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

SECTION 5.2 | z-Scores and Locations in a Distribution 137
When you use the z-score formula, it can be useful to pay attention to the definition of a
z-score as well. For example, we used the formula in Example 5.3 to calculate the z-score
corresponding to X = 95, and obtained z = 1.29. Using the z-score definition, we note
that X = 95 is located above the mean by 9 points, which is slightly more than one standard
deviation (σ = 7). Therefore, the z-score should be positive and have a value slightly
greater than 1.00. In this case, the answer predicted by the definition is in perfect agreement
with the calculation. However, if the calculations produce a different value, for example
z = 0.78, you should realize that this answer is not consistent with the definition of a
z-score. In this case, an error has been made and you should check the calculations.
■ Determining a Raw Score (X) from a z-Score
Although the z-score equation (Formula 5.1) works well for transforming X values into
z-scores, it can be awkward when you are trying to work in the opposite direction and
change z-scores back into X values. In general, it is easier to use the definition of a z-score,
rather than a formula, when you are changing z-scores into X values. Remember, the z-score
describes exactly where the score is located by identifying the direction and distance from
the mean. It is possible, however, to express this definition as a formula, and we will use
a sample problem to demonstrate how the formula can be created.
For a distribution with a mean of μ = 60 and σ = 8, what X value corresponds to
a z-score of z = −1.50?
To solve this problem, we will use the z-score definition and carefully monitor the stepby-step
process. The value of the z-score indicates that X is located below the mean by
a distance equal to 1.5 standard deviations. Thus, the first step in the calculation is to
determine the distance corresponding to 1.5 standard deviations. For this problem, the standard
deviation is σ = 8 points, so 1.5 standard deviations is 1.5(8) = 12 points. The next
step is to find the value of X that is located below the mean by 12 points. With a mean of
μ = 60, the score is
X = μ − 12 = 60 − 12 = 48
The two steps can be combined to form a single formula:
X = μ + zσ (5.2)
In the formula, the value of zσ is the deviation of X and determines both the direction
and the size of the distance from the mean. In this problem, zσ = (−1.5)(8) = −12, or
12 points below the mean. Formula 5.2 simply combines the mean and the deviation from
the mean to determine the exact value of X.
Finally, you should realize that Formula 5.1 and Formula 5.2 are actually two different
versions of the same equation. If you begin with either formula and use algebra to shuffle
the terms around, you will soon end up with the other formula. We will leave this as an
exercise for those who want to try it.
LEARNING CHECK
1. Of the following z-score values, which one represents the most extreme location
on the left-hand side of the distribution?
a. z = +1.00
b. z = +2.00
c. z = −1.00
d. z = −2.00