=0.25 0.09375

3 1/8 3/8 (3-1.5)²=2.25 0.28125

Total 12/8 = 1.5 0.75

The total in the last column is called the variance of the random variable, and the square root, 0.866, is the standard deviation.

Now, define the variance of a random variable as the weighted average of squared deviations of the values of X from the mean, where the weights are the respective probabilities. The variance, usually denoted by the symbol , is also denoted as and formally defined as

The variance gives a measure of how far the values of X are from the mean. Ask learners what the variance is when X is a constant with probability of one. Learners should be able to say that it is zero in this case (just as it was in the case of a set of data that does not vary). Inform learners that in nontrivial cases (i.e. when there is more than one possible distinct value of X), the variance will be a positive value.

The bigger the value of the variance, the farther the values of X get from the mean.

Define the standard deviation as the square root of the variance of X. That is,

III. Example of Gains in Life Insurance

Show the following calculations for “deviations” formed from subtracting the mean from the gains, as well as squared deviations, and weighted squared deviations.

Gains Probability Deviations Squared

Deviations Weighted Squared Deviations

10,400 0.999 10388.73 107925794 107817868

-489,600 0.001 -490.466 240556.92 240.55692

The variance is the sum of the entries on the last column, i.e., s²=107817868+240.55692 =107818108

while the standard deviation is the square root of the variance s = 10383.55

Remind learners that the standard deviation is the more understandable of the two measures of spread, since the standard deviation is in the same units as X. For example, if X is a random variable representing the number of heads in three tosses of a fair coin, then the units for standard deviation is “heads,” while the variance is in square heads (heads²).

Unlike the mean, there is no simple interpretation for the variance or standard deviation. The variance though is analogous to the moment of inertia in Physics, but that is not necessarily widely understood by learners. Stress that, in relative terms,

• a small standard deviation (and variance) means that the distribution of the random variable is quite concentrated around the mean

• a large standard deviation (and variance) means that the distribution is rather spread out, with some chance of observing values at some distance from the mean

Inform learners that, in practice, the variance is not computed with the definition, but rather using the following result:

Thus, the variance is the difference between the expected value of X2 and the square of the mean.

Explanatory Note: This can be derived from the definition, some algebraic

expansion of a binomial expression, and some properties of expected values (such as the mean of a constant is the constant):

It is suggested though that this derivation not be discussed in class. It may be helpful though to use this computational formula, and to use computers whenever possible.