COURSE: JUST 3900
INTRODUCTORY STATISTICS
FOR CRIMINAL JUSTICE
Test Review: Ch. 4-6
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Instructor:
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Chapter 4: Key Concepts



Know that this chapter is on variability, which measures
the differences between scores and describes the
degree to which the scores are spread out or clustered
together.
Variability also measures how well an individual score
represents the entire distribution.
The 3 measures of variability: the range, standard
deviation, and the variance.
Chapter 4: The Range


The range is the distance covered by the scores in a
distribution, from the smallest to the largest score.
There are 2 formulas for the range that you should be
aware of for the test:


range = URL for Xmax – LRL for Xmin
range = Xmax – Xmin
Remember: URL stands
for upper real limit, and
LRL stands for lower real
limit.
This is the definition
used by many computer
programs.
Chapter 4: The Range

Find the range for the following set of scores using both
formulas:

8, 1, 5, 1, 5
Chapter 4: The Range

range = URL for Xmax – LRL for Xmin


range = 8.5 – 0.5 = 8
range = Xmax – Xmin

range = 8 – 1 = 7
Chapter 4: Sum of Squares


Sum of Squares (SS) is the sum of the squared
deviations.
There are 2 formulas you need to know to compute SS:




Definitional Formula (Population): 𝑆𝑆 =
𝑋− 𝜇
Definitional Formula (Sample): 𝑆𝑆 =
𝑋−𝑀 2
𝑋2
Computational Formula (Population): 𝑆𝑆 =
Computational Formula (Sample): 𝑆𝑆 =
2
𝑋 −
2
−
Use when mean is
a whole number.
𝑋 2
𝑁
𝑋 2 Use for fractional
means.
𝑛
Chapter 4: Variance

Variance is the mean squared deviation, or the average
squared distance from the mean.
𝜎2
=
𝑆𝑆
𝑁

Population Variance:

2
Sample Variance: 𝑠 =

The variance will play a bigger role as we move into
inferential statistics in the coming chapters. For now, just
know that it is a necessary step to finding the standard
deviation.
𝑆𝑆
𝑛−1
Chapter 4: Standard Deviation




Standard deviation provides a measure of the standard,
or average, distance from the mean.
A large standard deviation tells us that our scores are
widely distributed.
A small standard deviation tells us that they are
clustered closely around the mean.
It is calculated by taking the square root of the variance.

Population Standard Deviation: 𝜎 =

Sample Standard Deviation: 𝑠 =
𝜎 2 or 𝜎 =
𝑠 2 or 𝑠 =
𝑆𝑆
𝑛−1
𝑆𝑆
𝑁
Chapter 4: Standard Deviation

Find the standard deviation for the following population
of N =7 scores:

8, 1, 4, 3, 5, 3, 4
Chapter 4: Standard Deviation

Step 1: Find the mean


𝜇=
𝑋
𝑁
=
8+1+4+3+5+3+4
7
=
28
7
=4
Step 2: Find SS

𝑆𝑆 =
𝑋− 𝜇
2
Because the mean is a whole number.
X
(X - µ)
(X - µ)2
8
8–4=0
(4)2 = 16
1
1 – 4 = -3
(-3)2 = 9
4
4–4=0
(0)2 = 0
3
3 – 4 = -1
(-1)2 = 1
5
5–4=1
(1)2 = 1
3
3 – 4 = -1
(-1)2 = 1
4
4–4=0
(0)2 = 0
SS = 8 + 9 + 0 +
1 + 1 + 1 + 0 = 28
Chapter 4: Standard Deviation

Step 3: Calculate the variance


𝜎2 =
𝑆𝑆
𝑁
=
28
7
=4
Step 4: Find the standard deviation

𝜎=
𝑆𝑆
𝑁
=
28
7
=
4=2
Chapter 4: Sample Standard
Deviation




A sample statistic is unbiased if the average value of the
statistic is equal to the population parameter.
A sample statistic is biased if the average value of the
statistic either overestimates or underestimates the
corresponding population parameter.
To avoid bias when calculating s2 or s, we use n – 1,
instead of n.
n – 1 is referred to as degrees of freedom, or df.
Chapter 4: Sample Standard
Deviation

Find the standard deviation for the following sample of
n = 4 scores:

7, 4, 2, 1
Chapter 4: Sample Standard
Deviation

Step 1: Find the mean


𝑀=
𝑋
𝑛
=
7+4+2+1
4
=
14
4
= 3.5
Step 2: Find SS

𝑆𝑆 =
𝑋2
−
𝑋 2
𝑛
Because the mean contains decimals.
X
X2
7
(7)2 = 49
4
(4)2 = 16
2
(2)2 = 4
1
(1)2 = 1
𝑋 = 14
𝑋 2 = 70
Chapter 4: Sample Standard
Deviation

Step 2: Find SS


𝑋 2
𝑛
−
= 70 −
14 2
4
= 70 −
196
4
= 70 − 49 = 21
Step 3: Calculate the variance


𝑆𝑆 =
𝑋2
𝑠2 =
𝑆𝑆
𝑛−1
=
21
4−1
=
21
3
=7
Use n-1 in the denominator because
we’re working with a sample.
Step 4: Find the standard deviation

𝑠=
𝑆𝑆
𝑛−1
=
21
3
=
7 = 2.65
Chapter 4: More About
Standard Deviation

It should be noted that adding a constant to each score
in a distribution does not change the standard deviation.
Remember that standard deviation is a measure of
variability, or the distance between scores. If the same
number were added to every score, the entire
distribution would shift to the right, but the distance
between each score would remain the same.
Chapter 4: More About
Standard Deviation




However, multiplying every score by a constant would
cause the standard deviation to be multiplied by the
same constant.
Imagine a distribution that included scores of X = 10 and
X = 11. The distance between these scores is only 1
point.
Now imagine we multiply every score in this distribution
by 2 points. Our new scores would be X = 20 and
X = 22, 2 points apart.
Because the distance between scores changes when we
multiply by a constant, the standard deviation also
changes by the same constant.
Chapter 5: Key Concepts




z-Scores specify the precise location of each X value
within a distribution.
The sign of the z-score (+ or -) signifies whether the
score is above the mean (positive) or below the mean
(negative).
The numerical value of the z-score describes the
distance from the mean by counting the number of
standard deviations between X and µ.
The z-score distribution will always have a mean of µ = 0
and a standard deviation of σ = 1.
Chapter 5: z-Scores


The formulas for calculating the z-score:
𝑋−𝜇
𝜎
𝑋−𝑀
𝑠

Population z-score: 𝑧 =

Sample z-score: 𝑧 =
Sometimes it’s necessary to calculate the X-value using
the z-score, mean, and standard deviation.


𝑋 = 𝜇 + 𝑧𝜎
𝑋 = 𝑀 + 𝑧𝑠
Chapter 5: z-Scores

For a population with µ = 50 and σ = 8, find the z-score
for each of the following X values:




X = 54
X = 42
X = 62
X = 48
Chapter 5: z-Scores

X = 54


=
54−50
8
=
4
8
𝑋−𝜇
𝜎
=
42−50
8
=
−8
8
= −1.00
𝑋−𝜇
𝜎
=
62−50
8
=
12
8
= 1.50
𝑋−𝜇
𝜎
=
48−50
8
=
−2
8
= −0.25
= 0.50
𝑧=
X = 62


𝑋−𝜇
𝜎
X = 42


𝑧=
𝑧=
X = 48

𝑧=
Chapter 5: Find X

Find the X value that corresponds with the following zscores: (µ = 50 and σ = 8)




z = -0.50
z = 0.75
z = -1.50
z = 0.25
Chapter 5: Find X

z = -0.50


z = 0.75


𝑋 = 𝜇 + 𝑧𝜎 = 50 + (0.75)(8) = 50 + 6 = 56
z = -1.50


𝑋 = 𝜇 + 𝑧𝜎 = 50 + (−.50)(8) = 50 − 4 = 46
𝑋 = 𝜇 + 𝑧𝜎 = 50 + (−1.50)(8) = 50 − 12 = 38
z = 0.25

𝑋 = 𝜇 + 𝑧𝜎 = 50 + (0.25)(8) = 50 + 2 = 52
Chapter 5: Standardized
Distributions



Standardized distributions are composed of scores that
have been transformed to create predetermined values
for µ and σ.
Standardized distributions are used to make dissimilar
distributions comparable.
A z-score distribution is an example of a standardized
distribution.
Chapter 5: Standardized
Distributions

A distribution with μ = 62 and σ = 8 is transformed into a
standardized distribution with μ = 100 and σ = 20. Find
the new, standardized score for each of the following X
values:




X = 60
X = 54
X = 72
X = 66
Chapter 5: Standardized
Distributions

Step 1: Find the z-scores

X = 60


𝑧=
𝑋−𝜇
𝜎
=
60−62
8
=
−2
8
= −0.25
Step 2: Find the X value for the new distribution

𝑋 = 𝜇 + 𝑧𝜎 = 100 + (−0.25)(20) = 100 − 5 = 95
Chapter 5: Standardized
Distributions

Step 1: Find the z-scores

X = 54


𝑋−𝜇
𝜎
=
54−62
8
=
−8
8
= −1.00
𝑋−𝜇
𝜎
=
72−62
8
=
10
8
= 1.25
𝑋−𝜇
𝜎
=
66−62
8
=
4
8
X = 72


𝑧=
𝑧=
X = 66

𝑧=
= 0.50
Chapter 5: Standardized
Distributions

Step 2: Find the X values for the new distribution

X = 54, z = -1.00


X = 72, z = 1.25


𝑋 = 𝜇 + 𝑧𝜎 = 100 + (−1.00)(20) = 100 − 20 = 80
𝑋 = 𝜇 + 𝑧𝜎 = 100 + (1.25)(20) = 100 + 25 = 125
X = 66, z = 0.50

𝑋 = 𝜇 + 𝑧𝜎 = 100 + (0.50)(20) = 100 + 10 = 110
Chapter 6: Key Concepts

For a situation in which several different outcomes are
possible, the probability for any specific outcome is
defined as a fraction of all the possible outcomes.



𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑐𝑙𝑎𝑠𝑠𝑖𝑓𝑖𝑒𝑑 𝑎𝑠 𝐴
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
Probability can be represented as a fraction, decimal,
or percent.


Probability of A =
p = 0.25 = ¼ = 25%
A random sample requires that each individual in the
population has an equal chance of being selected.
An independent random sample requires that each
individual has an equal chance of being selected and
that the probability of being selected remains constant
from one selection to the next.
Chapter 6: Random Sampling




Sampling with replacement requires that selected
individuals be returned to the population before the next
selection is made.
This ensures that the probability of selection remains
constant from one selection to the next.
Unless otherwise specified, random sampling assumes
replacement.
Probability does not remain constant when sampling
without replacement.
Chapter 6: Random Sampling

A psychology class consists of 14 males and 36 females.
If the professor selects names from the class list using
random sampling,
a)
b)
What is the probability that the first student selected will be a
female?
If a random sample of n = 3 students is selected and the first
two are both females, what is the probability that the third
student selected will be a male?
Chapter 6: Random Sampling
a)
b)
p = 36/50 = 0.72
p = 14/50 = 0.28
Because this is a random sample, replacement is assumed.
Therefore, the probability of selection remains constant.
Chapter 6: The Normal
Distribution

When dealing with probability in this chapter, we are
dealing with normal distributions.
34.13%
Normal distributions
are symmetrical.
13.59%
Percentage of the
population located
between z = 0 and
Z = 1.
2.28%

The unit normal table lists proportions of the normal
distribution for a full range of possible z-score values.
Chapter 6: The Unit Normal
Table


The first column (A) in the table lists z-scores
corresponding to different positions in a normal
distribution.
Column B presents the proportion in the body.


Column C presents the proportion in the tail.


The body is always the larger part of the distribution.
The tail is always the smaller part of the distribution.
Column D identifies the proportion located between the
mean and the z-score.
Note: The normal distribution is symmetrical, so the proportions on the right are the
same as the proportions on the left. And although the z-score values change signs
(+ and -), the proportions are always positive.
Chapter 6: The Unit Normal
Table

Find each of the probabilities for a normal distribution:





p(z > 0.25)
p(z > -0.75)
p(z < 1.20)
p(-0.25 < z < 0.25)
p(-1.25 < z < 0.25)
Chapter 6: The Unit Normal
Table

p(z > 0.25)
p = 0.4013

p(z > -0.75)
p = 0.7734

p(z < 1.20)
p = 0.8849

p(-0.25 < z < 0.25)

p(-1.25 < z < 0.25)
Find the proportion in
column D for both z-scores
and add them together.
p = 0.0987
p = 0.4931
Chapter 6: Binomial
Distributions




When a variable is measured on a scale consisting of
exactly two categories, the resulting data are called
binomial.
The normal distribution can be used to compute
probabilities with binomial data.
The two categories are defined as A and B.
The probabilities associated with each category are:




p = p(A) = the probability of A
q = p(B) = the probability of B
p + q = 1.00
The number of individuals or observations in the sample
is identified by n.
The variable X refers to the number of times category A
occurs in the sample.
Chapter 6: Binomial
Distributions

Binomial distributions tend to approximate a normal
distribution when two conditions are met:



pn ≥ 10
qn ≥ 10
Under these circumstances, the binomial distribution has
the following parameters:

Mean: µ = pn
Standard deviation: σ = 𝑛𝑝𝑞

z=

𝑋−𝜇
𝜎
=
𝑋 −𝑝𝑛
𝑛𝑝𝑞
Chapter 6: Binomial
Distributions
Notice that each X value
is represented by a bar in
the histogram. This means
a score of X = 8 spans the
interval of 7.5 to 8.5.
Chapter 6: Binomial
Distributions

A multiple choice test has 48 questions, each with four
response choices. If a student is simply guessing at the
answers,
a)
b)
c)
d)
What is the probability of guessing correctly for any question?
On average, how many questions would a student get correct
for the entire test?
What is the probability that a student would get more than15
answers correct simply by guessing?
What is the probability that a student would get 15 or more
answers correct simply by guessing?
Chapter 6: Binomial
Distributions
a)
b)
p = ¼ = 0.25
pn = 0.25(48) = 12
Chapter 6: Binomial
Distributions
c)
Step 1: Find pn and qn




Both are greater than 10,
so our distribution approximates
a normal distribution.
Step 2: Find μ and σ

µ = pn = 12

σ = 𝑛𝑝𝑞 =
48(0.25)(0.75) = 9 = 3
URL because we are
excluding the score X = 15.
Step 3: Find Z


pn = (0.25)(48) = 12
qn = 0.75(48) = 36
z=
𝑋−𝜇
𝜎
=
𝑋 −𝑝𝑛
𝑛𝑝𝑞
=
15.5 −12
3
=
3.5
3
Step 4: Find the proportion

p(z > 1.17) = 0.1210
= 1.17
Chapter 6: Binomial
Distributions
d)
Step 1: Find pn and qn




Step 2: Find μ and σ

µ = pn = 12

σ = 𝑛𝑝𝑞 =
48(0.25)(0.75) = 9 = 3
LRL because we are
including the score X = 15.
Step 3: Find Z


pn = (0.25)(48) = 12
qn = 0.75(48) = 36
z=
𝑋−𝜇
𝜎
=
𝑋 −𝑝𝑛
𝑛𝑝𝑞
=
14.5 −12
3
=
2.5
3
Step 4: Find the proportion

p(z > 0.83) = 0.2033
= 0.83