Statistics for the Behavioral Sciences by Frederick J. Gravetter, Larry B. Wallnau ISBN 10: 1305504917 ISBN 13: 9781305504912

104 CHAPTER 4 | Variability
In simple terms, the standard deviation provides a measure of the standard, or average,
distance from the mean, and describes whether the scores are clustered closely around the
mean or are widely scattered.
Although the concept of standard deviation is straightforward, the actual equations tend
to be more complex. Therefore, we begin by looking at the logic that leads to these equations.
If you remember that our goal is to measure the standard, or typical, distance from
the mean, then this logic and the equations that follow should be easier to remember.
STEP 1
The first step in finding the standard distance from the mean is to determine the deviation,
or distance from the mean, for each individual score. By definition, the deviation for each
score is the difference between the score and the mean.
DEFINITION
Deviation is distance from the mean:
deviation score 5 X 2 µ
A deviation score is
often represented by a
lowercase letter x.
For a distribution of scores with µ 5 50, if your score is X 5 53, then your deviation
score is
X 2 µ 5 53 2 50 5 3
If your score is X 5 45, then your deviation score is
X 2 µ 5 45 2 50 5 25
Notice that there are two parts to a deviation score: the sign (1 or 2) and the number.
The sign tells the direction from the mean—that is, whether the score is located above (1)
or below (2) the mean. The number gives the actual distance from the mean. For example,
a deviation score of 26 corresponds to a score that is below the mean by a distance of
6 points.
STEP 2
EXAMPLE 4.1
Because our goal is to compute a measure of the standard distance from the mean, the obvious
next step is to calculate the mean of the deviation scores. To compute this mean, you
first add up the deviation scores and then divide by N. This process is demonstrated in the
following example.
We start with the following set of N 5 4 scores. These scores add up to SX 5 12, so the
mean is µ 5 12
4 5 3. For each score, we have computed the deviation.
X
X 2 m
8 15
1 22
3 0
0 23
0 5 S(X 2 m)
■
Note that the deviation scores add up to zero. This should not be surprising if you
remember that the mean serves as a balance point for the distribution. The total of the
distances above the mean is exactly equal to the total of the distances below the mean