Suppose that a school has 20 classes: 16 with 25 students in each, three with 100 students in each, and one with 300 students, for a total of 1000 students.

The average class size is simply

$$\bar{x} = 50$$

If we let the random variable $X$ be the class size for the support $S = \{25, 100, 300 \}$. The p.m.f for $X$ is

$$f(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} \frac{400}{1000} & \text{if $x=25$}\\ \frac{300}{1000} & \text{if $x=100,300$}\\ \end{array}\right. $$

Thus the mathematical expectation $\mu$ is

$$\mu = 25(400/1000) + 400(300/1000) = 130$$

My questions are as follows: Which average is more correct? Which average would I choose and for what circumstance if they are both correct? Why is there a difference in the first place?

  • $\begingroup$ They're defined exactly the same. The reason they didn't evaluate the same is that your PMF is wrong. The probability $P(X = 25)$ isn't $0.4$, and the probabilities $P(X = 100)$ and $P(X = 300)$ are not $0.3$. Where did you get that PMF from? $\endgroup$ – Brian Tung May 19 '15 at 17:48
  • $\begingroup$ solutions manual $\endgroup$ – Benedict Voltaire May 19 '15 at 17:51
  • 2
    $\begingroup$ Is it possible that there is a question in there that asks not what the average class size is, averaged over classes, but rather what the average class size is, averaged over students? Because that's what that second PMF would be for: If you selected one of the $1000$ students at random and recorded the size of the class they were in, the expected value would be $130$. So it's still not that average and expected value are different in any way, but that you can calculate them over different sample spaces. $\endgroup$ – Brian Tung May 19 '15 at 17:53
  • 1
    $\begingroup$ For instance, suppose there are two families. One has a single child, Albert. The other has two children, Beth and Christine. If you select a family uniformly at random, that number of children in that family has an expected value of $(1+2)/2 = 3/2$. But if you select a child uniformly at random, the number of children in that child's family has an expected value of $(1+2+2)/3 = 5/3$. Do you see the distinction there? In either case, you can use the term "average" or "expected value"; they mean the same thing in these cases. $\endgroup$ – Brian Tung May 19 '15 at 17:56
  • $\begingroup$ Well, the frist question is (a) What isthe average class size, and then (b) Select a studnet randomly out of the 1000 students. Let the random variable $X$ equal the size of the class to which this student belongs, and define the p.m.f. of $X$. So, I think what you're saying is what's happening in this problem. $\endgroup$ – Benedict Voltaire May 19 '15 at 18:07

Both averages are calculated correctly. They differ because there are two sampling models being employed -- in one approach we sample a class at random and ask what is the average; in the other approach we sample a student at random and ask what is the average class size the student sees.

As for which number you would choose to represent the 'average' class size at the school, notice that the class-sampling approach makes the class sizes look rather small, while the student-centric approach gives a considerably larger number. If I were a prospective student I would want to know the student-centric average, because this is a more honest representation of the class size that your 'average' student will experience. OTOH the school administrators would love to brag that, according to their calculations, the average class size at the school is 50.

  • $\begingroup$ On the third hand, if you ask a randomly selected student how much of the instructor's attention they're getting (under a naive interpretation of "fraction of attention"), the expected value of that is the reciprocal of the instructor-centric average again. :-) $\endgroup$ – Brian Tung May 19 '15 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.