Standard Scores

Dr. Richard Jackson jackson_r@mercer.edu

© Mercer University 2005© Mercer University 2005

All Rights ReservedAll Rights Reserved

Standard Scores (SS) and the Unit Normal Curve

Example: SAT and GRE

Standard Scores (SS) and the Unit Normal Curve

SS is any measurement (score) that has been transformed from a raw score to a more meaningful score

Example: SAT and GRE

SAT Scores Example: You scored 600 on Math

section

X ± 1 SD = 68% of subjects

You scored at the 84th centile (50% + 34%)

f

600400

34%34%

500

34%34%

50%50%

+1 SD- 1 SD

X = 500

SD = 100

Z Score

Special type of standardized score Represents measures that have been

transformed from raw scores /measures Represents the number of standard

deviations a particular measure/score is above or below the mean

Z Score

X - Xs

Z = Formula:

50 60 704030

0 +1 +2-1-2

Raw

Z

X X - X Z

7070

6060

5050

4040

3030

20

10

0

-10

-20

+2.00

+1.00

0

-1.00

-2.00

x

X = 50SD = 10

Z Scores The mean of all Z

scores is 0 The SD of all Z scores

is 1 All GRE scores are

transformed into scores with mean of 500 and SD of 100 to make them more meaningful

500

f

600 700 800400300200

0 +1 +2 +3-1-2-3

68%68%

99%99%95%95%

X = 500SD = 100

Transforming Raw Scores into SS

Formula:

SS = what you want your X to be + (Z) what you want

your SD to be( )( )

Transforming Raw Scores into SS

Example

X = 70

80 - 7010Z =

X = 500

SS = 500 + 1 (100) = 600

Z = +1.00

SD = 100

SD = 10

70

f

8060RAW

0 +1-1Z

500 600400SS

Converting raw score of 80 to SS with a X of 500 and SD of 100

StepsCalculate Z Score

Choose what you want your mean and SD to be

Plug into the SS equation

Other Example of SS IQ Scores

X = 100SD = 15

IQ of 130 is 2 SD’s above the mean and it places you at the 97.5 centile

Only 2.5% of people scored higher than you

100 115 1308570SS

95%95%

2.5%2.5% 2.5%2.5%

Normal Curve

Bell Shaped Has its max y value at its

mean Includes approximately 3

SD’s on each side Not skewed Mesokurtic Unit Normal Curve

Total Area Under a Curve (AUC) is regarded as being equal to Unity (or 1)X = 0

SD = 1

f

0x

y

Relationship of AUC to Proportion of Subjects in Study

f

0

x

y

Relationship of AUC to Proportion of Subjects in Study

Table IV Normal curve area

The numbers in body of table represent the AUC between the mean and a particular Z Score value

0.00.10.20.30.4

Z

.0000

.0398

.0793

.1179

.1554

.00

1.31.41.5

1.61.71.81.92.0

.4032

.4192

.4332

.4452

.4554

.4641

.4713

.4772

.0040

.0438

.0832

.1217

.1591

.01

.4049

.4207

.4345

.4463

.4564

.4649

.4719

.4778

Examples Z = +1.50

0.00.10.20.3

Z

.0000

.0398

.0793

.1179

.00

1.31.41.5

.4032

.4192

.4332

.0040

.0438

.0832

.1217

.01

.4049

.4207

.4345

0 +1.50

0.4332 (43.32%)

Table IV Normal Curve Areas

50%50%

1.50from Table

IV

0.4332

What % of subjects fall below Z score of 1.5?

50% + 43.32% = 93.32%

C93.32

Examples Z = +2.00

0.00.10.2

Z

.0000

.0398

.0793

.00

1.71.81.92.0

.4554

.4641

.4713

.4772

.0040

.0438

.0832

.01

.4564

.4649

.4719

.4778

0 +2.00

0.4772 (47.72%)

Table IV Normal Curve Areas

0.5000.500

2.00 from Table

IV

0.4772

+2.0

0.0440 (4.4%)

0.5000.500

Z +1.5

Examples Find the AUC between Z=1.50 and

Z=2.00

2.00from Table

IV

0.4772

1.50from Table

IV

0.4332

(4.4%)0.4772 - 0.4332 = 0.0440+2.0

0.0440 (4.4%)

0.5000.500

Z +1.5

ExampleAssume that among diabetics the fasting blood level of glucose is approximately normally distributed with a mean of 105 mg per 100 ml and an SD of 9 mg per 100 ml.

1. What proportion of diabetics have levels between 90 and 125mg per 100ml?

2. What level cuts off the lower 10 percent (10th centile) of diabetics?

3. What levels equidistant from the mean encompass 95 percent of diabetics?

Active Learning Exercise:SS and the Normal Curve

1. What proportion of diabetics have levels between 90 and 125mg per 100ml?

X = 105SD = 9

90 - 1059

Z90 = = -1.67

125 - 1059

Z125 = = +2.22

2.22from

Table IV 0.4868

1.67from

Table IV 0.4525

(93.93%)0.4525 + 0.4868 = 0.9393

X=105 12590

0.48680.4525

Active Learning Exercise:SS and the Normal Curve

2. What level cuts off the lower 10 percent (10th centile) of diabetics?

X = 105

0.1000.100

0.4000.400

X - 1059

-1.28 =

X = 93.5

X - X

sZ =

Z1.28 from Table IV 0.4000

X = 105SD = 9

?

Active Learning Exercise:SS and the Normal Curve

3. What levels equidistant from the mean encompass 95 percent of diabetics?

X = 105

0.02500.0250

0.47500.4750

0.02500.0250

0.47500.4750

Z = +1.96Z = -1.96

X = 105SD = 9

X - 1059

-1.96 =

X = 87.4

X - X

sZ =

X - 1059

+1.96 =

X = 122.6

X - X

sZ =

Z +1.96from

Table IV 0.4750

Z -1.96from

Table IV 0.4750