COURSE: JUST 3900
INTRODUCTORY STATISTICS
FOR CRIMINAL JUSTICE
Chapter 4: Variability
Peer Tutor Slides
Instructor:
Mr. Ethan W. Cooper, Lead Tutor
© 2013 - - PLEASE DO NOT CITE, QUOTE, OR REPRODUCE WITHOUT THE
WRITTEN PERMISSION OF THE AUTHOR. FOR PERMISSION OR QUESTIONS,
PLEASE EMAIL MR. COOPER AT THE FOLLWING: coopere07@students.ecu.edu
Key Terms: Don’t Forget
Notecards
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Variability (p. 104)
Range (p. 106)
Deviation (p. 107)
Population Variance (p. 108)
Standard Deviation (p. 108)
Sum of Squares (p. 111)
Sample Variance (p. 115)
Degrees of Freedom (p. 117)
Unbiased (p. 119)
Biased (p. 119)
Think Notecards: For Formulas
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Sum of Squares (definitional): 𝑆𝑆 =
𝑋−𝜇
Sum of Squares (computational): 𝑆𝑆 =
Population Variance:
𝜎2
=
2
2
𝑋 −
𝑆𝑆
𝑁
Population Standard Deviation: 𝜎 =
𝜎2
=
𝑆𝑆
𝑁
𝑋 2
𝑁
Think Notecards: For Formulas

Sum of Squares (definitional): 𝑆𝑆 =
𝑋−𝑀

Sum of Squares (computational): 𝑆𝑆 =

𝑠2

Sample Variance:
=
2
𝑋 −
𝑆𝑆
𝑛−1
Sample Standard Deviation: 𝑠 =
𝑠2
=
2
𝑆𝑆
𝑛−1
𝑋 2
𝑛
Range
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Question 1: What is the range for the following set of
scores? (Use all 3 definitions)

1, 9, 5, 8, 7
The Range
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Question 1 Answer:
1.
Range = URL for Xmax – LRL for Xmin
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
2.
Range = Xmax – Xmin + 1
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
3.
Range = 9.5 – 0.5
Range = 9
Range = 9 – 1 + 1
Range = 9
Range = Xmax – Xmin
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Range = 9 – 1
Range = 8
Note: This formula is
an alternative to using
real limits, but it only works
for whole numbers. Do
NOT use for scores with
decimals!
Standard Deviation and
Variance for a Population
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Question 2: Briefly explain what is meant by the standard
deviation and what is measured by the variance.
Standard Deviation and
Variance for a Population
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Question 2 Answer:

Standard deviation measures the standard distance from the
mean and variance measures the average squared distance
from the mean.
Standard Deviation and
Variance for a Population

Question 3: What is the standard deviation for the
following set of N = 5 scores: 10, 10, 10, 10, and 10?
(Note: This can be done without using any calculations)
Standard Deviation and
Variance for a Population

Question 3 Answer:
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Because there is no variability (the scores are all the same), the
standard deviation is zero.
Standard Deviation and
Variance for a Population

Question 4: Calculate the variance and standard
deviation for the following population of N = 5 scores:
4, 0, 7, 1, 3.
Standard Deviation and
Variance for a Population
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Question 4 Answer:
1.
Find the mean.

2.
𝜇=
𝑋
𝑁
=3
Find SS: Definitional Formula 𝑆𝑆 =
𝑋−𝜇
2
X
(X - µ)
(X - µ)2
4
1
1
0
-3
9
7
4
16
1
-2
4
3
0
0
Standard Deviation and
Variance for a Population
3.
Find the sum of the squared deviations
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4.
Find the variance
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5.
SS = 1 + 9 + 16 + 4 +0
SS = 30
σ2 = SS/N
σ2 = 30/5
σ2 = 6
Find the standard deviation
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𝜎 = 𝜎2
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𝜎= 6
𝜎 = 2.45
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Standard Deviation and
Variance for a Sample
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Question 5: Calculate the variance and standard
deviation for the following sample of N = 5 scores:
4, 0, 7, 1, 3. (Use the computational formula for SS)

Note: Although the mean is a whole number, please use the
computational formula so that you can see that the formulas are
mathematical equivalents. However, on the test and on Aplia,
use the definitional formula when dealing with whole number
means and the computational formula with means containing
decimals. This will save lots of time.
Standard Deviation and
Variance for a Sample
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Question 5 Answer:
1.
Find the mean

2.
Find df
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3.
M=3
df = n - 1
df = 5 – 1
df = 4
Find SS.
X
X2
4
16
0
0
7
49
1
1
3
9
∑X = 15
∑X2 = 75
Standard Deviation and
Variance for a Sample
3.
Find SS.
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4.
Find s2
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5.
SS = 75 – [(152 )/5]
SS = 75 – (225/5)
SS = 75 – 45
SS = 30
s2 = SS/df = SS/(n-1)
s2 = 30/4
s2 = 7.5
Find s
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𝑠 = 𝑠2
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𝑠 = 7.5
𝑠 = 2.74
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Standard Deviation and
Variance for a Sample
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Question 6: Explain why the formula for sample variance
divides SS by n – 1 instead of dividing by n.
Standard Deviation and
Variance for a Sample
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Question 6 Answer:
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Without some correction, sample variability consistently
underestimates the population variability. Dividing by a small
number (n – 1 instead of n) increases the value of the sample
variance and makes it an unbiased estimate of the population
variance.
Things to Consider
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Correct!
Incorrect, but will be an
answer choice!
Things to Consider

On this week’s Aplia assignment, Aplia asks you to
calculate SS using the definitional formula for a set of
scores in which the mean has 3 decimal places. This
makes the ensuing calculations more complicated. Use
the computational formula to avoid this difficulty.
These calculations can
get complicated!
Save yourself some trouble by using the computational formula for SS.
Transformations of Scale
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Question 7: A population has a mean of µ = 70 and a
standard deviation of σ = 5.
a)
b)
If 10 points were added to every score in the population, what
would be the new values for the population mean and standard
deviation?
If every score were in the population were multiplied by 2, what
would be the new values for the population mean and standard
deviation?
Transformations of Scale
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Question 7 Answer:
a)
b)
The new mean would be µ = 80 but the standard deviation
would still be σ = 5.
The new mean would be µ = 140 and the new standard
deviation would be σ = 10.
In part a),the distribution moves, but
the distance between scores remains
constant. Thus, the standard deviation
remains the same.
Ex: Suppose that our original distribution
contained scores of X = 71 and X = 72. After
adding 10 points to every score, these scores
would become X = 81 and X = 82, respectively.
While the scores themselves have changed, the
distance between them remained the same, one
point. Thus, the standard deviation remains σ = 5.
In part b), the distribution moves, and
the distance between scores is doubled.
Thus, the standard deviation is also
doubled.
Ex: Again, suppose our distribution contained
scores of X = 71 and X = 72. If we multiplied every
score by 2, these scores would become X = 142
and X = 144, respectively. In this instance, the
distance between scores has increased from 1 point
to 2 points. This increase in variability increases the
standard deviation from σ = 5 to σ = 10.
Transformations of Scale
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Question 8: In a population with a mean of µ = 80 and a
standard deviation of σ = 8, would a score of X = 87 be
considered an extreme value (far out in the tail of the
distribution)? What if the standard deviation were
σ = 3?
Transformations of Scale
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Question 8 Answers:
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With σ = 8, a score of X = 87 would be located in the central
section of the distribution (within one standard deviation). With
σ = 3, a score of X = 87 would be an extreme value, located
more than two standard deviations above the mean.
X = 88
X = 87
X = 83
3
X = 86
X = 87
Frequently Asked Questions:
FAQs
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Frequently Asked Questions:
FAQs

How do I use the definitional formula when calculating
sum of squares?

The best way to use this formula is to create a chart:
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Find SS for the following population: 4, 0, 7, 1, 3
µ =ΣX/N = 15/5 = 3
X
(X - µ)
(X - µ)2
4
4-3=1
12 = 1
0
0 – 3 = (-3)
(-3)2 = 9
7
7–3=4
42 = 16
1
1 – 3 = (-2)
(-2)2 = 4
3
3–3=0
02 = 0
Frequently Asked Questions:
FAQs

How do I use the computational formula when
calculating sum of squares?

The best way to use this formula is to create a chart:

Find SS for the following sample: 4, 0, 7, 1, 3
X
X2
4
16
0
0
7
49
1
1
3
9
∑X = 15
∑X2 = 75
Frequently Asked Questions:
FAQs
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