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Learning Math Home
Data Session 5: Solutions
 
Session 5 Part A Part B Part C Part D Part E Homework
 
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Solutions for Session 5, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5


Problem B1

Selecting the most fair allocation should be fairly easy. Most people immediately pick Allocation A because it has 5 stacks of size 5, and the other 4 stacks are of size 4 or 6. It is the most similar to the fair allocation.

Selecting the least fair allocation is also relatively easy. Most people immediately pick Allocation B as the least fair. Its stacks are only of size 1 or 10, which is quite different from the fair allocation of 5 for each stack.

Otherwise, answers will vary depending on how people measure "fairness." One method is to find the allocation where, on average, the stack sizes are closest to 5. This is accomplished by finding, in total, how far away the stacks are from 5. Using this criterion, Allocation A is the most fair, followed by Allocations D, C, E, and B.

<< back to Problem B1


 

Problem B2

 

Allocation A requires 2 moves.

 

Allocation B requires 20 moves.

 

Allocation C requires 7 moves.

<< back to Problem B2


 

Problem B3

Allocation A is the most fair, and Allocation B is the least fair. The number of moves required to make an allocation fair is one way to measure how close to being fair the allocation is. That is, the more moves required to make an allocation fair, the more unfair the allocation is.

<< back to Problem B3


 

Problem B4

 

Four stacks are above the mean. Each of these has 10 coins, an excess of 5 coins (+5) above the average. The total excess of coins above the average is 20.

 

Five stacks are below the mean. Each of these has 1 coin, a deficit of 4 coins (-4) below the average. The total deficit of coins below the average is 20.

 

No stacks are exactly average.

<< back to Problem B4


 

Problem B5

 

Four stacks are above the mean. The total excess of coins above the average is 7 ([+1] + [+1] + [+2] + [+3]).

 

Four stacks are below the mean. The total deficit of coins below the average is 7 ([-3] + [-2] + [-1] + [-1]).

 

One stack is exactly average and has no excess or deficit (0).

<< back to Problem B5


 

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