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

# Variation About the Mean Part D: Deviations from the Mean (30 minutes)

**In This Par****t: Tallying Excesses and Deficits**

In Part B, we looked at excesses and deficits when we moved coins in the stacks to obtain the fair allocation. In Part C, we used a line plot to represent these excesses and deficits. We are now going to explore a new way to consider excesses and deficits. Let’s look at another line plot:

For this line plot, here is the corresponding allocation of our 45 coins:

Remember that the total of the excesses above the mean must equal the total of the deficits below the mean. In this case, each adds up to 7.

If you denote the values of excesses as positive numbers and deficits as negative numbers, then the total of the excesses is:

(+2) + (+2) + (+3) = +7

The total of the deficits is:

(-4) + (-2) + (-1) = -7

Statisticians refer to these excesses and deficits as deviations from the mean. For this allocation, the deviations from the mean are recorded in the table below.

Note that the deviations always sum to 0 because the total excesses (positive deviations) must be the same as the total deficits (negative deviations).

**Problem D1
**Here is another allocation of our 45 coins divided into 9 stacks:

**a. **Draw the corresponding line plot. Indicate the deviation for each dot as was done in the line plot that opened Part D

**b. **Complete the following table of deviations

**In This Part: Line Plot Representations
**

Use the Interactive Activity or your paper/poster board to display the line plots below (again, for 9 stacks of coins, with a mean of 5).

**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)

**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)

**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)?

When the positive and negative deviations are added together, the total is always 0. This property illustrates another way to interpret the mean: The mean is the balance point of the distribution when represented in a line plot, since the total deviation above the mean must equal the total deviation below the mean.

### Solutions

**Problem D1
a. **Here is the corresponding line plot, with the deviation for each dot included under the line:

**b. **Here is the completed table:

Problem D2

Here is the line plot:

**
Problem D3
**Here is the line plot:

**Problem D4
**The total of the deficits is (-4) + (-3) + (-3) + (-1) + (-1) = -12, so the remaining 4 stacks will need a total excess of +12 to maintain the balance required by the mean. One possible answer is for the remaining four values to be (+1), (+3), (+4), (+4) although many other answers are also possible, which corresponds to the following line plot:

**Problem D5**

The line plots would all move one unit to the right. This would not change the degree of fairness, since the degree of fairness relies on the deviations from the mean. Since the mean will move right along with the values in the line plot, the deviations will not change, and likewise the degree of fairness will not change.