Let's now contrast the mean and the median as summary measures. You previously found that when 45 coins were allocated to 9 stacks, the mean stack size was always 5 because the number of stacks and the sum of the stack sizes were constant, 9 and 45 respectively. Note 2
Do you expect that the median stack size for the 9 stacks will always be the same for any allocation? Why or why not? You might want to look back at the allocations you created for Problems A1-A3.
Problem A5
Put your 45 coins into this allocation:
a.
Why is the median not the fifth stack in the allocation above?
b.
Arrange your stacks in ascending order from left to right. Then find the median stack size using one of the methods you learned in Session 4, Part B.
The median is the value in the center of an ordered data set. Since there are 9 stacks, the median is in the position (9 + 1) / 2, that is; the median is in position (5) after ordering. Close Tip
Problem A6
Create a new allocation of the 45 coins into 9 stacks so that the median is equal to 5. (Do not use the allocation with 5 coins in each stack.)
Problem A7
a.
Create a new allocation of the 45 coins into 9 stacks so that the median is not equal to 5.
b.
What is the mean for your new allocation?
As you manipulate the coins, remember that the median can only be found in an ordered data set, so keep your 9 stacks in ascending order. Try to manipulate the stacks so that the fifth stack does not contain 5 coins. Close Tip
Problem A8
Find a third allocation that has a median different from the ones in Problems A6 and A7.
What is the smallest possible value for the median? What is the largest possible value for the median? Remember that there must be 9 stacks for the 45 coins, and each stack must contain at least 1 coin.
