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

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

Number of Coins in Stack Deviation from the Mean 2 -3 3 3 4 5 6 6 8 8 +3 + _______ _______ 45 0

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

Learning Math Home | Data Home | Register | Glossary | Map | ©

Session 5: Index | Notes | Solutions | Video