## Learning Math: Data Analysis, Statistics, and Probability

# Variation About the Mean Part C: Using Line Plots (30 minutes)

**In This Part:** **Creating a Line Plot **

Remember the line plots you worked with in previous sessions? If you reshuffle your stacks of coins just a bit, you can create a line plot representation that corresponds to the number of coins in each of the nine stacks, which will allow us to explore other interpretations of the mean. See Note 5, below.

To do this yourself, create a line plot on your paper or poster board. Across the bottom of the page, draw a horizontal line with 10 vertical tick marks numbered from 1 to 10 (placed far enough apart for an adhesive dot or note to fit between each). Your number line should look like this:

Set up your 45 coins in this ordered allocation:

Now arrange the stacks of coins on the paper above the number that corresponds to the height of each stack, like this:

Note that the 2 stacks of size 4 and the 2 stacks of size 6 are placed above the same number.

To form your line plot, replace each of the stacks with an adhesive dot or note. You should now have a line plot that looks like this:

Each dot in the line plot corresponds to a stack of that specified size.

**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:

**Video Segment**

In this video segment, participants compare their ordered stacks of snap-cubes to a line plot of the same data. Watch this segment after completing Problems C1-C3 to observe the transition from a physical representation of the data to a graphical representation.

What is one stack of snap-cubes equivalent to on the line plot of the data?

**In This Part:**** Means from the Line Plots**

In the previous examples, you explored line plot representations for sets of 45 coins, each in 9 stacks. For each allocation the mean was 5 coins. Now let’s use these line plot representations to explore another way to interpret the mean.

**Problem C4
**Here is a line plot corresponding to an allocation of 45 coins in 9 stacks:

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:

**a. **Form a different line plot with a mean equal to 5.**
b. **Form a line plot with a mean equal to 5 that has exactly 2 stacks of 5 coins.

**c.**Form a line plot with a mean equal to 5 but a median not equal to 5.

**d.**Form a line plot with a mean equal to 5 that has no 5-coin stacks.

**e.**Form a line plot with a mean equal to 5 that has two 5-coin stacks, 4 stacks with more than 5 coins, and 3 stacks with fewer than 5 coins.

**f.**Form a line plot with a mean equal to 5 that has two 5-coin stacks, 5 stacks with more than 5 coins, and 2 stacks with fewer than 5 coins.

**g.**Form a line plot with a mean equal to 5 that has two 5-coin stacks, two 10-coin stacks, and 5 stacks with fewer than 5 coins.

Don’t forget that the mean must always be equal to 5. If you move a dot to the right, it will increase the mean. Each time you move a dot to the right, you must balance this by moving another dot an equal distance to the left. Also, keep in mind that each dot represents a stack of coins, and that by moving the position of the dot, you change the number of coins in the stack. The total number of coins must remain 45.

**In This Part: Balancing Excesses and Deficits
**Regardless of the strategy you used in Problem C4, you must end up with an arrangement in which the sum of the 9 values is equal to 45. Let’s look at one possible strategy more closely. See Note 6 below.

For the sake of simplicity, we will begin with the line plot that corresponds to the fair allocation, 9 stacks of 5 coins each:

For this line plot, the sum is 45 and the mean is 5.

If we change one of the stacks of 5 coins to a stack of 8, the sum will increase by +3 to 48 and the mean will increase by +3/9 to 5 3/9. The line plot now looks like this:

**Problem C5
**How could you change another stack of 5 coins to reset the mean to 5?

**Problem C6**

If you could change the value of more than one stack, could you solve Problem C5 another way?

**Problem C7**

Now suppose that we change one of the stacks of 5 to a stack of 1, which reduces the total by 4. Here is the resulting line plot:

Describe at least three different ways to return the mean to 5.

**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.

For a non-interactive version, use your dots and coins on paper or poster board.

**a. **Form a line plot with a mean equal to 5 that has exactly 2 stacks of 5 coins.

**b. **Form a line plot with a mean equal to 5 but a median not equal to 5.

**c. **Form a line plot with a mean equal to 5 that has no 5-coin stacks.

**d. **Form a line plot with a mean equal to 5 that has two 5-coin stacks, 4 stacks with more than 5 coins, and 3 stacks with fewer than 5 coins.

**e. **Form a line plot with a mean equal to 5 that has two 5-coin stacks, 5 stacks with more than 5 coins, and 2 stacks with fewer than 5 coins.**
f. **Form a line plot with a mean equal to 5 that has two 5-coin stacks, two 10-coin stacks, and 5 stacks with fewer than 5 coins.

### Notes

**Note 5**

The transition from the physical representation of coins to a graphical representation (the line plot) can be challenging. It is easy to confuse the stacks of coins with the stacks of dots in the line plot, but in fact they are very different representations of the data. Each stack of coins corresponds to one dot on the line plot, and each dot on the line plot represents a stack of coins, rather than an individual coin (which it more closely resembles).

### Solutions

**Problem C1**

The arrangement of the coins should look like this:

Here is the corresponding line plot:

Note that there are nine dots, one for each stack, which are placed in the line plot according to the number of coins in each stack.

**Problem C2
**The arrangement of the coins should look like this:

Here is the corresponding line plot:

**Problem C3
**

The arrangement of the coins should look like this:

Here is the corresponding line plot:

**Problem C4**

There are multiple solutions to each of these problems, so answers will vary. Here are some possible solutions:

**a.** Here is a different line plot with a mean equal to 5:

**Problem C5**

You would need to remove 3 coins from one of the stacks of 5, resulting in a stack of 2 coins. This would reset the mean, since adding 3 coins to a stack changed the mean.

**Problem C6**

Yes, you could remove 1 coin from one stack and 2 coins from another.

Alternately, you could remove 1 coin from each of three stacks. You could even add 1 coin to a stack and remove 4 from another. The only requirement is that you subtract a total of 3 coins from the total number of coins (to counterbalance the 3 coins you added previously).

**Problem C7
**Answers will vary. Here are some possible solutions:

**•**You could add 4 coins to the stack of 3 to create a stack of 7 coins.

**•**You could add 2 coins to each of 2 stacks of 5 to create 2 stacks of 7 coins each.

**•**You could add 1 coin apiece to 4 of the stacks of 5 to create 4 stacks of 6 coins each.

**Problem C8
**The only requirement is that you must add 4 coins to the total number of coins to return the sum to 45, which will return the mean to 45 / 9 = 5.

As with Problem C4, answers will vary. See Problem C4 for some sample solutions.