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

138 CHAPTER 5 | z-Scores: Location of Scores and Standardized Distributions
2. Of the following z-score values, which one represents the location closest to
the mean?
a. z = +0.50
b. z = +1.00
c. z = −1.00
d. z = −2.00
3. For a population with μ = 100 and σ = 20, what is the z-score corresponding to
X = 105?
a. +0.25
b. +0.50
c. +4.00
d. +5.00
ANSWERS
1. D, 2. A, 3. A
5.3 Other Relationships Between z, X, m, and s
LEARNING OBJECTIVE
4. Explain how z-scores establish a relationship among X, μ, σ, and the value of z,
and use that relationship to find an unknown mean when given a z-score, a score,
and the standard deviation, or find an unknown standard deviation when given
a z-score, a score, and the mean.
In most cases, we simply transform scores (X values) into z-scores, or change z-scores back
into X values. However, you should realize that a z-score establishes a relationship between
the score, mean, and standard deviation. This relationship can be used to answer a variety
of different questions about scores and the distributions in which they are located. The
following two examples demonstrate some possibilities.
EXAMPLE 5.4
EXAMPLE 5.5
In a population with a mean of μ = 65, a score of X = 59 corresponds to z = −2.00. What
is the standard deviation for the population?
To answer the question, we begin with the z-score value. A z-score of −2.00 indicates
that the corresponding score is located below the mean by a distance of 2 standard deviations.
You also can determine that the score (X = 59) is located below the mean (μ = 65)
by a distance of 6 points. Thus, 2 standard deviations correspond to a distance of 6 points,
which means that 1 standard deviation must be σ = 3 points.
■
In a population with a standard deviation of σ = 6, a score of X = 33 corresponds to
z = +1.50. What is the mean for the population?
Again, we begin with the z-score value. In this case, a z-score of +1.50 indicates that the
score is located above the mean by a distance corresponding to 1.50 standard deviations.
With a standard deviation of σ = 6, this distance is (1.50)(6) = 9 points. Thus, the score is
located 9 points above the mean. The score is X = 33, so the mean must be μ = 24. ■
Many students find problems like those in Examples 5.4 and 5.5 easier to understand
if they draw a picture showing all of the information presented in the problem. For the