Problem A1: Record the number of coins in each of your 9 stacks. What is the mean number of coins in the 9 stacks?
This one was simple. All I did was add the number of coins (45) and divide it by the number of stacks (9). 45/9 = 5.
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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.
For this part, I broke up each stack and put a different number in each. Despite the different amount in each stack, the mean still remained 5.
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| b. | Why is the mean of this allocation equal to the mean of the first allocation?
The new allocation stays the same because you still have the same number of coins and the same number of stacks. You are not adding or subtracting any coins or stacks.
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| c. | Describe two things that you could do to this allocation that would change the mean number of coins in the stacks.
Again, the mean would change if you increased/decreased the number of coins in the collection or increased/decreased the number of stacks you want to separate.
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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.
Although this new arrangement looks different, I still have 45 coins and 9 stacks. The mean is still 5.
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| b. | What observations can you make about the mean in this special allocation?
Clearly, the mean remains the same (5) but it is interesting to see how there is an even number of coins in each stack. |
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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.
This is a rather tricky question. I think it depends on how many coins you have. If you have a large number of coins, you will have a lager median stack size than you would for say a group of 20 or 30 coins.
Problem A5:
| a. | Why is the median not the fifth stack in the allocation above?
The stacks of coins are not in order from least to greatest. According to the picture above, the median would be 4 (since it is the value in stack #5). This would be inaccurate.
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| 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.
After putting the stacks in ascending order, you can see that the median
value is 5.
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.)
I will admit that this was harder than it really seems to be. It took me a few minutes to make sure I had my numbers in order and that they added up to 45 coins. I basically figured it out by trial and error. I first made sure that I had 5 in the middle and then included numbers more and less than 5 into the set. Here's what I came up with. It is slightly different from the solution given in the Annenberg session.
| a. | Create a new allocation of the 45 coins into 9 stacks so that the median is not equal to 5.
Again, this task took a lot of trial and error! But I finally came up with this order of coins: 1,1,2,3,4,7,8,9,10. The ne median is 4. |
| b. | What is the mean for your new allocation?
The median of my new set of coins is 4.
Problem A8: Find a third allocation that has a median different from the ones in Problems A6 and A7.
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.
Part of this question was tricky for me. I figured that the lowest possible median would have to be 1 since you can have more than one of the same number. However, I was confused on what the highest mean could be. The solution given on the website was helpful. So I took 45-5 = 40 (since you need 5 numbers to figure the median. Next, I took 40 and divided it by the 5 open numbers and got 8. As you can see, I also had quite a time with this one.
Part B--Unfair Allocations
I will say that the idea of unfair allocations is new to me. However, the concept makes sense.
Problem B1: Rank the five allocations on their "fairness," from most fair to least fair. Also, explain how you decided on the level of "unfairness" in an allocation.
It was a little easy to look at these stacks and determine the "fairness" ranking. Without looking the solution, I came up with A,D,E,C,B. I almost had the correct ranking but was off. The answer is A,D,C,E,B.
The tip given in this section was super helpful--to put the stacks in order in ascending order and adding coins until you reach a fair amount. This makes it easier to determine big differences.
Problem B2: Below are three allocations of 45 coins in 9 stacks. For each allocation, find the minimum number of moves required to change the allocation into a fair allocation (i.e., one with 5 coins in each stack).
This was tough at first but as I worked through each allocation, I realized that I was thinking about it too hard. I started by subtracting 1 from each "high value" and adding it to "low values" to make things "fair". I hope this makes sense. In Allocation B, I was in the middle of writing out the math for each subtraction so that I wouldn't lose track. But then I realized that I simply needed to figure out how many I numbers I need to get to the "fair 5". In this case, there are 5 1s so I needed to go up 4 points per number. So, 4 x 5 = 20 moves.
Problem B3: Based on the minimum number of moves, which of the allocations in Problem B2 is the most fair? Which is the least fair?
Allocation A is the most fair because it required the least number of moves. Only 2.
Problem B4:Perform the calculations above to show that 20 moves are required to obtain a fair allocation for Allocation B, which is arranged in ascending order below:
a. Four stacks are above the mean. Each of these has 10 coins, +5 above the average. The total excess of coins above the average is 20 coins.
b. Five stacks are below the mean. Each of these has 1 coin, 4 below the average. The total deficit coins is 20.
c. No stacks are exact average.
Problem B5: Perform the same calculations to find the number of moves required to obtain a fair allocation for Allocation C, which is shown in ascending order below:
a. 4 stacks are above the mean. Two stacks a +1, one stack is +2, one stack is +3. Total excess is 7.
b. 4 stacks are below the mean. One stack is -3, one stack is -2, and two stacks are -2. Total deficit is 7.
c. One stack is an exact average and has excess or deficit.
Part C--Using Line Plots
I thought this was an interesting section. I never really thought about putting the stacks of coins into a line plot form.
Problem C1: Use the method described above to create a line plot for the following ordered allocation of 45 coins.
Problem C2: Create a line plot for this allocation of 45 coins:
Problem C3: Create a line plot for this equal-shares allocation of 45 coins:
Problem C4:
From this line plot, we can see that there are 3 stacks containing exactly 5 coins each, and 1 stack containing 6 coins. The maximum number of coins in a stack is 8, and the minimum is 2.
Rearrange the nine dots to form a line plot with each of these requirements:
I have provided images of my work to C4a-g below. Clearly, you will see that I had a few problems. I did this with a lot of trial and error and had to copy the answers to E. because I was getting super frustrated.
Problem C5: How could you change another stack of 5 coins to reset the mean to 5?
I must admit that I was completely lost when I first *attempted* to work through this part. I think was stuck because the line plot made it a little harder for me to visualize this new process. So, I wrote out the numbers: 
Problem C6: If you could change the value of more than one stack, could you solve Problem C5 another way?
Yes. I think it ultimately depends on how many stacks you are changing. There are many way to take and add coins to other stacks.
Problem C8: Applying the strategy you developed in Problems C5-C7, use the following Interactive Activity or your paper/poster board to revisit the allocations you worked with in Problem C4. You should begin with the fair allocation of the 45 coins; that is, 9 dots at the mean of 5. Try to come up with answers for the questions below that are different from the ones you found in Problem C4.
a.
b. 
c.
d. 
e. 
f. 
Part D--Deviations From the Mean
Problem D1:
| a. | Draw the corresponding line plot. Indicate the deviation for each dot as was done in the line plot that opened Part D.
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| b. | Complete the following table of deviations:
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Problem D2: Create a line plot with these deviations from the mean = 5: (-4), (-3), (-2), (-1), (0), (+1), (+2), (+3), (+4) 
Problem D3: Create a line plot with these deviations from the mean = 5: (-4), (-2), (-2), (-1), (0), (+1), (+2), (+2), (+4)
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Problem D4:
Create a line plot with these deviations from the mean = 5, and specify a set of four remaining values: (-4), (-3), (-3), (-1), (-1)
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Problem D5: How would the line plots you created in Problems D2-D4 change if you were told that the mean was 6 instead of 5? Would this change the degree of fairness of these allocations (as described in Problem B2)?
Since 6 is only one away from 5, there will only be a slight change. The plots would shift a little due to this but I don't think it would change the degree of fairness, as long as you consider the deviations.
Part E--Measuring Variations
Problem E1:Of the three, which line plot's data has the least variation from the mean? Which has the most variation from the mean?
The tip given with this question was helpful. Essentially, I learned that the plots with the least variation appears to have points that do not appear too far away from the mean. On the other hand, plots with the most variation appear far away from the mean.
Considering this, Plot B has the least variation while Plot C has the most variation.
Problem E2: Below is Line Plot B from Problem E1. Create a table like the one above, find the MAD for this allocation, and compare it to the MAD of Line Plot A from the same problem.
Work is shown below:
Problem E3: Below is Line Plot C from Problem E1. Create another table, find the MAD for this allocation, and compare it to the MADs of Line Plots A and B.
Work shown below:
Problem E4: Create a line plot with a MAD of 24 / 9.
I will say that this activity was hard to complete without revealing the Mean Absolute Deviation (MAD) as I worked.
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Problem E5: Create a line plot with a MAD of 22 / 9, with no 5s.
Problem E6: Create a line plot with a MAD of 12 / 9, with exactly two 5s, 5 values larger than 5, and 2 values smaller than 5.
Problem E7: Explain why it is not possible to create a line plot with 9 values that has a MAD of 1.
This was a tricky question for me. In checking the solution, I understand that it is impossible to split up coins to be 4.5 each. I guess I'm a little confused on how to get the 9/9 deviation that leads to this answer. Did anyone else experience a problem with this question?
Problem E8: Below is Line Plot B from Problem E1. Create a table like the one above, and find the variance and standard deviation for this allocation. Compare the standard deviation to the MAD of Line Plot B you found in Problem E2 and to the standard deviation of Line Plot A.
Work shown below:
Problem E9:Below is Line Plot C from Problem E1. Create another table, and find the variance and standard deviation for this allocation. Compare the standard deviation to the MAD of Line Plot C you found in Problem E3 and to the standard deviations of Line Plots A and B.
Work is shown below:
Problem E10
a. What would happen to the mean of a data set if you added 3 to every number in it?
The mean would increase by 3 points.
b. What would happen to the MAD of a data set if you added 3 to every number in it? .
The MAD would stay the same. When adding 3 to every number you also add three to the mean.
c. What would happen to the variance of a data set if you added 3 to every number in it?
The variation of the data would also stay the same. If the MAD does not change, neither would the variation.
d. What would happen to the standard deviation of a data set if you added 3 to every number in it?
The same goes for the standard deviation. This number will stay the same since the MAD and variation did not change.
e. What would happen to the mean of a data set if you doubled every number in it?
If you double every number in the, then the mean will also be doubled.
f. What would happen to the MAD of a data set if you doubled every number in it?
The MAD would also be doubled, thanks to the doubled deviations.
g. What would happen to the variance of a data set if you doubled every number in it?
This was a tricky question but reading the solution was helpful. It makes sense that since the variation is the squaring of the deviation, the number would be quadrupled.
h. What would happen to the standard deviation of a data set if you doubled every number in it?
The standard deviation would also be doubled.
Generating Meaning for Range, Mode, Median, and Mean TCM Article
I really enjoyed reading this article! I think the lessons described would do a great job of helping students truly learn about the concepts of finding the mean, median, and mode of a set of data. Moreover, I like how the lesson emphasizes how these terms are defined and how they represent the data. I love how it also guides the children to discover and experience what the numbers really mean. Thinking back to my memories with learning about mean, median, and mode, I can’t say that I had such a revealing experiences. Essentially, my teachers usually stuck with telling us or teaching us the information and then spending a few days doing some review activities. The article depicts a much different image because students are actually involved in the learning of the new terms—the teacher is not “feeding” them the information, they are working together to figure it out for themseleves. What a great way to engage students and allow for information to “stick” with them!
I love the way the lesson calls for teachers to explain the reason we use the mean, median, and mode. To begin, the terms are not introduced but children learn that there are certain numbers we use that give us a “snapshot” of the data. I also like how this explaination is compared to the idea that just as we use one-word adjectives to describe ourseleves, we use single numbers to describe a set of data. This is an interesting way to look at it!
The lesson worksheets and tables included in the article were very helpful to view as I read through the article. I think these helped guide students as they “investigated” the new terms. I liked reading about the clever strategies and definitions that were used by students to describe and figure out what mean, median, and mode meant. For example, students descibed the mode being “the number that shows up the most”. The significant thing about this definition is that they students came up with this on their own—the teacher did not give it to them. When they worked with the median, students discovered that simply crossing out low and high numbers in a data set would reveal the middle number—or the mean. Interestingly, they described this number to be “in the middle”. If there was an even number of data, they deemed that “the middle median is the number in the middle of the two middle numbers”. I can’t say that I had the opportunity in my elementary years to provide my own definitions to mathematical terms; something that probably would have helped me so much! I think that it’s important to allow students to work through problems and express their own thinking instead of teachers giving them answers and information. After all, they know what makes sense to themseleves. You can’t always guaruntee that a teacher’s description will reach every student. Although I understood what these terms mean and what they were used for, I wish I had more meaningful experiences in developing my own understanding—not my teachers’. It might seem like this way could be a little more time consuming and complex, but I like how the article ends with saying that it is all worth it when students appear eager to extend their learning of these new terms and concepts.
Working With the Mean Activity
How did you see the cubes to figure out the problems?
At first, this was a tricky activity but working with the cubes did help. I
started out by breaking up the cubes. I did not add the total of the 5 bags. Since I know the mean is 8, I took the cubes
and made it so that each bag had 8 peanuts. Now I have 5 bags with 8 peanuts
each (except for bag #1 that has an extra peanut). Next, I placed 8 cubes in
bag #6 to satisfy the mean of 8. Now I know that 6 bags X 8 peanuts per bag =48
(+ the one extra peanut) is 49. Since I need to have 7 bags, I know that 56-41
= 15-8 = 7. This means that the total amount of peanuts for the remaining bag
is 7. There are a number of other ways this can be broken up. Here’s what I came
up with:
Using the cubes was very helpful because they gave me a visual representation of the “peanuts” in the bags. It was helpful to be able to take apart the cubes and rearange them to figure out the problem.
How does this model demonstrate what the mean represent?
I think my images clearly represent the mean of this set of data. It is easy to look at each stack of cubes and see that most bags have at least 8 or close to 8 peanuts. It’s almost as if you can draw a line through the stack at the “8 mark” and see that the average number for each bag is 8.
How did you use the line plot to figure out the problem?
I will say that it was a little harder for me to use the line plot to figure out the problem because I wasn’t able to physically break apart the line plot like I could with the cubes. However, you can look at the line plot and see that there are at least 8 or close to 8 peanuts in each bag.
How does this model help demonstrate what the mean represents?
Again, the mean is visible on the line plot, showing that 8 is an average number of peantus for each bag.
What does an average tell us about the whole data set?
In this case, your can look at the “mean” value of 8 and know
to expect at least 8 peanuts in most bags.
Video--How Much Taller?
I really enjoyed listening and watching the discussion this class had about the collection of height measurements for 1st and 4th graders. There was a variety of great answers that students gave. Moreover, I think it is clear that some students are well on their way to understanding important concepts and features of data analysis. Below, I have provided the notes I took during the video.
Case Studies--Focus on the Mean
I thought these cases brought forth some interesting ideas. The students seemed to have very different understandings of what the mean is and what other features that data analysis can highlight. I thought it was great to see many students use terminology like "middle" and "average" to describe the features of median and mean. From the video, Grady and Samantha touched on these ideas. In case #27, I think students learned to not be quick to generalize data as it appears. Saying things like "most" or "all" students in one class are taller than the other would not be a fair assumption. Instead, it would be more appropriate to say "as a group". I thought this was interesting. Students in this case got very close to better understanding how to find the mean of a set of data. Leah mentioned that adding up the total number of measurements would show you what grade was "taller". All there is left to do is divide by the total number of students.
I found Trudy's decision (Phoebe's case) to redistribute the measurements was impressive. Essentially, this is one way to find the mean. Many students seemed to struggle with finding or even understanding the mean. Kayla (case #28) was on the right track but divided by the wrong number. Instead of dividing by the number of student names, she divided by the sets of numbers found in each name. Linnea (case #26) also struggled with the concept of the mean. She thought it represented the "most common" height. We know this is not always the case.
Other Questions to Consider
*Do some reading and thinking about the concept of the average or mean and its application in schools through the bell curve. What does the mean suggest in terms of grades and achievement? Why is the concept represented with a bell curve? What are the implications for grading on the curve? Is it fair? Why or why not?
Education Portal ( http://education-portal.com/academy/lesson/bell-curve-definition-impact-on-grades.html) explains that grading on a bell curve "is a statistical concept that is designed to establish a normal distribution". Grading on a bell curve means that the grades given to each student is placed on a graph that forms the shape of a bell (symmetrical). In this case, the mean suggests that students scored typically scored in a certain grade range. On a A-F grading scale, the average score might be a C. However, that doesn't always mean that most students scored a C. I don't think grading on the curve is fair because it doesn't give a realistic overview of how students really scored. On curve, a student might score a B when they really deserved an A.
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