2021 AMC Fall 10A Problem 10

For $t$, each of the $5$ teachers is equally likely, so $t$ is the ordinary mean of the five class sizes:
$$t = \frac{50 + 20 + 20 + 5 + 5}{5} = \frac{100}{5} = 20.$$

For $s$, each of the $100$ students is equally likely, and a student in a class of size $n$ reports $n$. There are $50$ students who report $50$, $40$ who report $20$, and $10$ who report $5$:
$$s = \frac{50 \cdot 50 + 20 \cdot 20 + 20 \cdot 20 + 5 \cdot 5 + 5 \cdot 5}{100} = \frac{2500 + 800 + 50}{100} = 33.5.$$

Therefore $t - s = 20 - 33.5 = -13.5$.

The answer is $\boxed{\textbf{(B)}\ {-}13.5}$.

Averaging over classes weights each class equally; averaging over students weights each class by its size, so large classes count more. This "size-biased" average is always at least the plain average (they agree only when all classes are equal), which tells you in advance that $t - s$ is negative. The same effect explains why "average class size as experienced by students" exceeds the school's official average.

• Assuming s equals t because 'the average class size is 20', ignoring that a random student is far more likely to be in the big class.
• Getting the sign backwards and answering $13.5$, choice (D); the question asks for $t - s$, and $s$ is the larger one.