Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 13 - Voting and Apportionment - Chapter Summary, Review, and Test - Review Exercises: 41

Answer

(a) Using Hamilton's method, each school is apportioned the following number of computers: School A is apportioned 8 computers. School B is apportioned 67 computers. School C is apportioned 75 computers. (b) When there were 150 computers, School A was apportioned 8 computers. After the total number of computers increased to 151 computers, School A was only apportioned 7 computers. Therefore, the Alabama paradox occurs.

Work Step by Step

(a) We can find the standard divisor when there are 150 computers. $standard~divisor = \frac{total~enrollment}{number~of~ computers}$ $standard~divisor = \frac{7500}{150}$ $standard~divisor = 50$ The standard divisor is 50 We can find the standard quota for each school. School A: $standard~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{370}{50}$ $standard~quota = 7.4$ School B: $standard~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{3365}{50}$ $standard~quota = 67.3$ School C: $standard~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{3765}{50}$ $standard~quota = 75.3$ Hamilton's method is an apportionment method that involves rounding each standard quota down to the nearest whole number. Surplus computers are given, one at a time, to the schools with the largest decimal parts in their standard quotas until there are no more surplus computers. Initially, each school is apportioned its lower quota. School A is apportioned 7 computers. School B is apportioned 67 computers. School C is apportioned 75 computers. The total number of computers which have been apportioned is 7 + 67 + 75 = 149 computers Since there is a total of 150 computers, there is one surplus computer. One more computer is given to School A because it has the largest decimal part (0.4) in its standard quota. Using Hamilton's method, each school is apportioned the following number of computers: School A is apportioned 7 + 1 = 8 computers. School B is apportioned 67 computers. School C is apportioned 75 computers. (b) Let's suppose the number of computers increases from 150 to 151. We can find the standard divisor when there are 151 computers. $standard~divisor = \frac{total~enrollment}{number~of~ computers}$ $standard~divisor = \frac{7500}{151}$ $standard~divisor = 49.67$ The standard divisor is 49.67 We can find the standard quota for each school. School A: $standard~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{370}{49.67}$ $standard~quota = 7.45$ School B: $standard~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{3365}{49.67}$ $standard~quota = 67.75$ School C: $standard~quota = \frac{enrollment}{standard~divisor}$ $standard~quota = \frac{3765}{49.67}$ $standard~quota = 75.80$ Hamilton's method is an apportionment method that involves rounding each standard quota down to the nearest whole number. Surplus computers are given, one at a time, to the schools with the largest decimal parts in their standard quotas until there are no more surplus computers. Initially, each school is apportioned its lower quota. School A is apportioned 7 computers. School B is apportioned 67 computers. School C is apportioned 75 computers. The total number of computers which have been apportioned is 7 + 67 + 75 = 149 computers Since there is a total of 151 computers, there are two surplus computers. The first computer is given to School C because it has the largest decimal part (0.80) in its standard quota. The second computer is given to School B because it has the second largest decimal part (0.75) in its standard quota. Using Hamilton's method, each school is apportioned the following number of computers: School A is apportioned 7 computers. School B is apportioned 67 + 1 = 68 computers. School C is apportioned 75 + 1 = 76 computers. When there were 150 computers, School A was apportioned 8 computers. After the total number of computers increased to 151 computers, School A was only apportioned 7 computers. Therefore, the Alabama paradox occurs.
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