Download Normal curve & Z scores

Measures of Central Tendency
Purpose is to describe a distribution’s
typical case – do not say “average” case




Mode
Median
Mean (Average)
MEASURES OF DISPERSION

Standard deviation



Uses every score in the distribution
Measures the standard or typical distance from the
mean
Deviation score = Xi - X

Example: with Mean= 50 and Xi = 53, the deviation
score is 53 - 50 = 3
The Problem with Summing Deviations From Mean
•2
parts to a deviation score: the sign and the number
Mean = 3

X
8
1
3
0
12
Xi - X
+5
-2
0
-3
0
•Deviation scores
add up to zero
•Because sum of deviations
is always 0, it can’t be
used as a measure of
dispersion
Average Deviation
(using absolute value)

Works OK, but…

X=3
AD =
 |Xi – X|
N
X |Xi – X|
8 5
1 2
3 0
0 3
12 10
AD = 10 / 4 = 2.5
Absolute Value to get rid
of negative values
(otherwise it would add to
zero)
Variance & Standard Deviation
1.
2.
Purpose: Both indicate
“spread” of scores in a
distribution
Calculated using deviation
scores
4.
5.
(Xi – X)
(Xi - X)2
Difference between the
mean & each individual
score in distribution
5
1
1
2
-2
4
To avoid getting a sum of
zero, deviation scores are
squared before they are
added up.
Variance (s2)=sum of
squared deviations / N
Standard deviation
6
2
4
5
1
1
2
-2
4
=0
 = 14

3.
Xi
•
Square root of the variance
 = 20
Terminology

“Sum of Squares” = Sum of Squared
Deviations from the Mean =  (Xi - X)2

Variance = sum of squares divided by sample
size =  (Xi - X)2 = s2
N
Standard Deviation = the square root of the
variance = s

Calculating Variance, Then
Standard Deviation

Number of credits a
sample of 8 students
is are taking:




Calculate the mean,
variance & standard
deviation
Mean = 112/8 = 14
S2 = 72/8 = 9
S=3
Xi
(Xi – X)
(Xi - X)2
10
-4
16
9
-5
25
13
-1
1
17
3
9
15
1
1
16
2
4
14
0
0
18
4
16
 = 112
0
72
Summary Points about the
Standard Deviation
1.
Uses all the scores in the distribution
2.
Provides a measure of the typical, or standard,
distance from the mean
Increases in value as the distribution becomes
more heterogeneous

3.
Useful for making comparisons of variation
between distributions
4.
Becomes very important when we discuss the
normal curve (Chapter 5, next)
Mean & Standard Deviation Together

Tell us a lot about the typical score & how the scores
spread around that score


Useful for comparisons of distributions:
Example:
 Class A: mean GPA 2.8, s = 0.3
 Class B: mean GPA 3.3, s = 0.6
 Mean & Standard Deviation Applet
Example Using SPSS Output

Hours watching TV for Soc
3155 students:
1.
2.
3.
4.
What is the range &
interquartile range?
Is there skew (positive or
negative) in this distribution?
What is the most common
number of hours reported?
What is the average squared
distance that cases deviate
from the mean?
Statistics
Hours watch TV in typical week
N
Valid
18
Missing 11
Mean
8.2778
Median 5.0000
Mode
5.00
Std. Deviation
Variance
Minimum
Maximum
Percentiles
25
50
75
7.97648
63.624
1.00
28.00
3.0000
5.0000
14.0000
The Normal Curve & Z Scores
THE NORMAL CURVE

Characteristics:






Theoretical distribution of
scores
Perfectly symmetrical
Bell-shaped
Y
Unimodal
axis
Continuous
 There is a value of Y for
every value of X, where X
is assumed to be
continuous variable
Tails extend infinitely in both
directions
1.2
1.0
.8
.6
.4
.2
0.0
-2.07
-1.21
-.36
.50
1.36
x AXIS
Normal Curve, Mean
= .5, SD = .7
2.21
3.07
THE NORMAL CURVE

Assumption of normality of
a given empirical
distribution makes it
possible to describe this
“real-world” distribution
based on what we know
about the (theoretical)
normal curve
1.2
1.0
.8
.6
.4
.2
0.0
-2.07
-1.21
-.36
.50
1.36
Normal Curve, Mean = .5, SD = .7
2.21
3.07
THE NORMAL CURVE

.68 of area under the
curve (.34 on each side
of mean) falls within 1
standard deviation (s) of
the mean

In other words, 68% of
cases fall within +/- 1 s
95% of cases fall within
2 s’s
 99% of cases fall within
3 s’s

Areas Under the Normal Curve


Because the normal
curve is symmetrical,
we know that 50% of its
area falls on either side
of the mean.
FOR EACH SIDE:



34.13% of scores in
distribution are b/t the
mean and 1 s from the
mean
13.59% of scores are
between 1 and 2 s’s
from the mean
2.28% of scores are
> 2 s’s from the mean
THE NORMAL CURVE

Example:

Male height = normally
distributed, mean = 70
inches, s = 4 inches
 What is the range of
heights that
encompasses 99% of
the population?



Hint: that’s +/- 3
standard deviations
Answer: 70 +/- (3)(4)
= 70 +/- 12
Range = 58 to 82
THE NORMAL CURVE & Z SCORES
– To use the normal
curve to answer
questions, raw scores
of a distribution must
be transformed into Z
scores
• Z scores:
Formula: Zi = Xi – X
s
– A tool to help
determine how a
given score measures
up to the whole
distribution
RAW SCORES: 66
Z SCORES:
-1
70
0
74
1
NORMAL CURVE & Z SCORES

Transforming raw scores to
Z scores


a.k.a. “standardizing”
converts all values of
variables to a new scale:




mean = 0
standard deviation = 1
Converting raw scores to Z
scores makes it easy to
compare 2+ variables
Z scores also allow us to
find areas under the
theoretical normal curve
Z SCORE FORMULA
Z = Xi – X
S
• Xi = 120; X = 100; s=10
– Z= 120 – 100 = +2.00
10
• Xi = 80, S = 10
Z= 80 – 100 = -2.00
10
• Xi = 112, S = 10
Z = 112 – 100 = 1.20
10
• Xi = 95; X = 86; s=7
Z= 95 – 86 = 1.29
7
USING Z SCORES FOR COMPARISONS
– Example 1:
• An outdoor magazine does an analysis that assigns
separate scores for states’ “quality of hunting” (MN =
81) & “quality of fishing” (MN =74). Based on the
following information, which score is higher relative to
other states?
• Formula: Zi = Xi – X
s
– Quality of hunting for all states: X = 69, s = 8
– Quality of fishing for all states: X = 65, s = 5
– Z Score for “hunting”:
81 – 69 = 1.5
8
– Z Score for “fishing”:
73 – 65 = 1.6
5
• CONCLUSION: Relative to other states, Minnesota’s
“fishing” score was higher than its “hunting” score.
USING Z SCORES FOR COMPARISONS
– Example 2:
• You score 80 on a Sociology exam & 68 on a
Philosophy exam. On which test did you do better
relative to other students in each class?
Formula: Zi = Xi – X
s
– Sociology: X = 83, s = 10
– Philosophy: X = 62, s = 6
– Z Score for Sociology:
80 – 83 = - 0.3
10
– Z Score for Philosophy:
68 – 62 = 1
6
• CONCLUSION: Relative to others in your classes, you
did better on the philosophy test
Normal curve table

For any standardized normal distribution,
Appendix A (p. 453-456) of Healey provides
precise info on:




the area between the mean and the Z score
(column b)
the area beyond Z (column c)
Table reports absolute values of Z scores
Can be used to find:


The total area above or below a Z score
The total area between 2 Z scores
THE NORMAL DISTRIBUTION

Area above or below a Z score

If we know how many S.D.s away from the mean a
score is, assuming a normal distribution, we know
what % of scores falls above or below that score

This info can be used to calculate percentiles
AREA BELOW Z
• EXAMPLE 1: You get a 58 on a Sociology test.
You learn that the mean score was 50 and the
S.D. was 10.
– What % of scores was below yours?
Zi = Xi – X = 58 – 50 = 0.8
s
10
AREA BELOW Z
•
What % of scores was below
yours?
Zi = Xi – X = 58 – 50 = 0.8
s
10
•
Appendix A, Column B -- .2881
(28.81%) of area of normal curve
falls between mean and a Z score
of 0.8
•
Because your score (58) > the
mean (50), remember to add .50
(50%) to the above value
•
.50 (area below mean) + .2881
(area b/t mean & Z score) = .7881
(78.81% of scores were below
yours)
•
YOUR SCORE WAS IN THE 79TH
PERCENTILE
FIND THIS AREA
FROM COLUMN B
AREA BELOW Z
– Example 2:
– Your friend gets a 44 (mean = 50 & s=10) on the same
test
– What % of scores was below his?
Zi = Xi – X = 44 – 50 = - 0.6
s
10
AREA BELOW Z
• What % of scores was
below his?
Z = Xi – X = 44 – 50= -0.6
s
10
• Appendix A, Column C -.2743 (27.43%) of area of
normal curve is under a Z
score of -0.6
• .2743 (area beyond [below]
his Z score) 27.43% of
scores were below his
• YOUR FRIEND’S SCORE
WAS IN THE 27TH
PERCENTILE
FIND THIS AREA
FROM COLUMN C
1.2
1.0
.8
.6
.4
.2
0.0
-2.07
-1.21
-.36
.50
1.36
Normal Curve, Mean = .5, SD = .7
2.21
3.07
Z SCORES: “ABOVE” EXAMPLE
– Sometimes, lower is better…
• Example: If you shot a 68 in golf
(mean=73.5, s = 4), how many scores
are above yours?
68 – 73.5 = - 1.37
4
– Appendix A, Column B -- .4147
(41.47%) of area of normal curve
falls between mean and a Z score
of 1.37
1.2
1.0
.8
.6
– Because your score (68) < the
mean (73.5), remember to add .50
(50%) to the above value
.4
.2
– .50 (area above mean) + .4147
(area b/t mean & Z score) = .9147
(91.47% of scores were above
yours)
0.0
-2.07
-1.21
-.36
68
.50
73.5
1.36
Normal Curve, Mean = .5, SD = .7
FIND THIS
AREA FROM
COLUMN B
2.21
3.07
Area between 2 Z Scores

What percentage of
people have I.Q. scores
between Stan’s score of
110 and Shelly’s score
of 125? (mean = 100, s
= 15)
 CALCULATE Z
SCORES
AREA BETWEEN 2 Z SCORES

What percentage of people
have I.Q. scores between
Stan’s score of 110 and
Shelly’s score of 125? (mean =
100, s = 15)

CALCULATE Z SCORES:





Stan’s z = .67
Shelly’s z = 1.67
Proportion between mean (0)
& .67 = .2486 = 24.86%
Proportion between mean &
1.67 = .4525 = 45.25%
Proportion of scores between
110 and 125 is equal to:

45.25% – 24.86% = 20.39%
0
.67
1.67
AREA BETWEEN 2 Z SCORES

EXAMPLE 2:

If the mean prison admission rate for U.S. counties
was 385 per 100k, with a standard deviation of 151
(approx. normal distribution)


Given this information, what percentage of counties fall
between counties A (220 per 100k) & B (450 per 100k)?
Answers:
 A: 220-385 = -165 = -1.09
151
151
B: 450-385 = 65 = 0.43
151 151
County A: Z of -1.09 = .3621 = 36.21%
County B: Z of 0.43 = .1664 = 16.64%
Answer: 36.21 + 16.64 = 52.85%
4 More Sample Problems

For a sample of 150 U.S. cities, the mean
poverty rate (per 100) is 12.5 with a standard
deviation of 4.0. The distribution is
approximately normal.

Based on the above information:
1.
2.
3.
4.
What percent of cities had a poverty rate of more than
8.5 per 100?
What percent of cities had a rate between 13.0 and
16.5?
What percent of cities had a rate between 10.5 and
14.3?
What percent of cities had a rate between 8.5 and
10.5?