Chapter 4
z scores and
Normal
Distributions
Computing a z score
z =
X -m
s
X -M
z=
s
Example:
X = 400 μ = 500 σ = 100
what is z?
z=
X -m
s
400 - 500
z=
100
-100
z=
100
z = -1.00
The score of 400 is
-1.00 standard
deviations from
the mean
Comparing/Combining
with z scores
Comparison - Joe has a measured IQ of
105, and received a 700 on the SAT
Verbal, how do these scores compare?
 IQ scores:
μIQ = 100
σIQ = 15
 SAT scores: μSATV = 500
σSATV = 100

Comparing Scores using z
transformations
X 
105  100 5
z IQ 

  0.3333

15
15
X   700  500 200
zSATV 


 2.0000

100
100

These scores suggest that Joe’s
SAT performance was better
than would be expected by his
general intellectual ability
Comparing Scores using z
transformations

Matt’s scores on three tests in Stats:
MX =
X – MX =
s =
zi =
=
Test 1
Test 2
Test 3
31
22.2
8.8
12.5
21
19.5
1.5
2.1
35
32.0
3.0
1.8
(31-22.2)/12.5
(21-19.5)/2.1
(35-32)/1.8
+0.70
+0.71
+1.67
Back to Distributions

What if we took a distribution of raw
scores and transformed all of them to
z-scores?
Positive skewed
Distribution
Of Raw scores
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
Positive skewed
Distribution
Of z-scores
X-M
z=
s
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
scores
z-scores
Bimodal,
Negatively
Skewed,
Asymmetric
Distribution
Of Raw Scores
Bimodal,
Negatively
Skewed,
Asymmetric
Distribution
Of z-Scores
X-M
z=
s
5
5
4
4
3
3
2
2
1
1
0
0
scores
z-scores
Normal
Distribution
Of Raw Scores
X-M
z=
s
Normal
Distribution
Of z Scores
0.45
0.45
0.4
0.4
0.35
0.35
0.3
0.3
0.25
0.25
0.2
0.2
0.15
0.15
0.1
0.1
0.05
0.05
0
0
scores
z-scores
A VERY VERY VERY Special
Distribution: Standard UnitNormal Distribution



A Normal Distribution of z-scores
Popular member of the family where:
μ = 0 and σ = 1
It is also known as
– Unit-Normal Distribution or
– The Gaussian
– Often Symbolized “zUN”
Transforming Normal
Distributions

ANY normal distribution can be
transformed into a unit-normal
distribution by transforming the raw
scores to z scores:
zUN =
zUN
X -m
s
X -M
=
s
Unit-Normal Distributions (zUN)
.02
.02
.14
-3
-2
.34
-1
.34
0
.14
1
2
3
Using Table A (and a zUN
score) to find a %tile Rank
To find the corresponding percentile
rank of a z = 1.87, Table A from your
text book is used
 Find z = 1.87
 The area between zUN = 0 and zUN =
1.87 is .9693

Using Table A to determine Percentile Rank
.9693
.0307
-3
-2
-1
0
1
zUN = 1.87 = .0307
Percentile rank = 1-.0307 = 96.93%
2
3
Procedure (in words)
(raw score to z to %tile rank)

Transform raw score to zUN (scores must be
normally distributed)
Look up the proportion (p) of scores
between -∞ and the the zUN of interest
 Multiply by 100
