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| − | ==Problem==
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_7]] |
| − | A school has <math>100</math> students and <math>5</math> teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are <math>50, 20, 20, 5, </math> and <math>5</math>. Let <math>t</math> be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let <math>s</math> 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 <math>t-s</math>?
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| − | <math>\textbf{(A)}\ {-}18.5 \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\
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| − | 13.5 \qquad\textbf{(E)}\ 18.5</math>
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| − | | |
| − | ==Solution==
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| − | The formula for expected values is <cmath>\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).</cmath>
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| − | We have
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| − | <cmath>\begin{align*}
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| − | t &= \frac15\cdot50 + \frac15\cdot20 + \frac15\cdot20 + \frac15\cdot5 + \frac15\cdot5 \\
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| − | &= \frac15\cdot(50+20+20+5+5) \\
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| − | &= \frac15\cdot100 \\
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| − | &= 20, \\
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| − | s &= \frac{50}{100}\cdot50 + \frac{20}{100}\cdot20 + \frac{20}{100}\cdot20 + \frac{5}{100}\cdot5 + \frac{5}{100}\cdot5 \\
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| − | &= 25 + 4 + 4 + 0.25 + 0.25 \\
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| − | &= 33.5.
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| − | \end{align*}</cmath>
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| − | Therefore, the answer is <math>t-s=\boxed{\textbf{(B)}\ {-}13.5}.</math>
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| − | ~MRENTHUSIASM
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