The formula for expected values is $$ \text { Expected Value }=\sum(\text { Outcome } \cdot \text { Probability }). $$ We have $$ \begin{aligned} t \& =50 \cdot \frac{1}{5}+20 \cdot \frac{1}{5}+20 \cdot \frac{1}{5}+5 \cdot \frac{1}{5}+5 \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =(50+20+20+5+5) \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =100 \cdot \frac{1}{5} \textbackslash\textbackslash{} \& =20, \textbackslash\textbackslash{} s \& =50 \cdot \frac{50}{100}+20 \cdot \frac{20}{100}+20 \cdot \frac{20}{100}+5 \cdot \frac{5}{100}+5 \cdot \frac{5}{100} \textbackslash\textbackslash{} \& =25+4+4+0.25+0.25 \textbackslash\textbackslash{} \& =33.5 . \end{aligned} $$  Therefore, the answer is $t-s=(\mathbf{B})-13.5$. 
A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50,20,20,5$, and $5$. Let $l$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$? 
Options:
A. $$-18.5$$ 
B. $$-13.5$$ 
C. $$0$$ 
D. $$13.5$$ 
E. $$18.5$$ 