Thinking Mathematically (6th Edition)

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

Chapter 13 - Voting and Apportionment - Chapter 13 Test - Page 890: 24

Answer

When the number of doctors increases from 10 to 12 as one more clinic is added, the three original clinics do not lose any apportioned doctors. Therefore, the new states paradox does not occur.

Work Step by Step

We can find the total patient load. total load = 119 + 165 + 216 = 500 We can find the standard divisor when there are 10 doctors. $standard~divisor = \frac{total~load}{number~of~ doctors}$ $standard~divisor = \frac{500}{10}$ $standard~divisor = 50$ The standard divisor is 50 We can find the standard quota for each clinic. Clinic A: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{119}{50}$ $standard~quota = 2.38$ Clinic B: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{165}{50}$ $standard~quota = 3.30$ Clinic C: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{216}{50}$ $standard~quota = 4.32$ Hamilton's method is an apportionment method that involves rounding each standard quota down to the nearest whole number. Surplus doctors are given, one at a time, to the clinics with the largest decimal parts in their standard quotas until there are no more surplus doctors. Initially, each clinic is apportioned its lower quota. Clinic A is apportioned 2 doctors. Clinic B is apportioned 3 doctors. Clinic C is apportioned 4 doctors. The total number of doctors which have been apportioned is 2 + 3 + 4 = 9 doctors Since there is a total of 10 doctors, there is one surplus doctor. One more doctor is given to Clinic A because it has the largest decimal part (0.38) in its standard quota. Using Hamilton's method, each clinic is apportioned the following number of doctors: Clinic A is apportioned 2 + 1 = 3 doctors. Clinic B is apportioned 3 doctors. Clinic C is apportioned 4 doctors. Let's suppose the number of doctors is increased from 10 to 12 as another clinic is added to the HMO. We can find the new standard divisor when there are 12 doctors. $standard~divisor = \frac{total~load}{number~of~ doctors}$ $standard~divisor = \frac{610}{12}$ $standard~divisor = 50.83$ The standard divisor is 50.83 We can find the standard quota for each clinic. Clinic A: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{119}{50.83}$ $standard~quota = 2.34$ Clinic B: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{165}{50.83}$ $standard~quota = 3.25$ Clinic C: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{216}{50.83}$ $standard~quota = 4.25$ Clinic D: $standard~quota = \frac{patient~load}{standard~divisor}$ $standard~quota = \frac{110}{50.83}$ $standard~quota = 2.16$ Hamilton's method is an apportionment method that involves rounding each standard quota down to the nearest whole number. Surplus doctors are given, one at a time, to the clinics with the largest decimal parts in their standard quotas until there are no more surplus doctors. Initially, each clinic is apportioned its lower quota. Clinic A is apportioned 2 doctors. Clinic B is apportioned 3 doctors. Clinic C is apportioned 4 doctors. Clinic D is apportioned 2 doctors. The total number of doctors which have been apportioned is 2 + 3 + 4 + 2 = 11 doctors Since there is a total of 12 doctors, there is one surplus doctor. One more doctor is given to Clinic A because it has the largest decimal part (0.34) in its standard quota. Using Hamilton's method, each clinic is apportioned the following number of doctors: Clinic A is apportioned 2 + 1 = 3 doctors. Clinic B is apportioned 3 doctors. Clinic C is apportioned 4 doctors. Clinic D is apportioned 2 doctors. When the number of doctors increases from 10 to 12 as one more clinic is added, the three original clinics do not lose any apportioned doctors. Therefore, the new states paradox does not occur.
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