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Variability
Variability

Variability


How tightly clustered or how widely dispersed
the values are in a data set.
Example



Data set 1: [0,25,50,75,100]
Data set 2: [48,49,50,51,52]
Both have a mean of 50, but data set 1 clearly
has greater Variability than data set 2.
Variability: The Range

The Range is one measure of variability


The range is the difference between the maximum
and minimum values in a set
Example




Data set 1: [1,25,50,75,100]; R: 100-0 +1 = 100
Data set 2: [48,49,50,51,52]; R: 52-48 + 1= 5
The range ignores how data are distributed and
only takes the extreme scores into account
RANGE = (Xlargest – Xsmallest) + 1
Quartiles

Split Ordered Data into 4 Quarters
25%
25%
 Q1 



25%
 Q2 
25%
Q3 
Q1 = first quartile
Q2 = second quartile= Median
Q3 = third quartile
Quartiles
25%
75%
Md
Q1
Q3
Variability: Interquartile Range

Difference between third & first quartiles

Interquartile Range = Q3 - Q1

Spread in middle 50%

Not affected by extreme values
Standard Deviation and Variance

How much do scores deviate from the mean?

deviation =
X 
X
X-
1
0
6
1
=2


(X -  ) 
Why not just add these all up and take the mean?
Standard Deviation and Variance

Solve the problem by squaring the deviations!
X-
(X-)2
1
-1
1
0
-2
4
6
+4
16
1
-1
1
X
=2
Variance =

2
( X  u)


N
2
Standard Deviation and Variance



Higher value means greater variability around 
Critical for inferential statistics!
But, not as useful as a purely descriptive statistic


hard to interpret “squared” scores!
Solution  un-square the variance!
Standard Deviation =

 ( X  u)
N
2
Variability: Standard Deviation







The Standard Deviation tells us approximately
how far the scores vary from the mean on average
estimate of average deviation/distance from 
small value means scores clustered close to 
large value means scores spread farther from 
Overall, most common and important measure
extremely useful as a descriptive statistic
extremely useful in inferential statistics
The typical deviation in a given distribution
Variability: Standard Deviation

Standard Deviation can be calculated with
the sum of squares (SS) divided by n
 
 ( X  )
N
2
SS

N
Sample variance and standard deviation



Sample will tend to
have less variability
than popl’n
if we use the
population fomula, our
sample statistic will be
biased
will tend to
underestimate popl’n
variance
Sample variance and standard deviation

Correct for problem by adjusting formula
s





2
(X  M )


2
n 1
Different symbol:
s2 vs. 2
Different denominator: n-1 vs. N
n-1 = “degrees of freedom”
Everything else is the same
Interpretation is the same
Definitional Formula:
s
Variance:




2
(X  X )


2
n 1

ss

n 1
SS
df
deviation
squared-deviation
‘Sum of Squares’ = SS
degrees of freedom
Standard
Deviation:
s 
(X  X )
n 1
2

ss

n 1
SS
df
Variability: Standard Deviation



let X = [3, 4, 5 ,6, 7]
M=5
(X - M) = [-2, -1, 0, 1, 2]
s
 subtract M from each number in X

(X - M)2 = [4, 1, 0, 1, 4]
 squared deviations from the mean

 (X - M)2 = 10
 sum of squared deviations from the mean (SS)

 (X - M)2 /n-1 = 10/5 = 2.5
 average squared deviation from the mean

 (X - M)2 /n-1 =
2.5 = 1.58
 square root of averaged squared deviation
2
(
X

X
)

n 1
Variability: Standard Deviation



let X = [1, 3, 5, 7, 9]
M=5
(X - M) = [-4, -2, 0, 2, 4 ]
s
 subtract M from each number in X

(X - M)2 = [16, 4, 0, 4, 16]
 squared deviations from the mean

 (X - M)2 = 40
 sum of squared deviations from the mean (SS)

 (X - M)2 /n-1 = 40/4 = 10
 average squared deviation from the mean

 (X - M)2 /n-1 = 10 = 3.16
 square root of averaged squared deviation
2
(
X

X
)

n 1
In class example

Work on handout
Standard Deviation & Standard Scores

Z scores are expressed in the following way
Z

X 

Z scores express how far a particular score is
from the mean in units of standard deviation
Standard Deviation & Standard Scores

Z scores provide a common scale to express
deviations from a group mean
Z
X 

X  Z  
Standard Deviation and Standard Scores

Let’s say someone has an IQ of 145 and is 52
inches tall




IQ in a population has a mean of 100 and a
standard deviation of 15
Height in a population has a mean of 64” with a
standard deviation of 4
How many standard deviations is this person
away from the average IQ?
How many standard deviations is this person
away from the average height?
Homework

Chapter 4

8, 9, 11, 12, 16, 17
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