Transcript Chapter 1: Introduction to Statistics

COURSE: JUST 3900
TIPS FOR APLIA
Chapter 5:
z-Scores
Developed By:
Ethan Cooper (Lead Tutor)
John Lohman
Michael Mattocks
Aubrey Urwick
Key Terms and Formulas: Don’t
Forget Notecards

Describing z-Scores

Question 1: Identify the z-score value corresponding to
each of the following locations in a distribution.




Below the mean by 3 standard deviations.
Above the mean by 1/4 standard deviations.
Below the mean by 2.50 standard deviations.
Question 2: Describe the location in the distribution for
each of the following z-scores.




z = - 1.50
z = 0.25
z = - 3.50
z = 1.75
Describing z-Scores

Question 1 Answer:




z = -3.00
z = 0.25
z = -2.50
Question 2 Answer:




Below the mean by 1.50 standard deviations.
Above the mean by ¼ standard deviations.
Below the mean by 3.50 standard deviations.
Above the mean by 1.75 standard deviations.
Describing z-Scores

𝑋−𝜇
The numerator in our z-score formula (𝑧 =
)
𝜎
describes the difference between X and µ. Therefore, if
a question asks you to calculate the z-score for a score
that is above the mean by 4 points and has a standard
deviation of σ = 2, you cannot calculate (X - µ) - - i.e.,
you cannot find the original values for X or µ and
calculate the difference between the two. In this case,
it has already been provided for you because the
question tells you the distance between X and µ
(X - µ) = 4. Thus, your formula should read z = 4/2,
which comes out to be z = 2.00.
Transforming X-Values into
z-Scores

Question 3: For a distribution of µ = 40 and σ = 12, find
the z-score for each of the following scores.
a)
b)
c)

X = 36
X = 46
X = 56
Question 4: For a population with µ = 30 and σ = 8, find
the z-score for each of the following scores.
a)
b)
c)
X = 32
X = 26
X = 42
Transforming X-Values into
z-Scores

Using z-Scores to Compare
Different Populations

Question 5: A distribution of English exam scores has
µ = 70 and σ = 4. A distribution of history exam scores
has µ = 60 and σ = 20. For which exam would a score of
X = 78 have a higher standing? Explain your answer.
Using z-Scores to Compare
Different Populations

Question 5 Answer:
𝑋−𝜇
𝜎
𝑋−𝜇
𝜎
=
78−70
4
78−60
20
𝑧=

𝑧=

For the English exam, X = 78 corresponds to z = 2.00, which is a
higher standing than z = 0.90 for the history exam.
=
=
8
= 2.00
4
18
= 0.90
20

=
Remember that 95% of all scores fall between ± 2.00. Thus, a score
of +2.00 means that over 95% of the class scored below 78 on the
English exam.
Using z-Scores to Compare
Different Populations

Question 6: A distribution of English exam scores has
µ = 50 and σ = 12. A distribution of history exam scores
has µ = 58 and σ = 4. For which exam would a score of
X = 62 have a higher standing? Explain your answer.
Using z-Scores to Compare
Different Populations

Question 6 Answer:
𝑋−𝜇
𝜎
𝑋−𝜇
𝜎
=
62−50
12
62−58
4
𝑧=

𝑧=

The score X = 62 corresponds to z = 1.00 in both distributions.
The score has exactly the same standing for both exams.
=
=
12
= 1.00
12
4
= 1.00
4

=
z-Scores and Standardized
Scores

Question 7: A population of scores has µ = 73 and σ = 8.
If the distribution is standardized to create a new
distribution with µ = 100 and σ = 20, what are the new
values for each of the following scores from the original
distribution?
a)
b)
c)
d)
X = 65
X = 71
X = 81
X = 83
z-Scores and Standardized
Scores

z-Scores and Standardized
Scores

Question 8: A population with a mean of µ = 44 and a
standard deviation of σ = 6 is standardized to create a
new distribution with µ = 50 and σ = 10.
a)
b)
What is the new standardized value for a score of X = 47 from
the original distribution?
One individual has a new standardized score of X = 65. What
was this person’s score in the original distribution?
z-Scores and Standardized
Scores

X = 47
z = 0.50
X = 55
Old Distribution
32
38
44
50
56
z-Score Distribution
New Standardized Distribution
z-Scores and Standardized
Scores

X = 53
z = 1.50
X = 65
32
38
44
Old Distribution
50
56
z-Score Distribution
New Standardized Distribution
Measure of Relative Location
and Detecting Outliers

Question 9: A sample has a mean of M = 30 and a
standard deviation of s = 8.
a)
b)
Would a score of X = 36 be considered a central score or an
extreme score in the sample?
If the standard deviation were s = 2, would X = 36 be central or
extreme?
Measure of Relative Location
and Detecting Outliers

Question 9 Answer:
a)
𝑧=
1.
b)
𝑧=
1.
𝑋−𝑀
𝑠
=
36−30
8
=
6
8
= 0.75
X = 36 is not an extreme score because it is within two standard
deviations of the mean.
𝑋−𝑀
𝑠
=
36−30
2
=
6
2
= 3.00
In this case, X = 36 is an extreme score because it is more than
two standard deviations above the mean.
WARNING!!!



The book defines an extreme score as being more than
TWO standard deviations away from the mean.
However, Aplia defines extreme scores as being more
than THREE standard deviations from the mean.
When using Aplia, use the THREE definition of standard
deviation for extreme scores.
On in class exercises and on the test, use the TWO
definition of standard deviation for extreme scores.
Frequently Asked Questions
FAQs

How do I find the z-scores from a raw set of scores?

1)
X = 11, 0, 2, 9, 9, 5
Find the mean:
1)
2)
𝜇=
𝑋
𝑁
=
11+0+2+9+9+5
6
=
36
6
=6
Find SS:
1)
𝑆𝑆 =
𝑋−𝜇
2
X
X-µ
(X - µ)2
11
11 – 6 = 5
(5)2 = 25
0
0 – 6 = -6
(-6)2 = 36
2
2 – 6 = -4
(-4)2 = 16
2
9–6=3
(3)2 = 9
9
9–6=3
(3)2 = 9
5
5 – 6 = -1
(-1)2 = 1
𝑆𝑆 =
𝑋−𝜇
2
= 96
Frequently Asked Questions
FAQs
3)
Find σ2:
1)
4)
𝑆𝑆
𝑁
=
96
6
= 16
Find σ:
1)
5)
𝜎2 =
𝜎=
𝑆𝑆
𝑁
=
96
6
=
16 = 4
Find z-score for each X:
1)
𝑧=
2)
𝑧=
3)
𝑧=
4)
𝑧=
5)
𝑧=
6)
𝑧=
𝑋−𝜇
𝜎
𝑋−𝜇
𝜎
𝑋−𝜇
𝜎
𝑋−𝜇
𝜎
𝑋−𝜇
𝜎
𝑋−𝜇
𝜎
=
=
=
=
=
=
11−6
5
=
= 1.25
4
4
0−6
−6
=
= −1.50
4
4
2−6
−4
=
= −1.00
4
4
9−6
3
=
= 0.75
4
4
9−6
3
=
= 0.75
4
4
5−6
−1
=
= −0.25
4
4