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Private: Learning Math: Data Analysis, Statistics, and Probability
Variation About the Mean Part A: Fair Allocations (25 minutes)
In This Part: The Mean
The term average is a popular one; it is often used, and often used incorrectly.
Although there are different types of averages, the typical definition of the word “average” when talking about a list of numbers is “what you get when you add all the numbers and then divide by how many numbers you have.” This statement describes how you calculate the arithmetic mean, or average. But knowing how to calculate a mean doesn’t necessarily tell you what it represents.
Let’s begin our exploration of the mean: Using your 45 coins, create 9 stacks of several sizes. You must use all 45 coins, and at least 1 coin must be in each of the 9 stacks. It’s fine to have the same number of coins in multiple stacks.
Here is one possible arrangement, or allocation, of the 45 coins:
Problem A1
Record the number of coins in each of your 9 stacks. What is the mean number of coins in the 9 stacks?
To find the mean, divide the number of coins (45) by the number of stacks.
Problem A2
Create a second allocation of the 45 coins into 9 stacks.
a. Record the number of coins in each of your 9 stacks, and determine the mean for this new allocation.
b. Why is the mean of this allocation equal to the mean of the first allocation?
c. Describe two things that you could do to this allocation that would change the mean number of coins in the stacks.
Problem A3
Create a third allocation of the 45 coins into 9 stacks in a special way:
• First take a coin from the pile of 45 and put it in the first stack.
• Then take another coin from the pile and put it in the second stack.
• Continue in this way until you have 9 stacks with 1 coin each, and 36 coins remaining.
• Take a coin from the pile of 36 and put it in the first stack.
• Then take another coin from the pile and put it in the second stack.
• Continue in this fashion until all of the remaining coins have been used.
a. Now record the number of coins in each of your 9 stacks, and determine the mean for this new allocation
b. What observations can you make about the mean in this special allocation?
This method produces what is called a fair or equal-share allocation. Each stack, in fact, contains the average number of coins. You might think of this as a fair allocation of the 45 coins among 9 people: Each person gets the same number of coins.
Video Segment
In this video segment, Professor Kader asks participants to create snap-cube representations of the number of people in their families. He then asks them to find a way of finding the mean without using calculation. Watch this segment for an exploration of the mean.
How does the mean relate to the fair allocation of the data?
In This Part: The Mean and the Median
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. See Note 2 below.
Problem A4
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.
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.
Problem A8
Find a third allocation that has a median different from the ones in Problems A6 and A7.
TAKE IT FURTHER
Problem A9
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.
Notes
Note 2
In Problems A4-A8, you examine the differences between various allocations of values. In particular, you focus on the median in each data set and its relation to the rest of the data. While the median is not the main focus of this session, it is important to emphasize the difference between the mean and the median, a distinction that is not always clear.
Solutions
Problem A1
The mean is 5. To calculate the mean, add the numbers in each stack. The sum is 45. Then divide by the total number of stacks, 9, to find the mean.
Problem A2
a. The mean is still 5.
b. Since the total number of coins remains 45, and the total number of stacks remains 9, the mean must still be 45 / 9 = 5.
c. You could increase or decrease the number of coins, or you could increase or decrease the number of stacks.
Problem A3
a. As before, the mean is 5.
b. Here, the mean is equal to the number of coins in each stack. Since all the stacks have the same number of coins, the average is equal to the number of coins in any given stack.
Problem A5
| a. | The median is not the fifth stack because the allocation is not in order. It must first be placed in order. |
| b. | Here’s the ordered allocation:
The stack in position (5), the middle, has 5 coins, so the median size is 5. For this allocation, the median is the same as the mean. |
Problem A6
Answers will vary. Here is one possible allocation:
Problem A7
| a. | Answers will vary. Here is one such allocation, with a median size of 4 coins:
|
| b. | As before, the mean must remain 5, since the number of coins and the number of stacks had not changed. |
Problem A8
Answers will vary. Here is one such allocation, with a median size of 6 coins:
Problem A9
The smallest possible value for the median is 1. It cannot be smaller, since each stack must contain at least 1 coin. One possible allocation with this median is {1, 1, 1, 1, 1, 10, 10, 10, 10}; many other allocations are possible if a stack may contain more than 10 coins.
The largest possible value for the median is 8. One way to think about this problem is to recognize that there must be 4 stacks below the median, and each must have at least 1 coin. This leaves 41 or fewer coins for the remaining five stacks, and the median must be the smallest of these stacks. Using an equal-shares allocation of the remaining 41 coins results in 4 stacks of 8 coins and 1 of 9, so the median cannot be greater than 8. There are only two possible allocations with this median: {1, 1, 1, 1, 8, 8, 8, 8, 9} and {1, 1, 1, 2, 8, 8, 8, 8, 8}.
