Self Check Answers

1. Convert all times to minutes, and note that 24 hours is 1440 minutes. Thus, the administrative assistant worked of the day. He commuted of the day, and spent of the day on personal activities. Rounding these percentages to their nearest whole numbers we get

   To make the sum 100%, one of the percentages must be rounded down. You have no guidance yet on how to do this, so there is no wrong way to choose which one.

2. North Carolina was apportioned 12 seats, so its district population is its population of 353,523 divided by 12, or 29,460, which is below the limit. New York’s population, 331,589, divided by its apportionment of 11 seats, is equal to 30,144, so New York’s district population does satisfy the requirement.
3. The states are the families (Browns, Joneses, and Robinsons); the house size is the number of candies to be apportioned (123); the populations are Browns (5), Joneses (3), and Robinsons (4), for a total of 12. The standard divisor is the total population divided by the house size, .
4. The quotas are: Browns, ; Joneses, ; Robinsons, .
5. Rhode Island’s quota was , and North Carolina’s quota was . The district populations are for Rhode Island, and for North Carolina.

6. Add the lower quotas: . The fractional parts are 0.67%, 0.72%, and 0.61%. Round the first two quotas—which have the greatest fractional parts—up to get the apportionment .

   610

7. The district populations for the subjects are the average number of students in a class: 26 for geometry; 16.5 for precalculus; and 15 for calculus.
8. Yes. The house size decreased, the populations did not change, and one subject, Calculus III, received an increased apportionment.

9. The following table is the apportionment. In any apportionment in which no state receives more than one representative per 30,000 population, the maximum apportionment permitted is

   These are the apportionments shown in the table. Thus, the maximum possible house size is 112.

   State	Population	Pop/30,000	Apportionment
   Virginia	630,560	21.02	21
   Massachusetts	475,327	15.84	15
   Pennsylvania	432,879	14.43	14
   North Carolina	353,523	11.78	11
   New York	331,589	11.05	11
   Maryland	278,514	9.28	9
   Connecticut	236,841	7.89	7
   South Carolina	206,236	6.87	6
   New Jersey	179,570	5.99	5
   New Hampshire	141,822	4.73	4
   Vermont	85,533	2.85	2
   Georgia	70,835	2.36	2
   Kentucky	68,705	2.29	2
   Rhode Island	68,446	2.28	2
   Delaware	55,540	1.85	1
   Totals	3,615,920	120.53	112

10. Table 14.11 shows that with the divisor 0.225, the apportionment quotient for Winnie was exactly 80. The apportionment quotients for Louise and Tim were 17.78 and 4.44, respectively. The divisor 0.227 reduced the apportionment quotient for Winnie to 79.30, while Louise and Tim had apportionment quotients of 17.62 and 4.41, respectively. Thus, the divisor 0.226 would produce apportionment quotients for Winnie between 79.30 and 80, for Louise between 17.62 and 17.78, and for Tim between 4.41 and 4.44. These quotas, rounded down, would be 79, 17, and 4 diamonds, respectively.
11. The maximum number of sections for a course would be the population for that course, divided by 24, and rounded down. This would be found by the Jefferson method, using the divisor 24. Referring to Table 14.13, where that divisor was employed, we see that 28 TAs are needed.

12. A state has population . Let s be the standard divisor and be the Jefferson divisor. The state’s apportionment quotient is and its quota is . Since ,

    It follows that rounding the apportionment quotient down results in a whole number not less than the lower quota, which we obtain by rounding the quota down.

13. The sum of the lower quotas is 98%. Divide each quota by 0.98, as suggested, to get apportionment quotients 99.18%, 1.94%, 0.92%. Rounding down we have

    This violates the quota condition, because the apportionment to the first percentage, 97.2%, is greater than its upper quota, 98%.

14. Compare the current priority numbers: Whigs, 394,670; Tories, 385,572; Liberals, 386,423; Centrists, 201,203. The Whigs have the highest priority and get seat #31.

15. With the Webster method, start by rounding each quota to the nearest whole number. This yields , which is the Webster apportionment. There is no violation of the quota condition.

16. Geometry		Precalculus		Calculus
    #1	52	#2	33	#4	15
    #3	17.33	5	11		5
    	10.4		6.6		3

    Geometry and precalculus get two sections, and calculus gets one section.

    611

17. With the current apportionment, the difference in district populations is in California’s favor by 289,851. If a seat had been transferred from California to Montana, California’s district population would increase to . Montana’s district population would be . The difference, 220,907, is less than the difference before the transfer. Therefore, this apportionment is not equitable by absolute difference in district population. (However, the percentage difference after the transfer, , is greater than the percentage difference before the transfer, . This was to be expected, because the Hill-Huntington method minimizes percentage differences.)
18. • The percentage difference in the records is .
    • Bolt’s average speed was . Lindberg’s average speed was .
    • The percentage difference in speeds is .

    The percentage difference in speeds is exactly equal to the percentage difference in the times taken to complete the run.

20. By Hill-Huntington, all numbers between 0 and 1 are rounded to 1, all numbers between 1.414 and 2 are rounded to 2, and 98.1 would be rounded to 98. Thus, the rounded percentages are . We have to use a divisor greater than the standard divisor (which is 1) to reduce the sum. As suggested, we’ll take 1.7 as the divisor. The apportionment quotients are . The rounding point for numbers between 97 and 98 is

    Thus, 97.42% is rounded down to 97%, and the other two percentages are rounded up. The Hill-Huntington rounding,

    violates the quota condition because the first percentage is apportioned less than its lower quota.