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```PSY 216
1. Problem 1 from the text
What information is provided by the sign (+ / -) of a z-score? What information is provided by the
numerical value of the z-score?
The sign tells you whether the score (X) is above (+) or below (-) the mean. The numerical value
of the z-score tells you how far (in standard deviation units) the score is from the mean.
2. Problem 2 from the text
A distribution has a standard deviation of σ = 12. Find the z-score for each of the following
locations in the distribution. (Problem 2 from the text)
a. Above the mean by 3 points
z = (X – M) / σ
z = (M + 3 – M) / 12 = 0.25
b. Above the mean by 12 points
z = (X – M) / σ
z = (M + 12 – M) / 12 = 1.00
c. Below the mean by 24 points
z = (X – M) / σ
z = (M - 24 – M) / 12 = -2.00
d. Below the mean by 18 points
z = (X – M) / σ
z = (M - 18 – M) / 12 = -1.50
3. Problem 6 from the text
For a population with a mean of μ = 100 and a standard deviation σ = 12,
a. Find the z-score for each of the following X values.
X = 106
X = 115
X = 130
X = 91
X - μ 106  100

 0.5
σ
12
X - μ 115  100
Z

 1.25
σ
12
Z
X - μ 130  100

 2.5
σ
12
X - μ 91  100
Z

 0.75
σ
12
Z
X = 88
Z
X - μ 88  100

 1.0
σ
12
X = 64
Z
X - μ 64  100

 3.0
σ
12
b. Find the score (X value) that corresponds to each of the following z-scores.
z = -1.00
z = -0.50
z = 2.00
z = 0.75
z = 1.50
z = -1.25
X  μ  z  σ  100 - 1.00  12  88
X  μ  z  σ  100 - 0.50  12  94
X  μ  z  σ  100  2.00  12  124
X  μ  z  σ  100  0.75  12  109
X  μ  z  σ  100  1.50  12  118
X  μ  z  σ  100 - 1.25  12  85
4. Problem 12 from the text
A score that is 6 points below the mean corresponds to a z-score of z = -0.50. What is the
population standard deviation?
z = (X – M) / σ
-0.50 = (M – 6 – M) / σ
σ = -6 / -0.50
σ = 12
5. Which of the following exam scores should lead to the better grade?
a. A score of X = 55 on an exam with μ = 60 and σ = 5.
b. A score of X = 40 on an exam with μ = 50 and σ = 20.
Because the distributions have different means and/or standard deviations, they are not directly
comparable. You should first convert each score to a z score so the scales of the distributions are
comparable:
X - μ 55  60

 1
σ
5
X - μ 40  50
zb 

 0.5
σ
20
za 
Because zb is larger than za, the score on exam a should lead to be a better grade.
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