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The mathematics department has 30 teaching assistants to be divided among three courses, according to their respective enrollments. The table shows the courses and the number of students enrolled in each course.$$ \begin{array}{|l|c|c|c|c|} \hline \text { Course } & \begin{array}{c} \text { College } \\ \text { Algebra } \end{array} & \text { Statistics } & \begin{array}{c} \text { Liberal Arts } \\ \text { Math } \end{array} & \text { Total } \\ \hline \text { Enrollment } & 978 & 500 & 322 & 1800 \\ \hline \end{array} $$a. Apportion the teaching assistants using Hamilton’s method. b. Use Hamilton’s method to determine if the Alabama paradox occurs if the number of teaching assistants is increased from 30 to 31. Explain your answer.

Short Answer

Expert verified
Using Hamilton's Method, the original distribution is 17 teaching assistants for College Algebra, 8 for Statistics, and 5 for Liberal Arts Math. When the number of teaching assistants increases by 1, College Algebra still gets 17 but Statistics gets 9 and Liberal Arts Math remains at 5. This demonstrates the Alabama Paradox because College Algebra doesn't get more resources when the total number of resources increases.

Step by step solution

01

Calculating Standard Divisors and Initial Quotas

The standard divisor is the total number of students divided by the number of teaching assistants (1800/30 = 60 students per assistant). Each course's initial quota is its number of students divided by the standard divisor. For College Algebra: 978/60 = 16.3, for Statistics: 500/60 = 8.3, and for Liberal Arts Math: 322/60 = 5.36.
02

Assigning Assistants Using Hamilton's Method

Using Hamilton's Method, assign each course teaching assistants equal to the whole number part of their quota. College Algebra gets 16, Statistics gets 8, and Liberal Arts Math gets 5. This totals 29 assistants. The remaining assistant goes to the quota with the largest fractional part, which is College Algebra. So College Algebra gets another one for a total of 17, giving a final distribution of 17 teaching assistants for College Algebra, 8 for Statistics and 5 for Liberal Arts Math.
03

Checking for the Alabama Paradox

To check for the Alabama Paradox, repeat the process with 31 instead of 30 teaching assistants. The new standard divisor is 58.06. Doing the same calculation as before, College Algebra's quota is 16.85, Statistics' is 8.61, and Liberal Arts Math's is 5.54. Using the whole numbers, we assign 16 teaching assistants to College Algebra, 8 to Statistics, and 5 to Liberal Arts Math for a total of 29. The remaining two will go to the quotas with the largest fractional parts at this point, which are College Algebra and Statistics. Now, each of College Algebra and Statistics get 1 more teaching assistant, so the final distribution is 17 for College Algebra, 9 for Statistics and 5 for Liberal Arts Math.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hamilton's Method
Hamilton's method is a fun and intuitive way to distribute resources, like teaching assistants, among several courses. It's based on ensuring each course gets an equal share, proportionally based on the number of students enrolled. To do this, we start by calculating what's called a "standard divisor". It's simply the total number of students divided by the total number of teaching assistants available, essentially telling us how many students each assistant should cater to.

Next, we take each course's enrollment and divide it by this standard divisor to get the initial quota. The whole number part of each quota tells us how many assistants are assigned right away. For example, if a course has a quota of 16.3, it gets 16 assistants initially.

Hamilton's method still leaves some assistants undistributed, because not all quotas are whole numbers. We give the leftover positions in order of the largest fractional parts. This ensures the fairest distribution by considering even the small parts each course may need to be complete.
Alabama Paradox
The Alabama Paradox is one of the quirkiest puzzles in mathematical distribution problems. It happens when increasing the total number of resources, like teaching assistants, results in a course receiving fewer than before the increase. This seems counterintuitive, doesn't it?

In our example, when moving from 30 to 31 teaching assistants, we saw how adding one more assistant can shuffle the exact fractional parts used to decide who gets the remaining spots. In some bizarre cases, instead of benefitting from an additional resource, a course might lose one they had initially received. Thus, even though there's one more assistant, the way they are fractionally distributed ends up changing who benefits, highlighting the unpredictable nature of the Alabama Paradox.
Teaching Assistant Distribution
Distributing teaching assistants fairly is pivotal in ensuring a balanced workload and effective learning environment for students. Scheme distribution using mathematical methods like Hamilton's method helps achieve equitable allocation based on actual needs.

The distribution takes into account not just the number of assistants available but also the specific enrollment figures for courses. This ensures that resources are optimized and courses with higher enrollments receive proportionately the support they need. By considering both whole quotas and resolving fractional leftover positions, the distribution becomes a calculated win for everyone, aligning with both fairness and efficiency.
College Mathematics Applications
College mathematics isn’t just about solving abstract problems on paper; it's also about applying these concepts to real-world scenarios like resource allocation. Situations like dividing teaching assistants among courses are perfect examples of such applications.

By delving into college mathematics, students learn not only the beauty of numbers but also their utility in decision-making and problem-solving. Methods like Hamilton’s are particularly valuable for understanding and implementing fair distribution systems – a skill useful in many fields, from universities to public policy. Such practical applications enhance the student’s ability to tackle diverse challenges, proving that mathematics is, indeed, everywhere.

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Most popular questions from this chapter

Three candidates, A, B, and \(\mathrm{C}\), are running for mayor. Election rules stipulate that the pairwise comparison method will determine the winner. In the event that the pairwise comparison method leads to a tie, the Borda count method will decide the winner. The election results are summarized in the following preference table. Under these rules, which candidate becomes the new mayor? $$ \begin{array}{|l|c|c|c|c|c|} \hline \text { Number of Votes } & \mathbf{6 0 , 0 0 0} & \mathbf{4 0 , 0 0 0} & \mathbf{4 0 , 0 0 0} & \mathbf{2 0 , 0 0 0} & \mathbf{2 0 , 0 0 0} \\ \hline \text { First Choice } & \text { A } & \text { C } & \text { B } & \text { A } & \text { C } \\ \hline \text { Second Choice } & \text { B } & \text { A } & \text { C } & \text { C } & \text { B } \\ \hline \text { Third Choice } & \text { C } & \text { B } & \text { A } & \text { B } & \text { A } \\ \hline \end{array} $$

Make Sense? In Exercises 18-21, determine whethereach statement makes sense or does not make sense, and explain your reasoning. The county hired seven new doctors to apportion among its three clinics. Although our local clinic has the same proportion of the county's patients as it did before the doctors were hired, it now has one fewer doctor.

Make Sense? In Exercises 28-31, determine whether each statement makes sense or does not make sense, and explain your reasoning. My candidate received a majority of first-place votes and lost the election.

A computer company is considering opening a new branch in Atlanta (A), Boston (B), or Chicago (C).Senior managers vote to decide where the new branch will be located. The winning city is to be determined by the plurality method. The preference table for the election is shown. $$ \begin{array}{|l|c|c|c|} \hline \text { Number of Votes } & 20 & 19 & 5 \\ \hline \text { First Choice } & \text { A } & \text { B } & \text { C } \\ \hline \text { Second Choice } & \text { B } & \text { C } & \text { B } \\ \hline \text { Third Choice } & \text { C } & \text { A } & \text { A } \\ \hline \end{array} $$ a. Which city is favored over all others using a head-tohead comparison? b. Which city wins the vote using the plurality method? c. Is the head-to-head criterion satisfied? Explain your answer.

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