problem 16

(a): From the student’s perspective, the average class size is $E (X) = \frac{200}{360} 100 + \frac{160}{360} 10 = 60$ . From the dean’s perspective, the average class size is $E (X) = \frac{16}{18} 10 + \frac{2}{18} 100 = 20$ . The discrepancy comes from the fact that when surveying the dean, there are only two data points with a large number of students. However, when surveying students, there are two hundred data points with a large number of students. In a sense, the student’s perspective overcounts the classes.
(b): Let $C$ be a set of $n$ classes with $c_{i}$ students for $1 \leq i \leq n$ . The dean’s view of average class size then is $E (X) = \sum_{i = 1}^{n} \frac{c_{i}}{n}$ . The students’ view of average class size is $E (X) = \sum_{i = 1}^{n} (c_{i} \frac{c_{i}}{\sum_{i = 1}^{n} c_{i}})$ . In the dean’s perspective, all $c_{i}$ are equally weighted - $\frac{1}{n}$ . However, in the students’ perspective, weights scale with the size of the class. Thus, the students’ perspective will always be larger than the dean’s, unless all classes have the same number of students.