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Learning Math Home
Session 5, Part D: Deviations from the Mean
 
Session5 Part A Part B Part C Part D Part E Homework
 
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Session 5, Part D:
Deviations from the Mean (30 minutes)

In This Part: Tallying Excesses and Deficits | Line Plot Representations

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.

Number of Coins in Stack

Deviation from the Mean

1

-4

3

-2

4

-1

5

0

5

0

5

0

7

+2

7

+2

8

+3

+ _______

_______

45

0

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

Solution  

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:

Number of Coins in Stack

Deviation from the Mean

2

-3

3

3

4

5

6

6

8

8

+3

+ _______

_______

45

0

show answers

Number of Coins in Stack

Deviation from the Mean

2

-3

3

-2

3

-2

4

-1

5

0

6

+1

6

+1

8

+3

8

+3

+ _______

_______

45

0


hide answers


Next > Part D (Continued): Line Plot Representations

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