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

SECTION 5.3 | Other Relationships Between z, X, m, and s 139
s
s
F I G U R E 5.4
A visual representation of the question in
Example 5.4. If 2 standard deviations
correspond to a 6-point distance, then
1 standard deviation must equal 3 points.
59
6 points
65
problem in Example 5.4, the picture would begin with a distribution that has a mean of
μ = 65 (we use a normal distribution that is shown in Figure 5.4). The value of the standard
deviation is unknown, but you can add arrows to the sketch pointing outward from the mean
for a distance corresponding to 1 standard deviation. Finally, use standard deviation arrows
to identify the location of z = −2.00 (2 standard deviations below the mean) and add X =
59 at that location. All of these factors are shown in Figure 5.4. In the figure, it is easy to see
that X = 59 is located 6 points below the mean, and that the 6-point distance corresponds to
exactly 2 standard deviations. Again, if 2 standard deviations equal 6 points, then 1 standard
deviation must be σ = 3 points.
A slight variation on Examples 5.4 and 5.5 is demonstrated in the following example.
This time you must use the z-score information to find both the population mean and the
standard deviation.
EXAMPLE 5.6
In a population distribution, a score of X = 54 corresponds to z = +2.00 and a score of
X = 42 corresponds to z = −1.00. What are the values for the mean and the standard
deviation for the population? Once again, many students find this kind of problem easier to
understand if they can see it in a picture, so we have sketched this example in Figure 5.5.
The key to solving this kind of problem is to focus on the distance between the two
scores. Notice that the distance can be measured in points and in standard deviations. In
points, the distance from X = 42 to X = 54 is 12 points. According to the two z-scores,
X = 42 is located 1 standard deviation below the mean and X = 54 is located 2 standard
deviations above the mean (see Figure 5.5). Thus, the total distance between the two
scores is equal to 3 standard deviations. We have determined that the distance between the
two scores is 12 points, which is equal to 3 standard deviations. As an equation
Dividing both sides by 3, we obtain
3σ = 12 points
σ = 4 points
Finally, note that X = 42 corresponds to z = −1.00, which means that X = 42 is located
one standard deviation below the mean. With a standard deviation of σ = 4, the mean must be
μ = 46. Thus, the population has a mean of μ = 46 and a standard deviation of σ = 4. ■
The following example is an opportunity for you to test your understanding by solving
a problem similar to the demonstration in Example 5.6.