#### 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.