Measures of Variability

Why to Measure of Variability?

Information given by various measures of central tendency is too limited; and

A statistical value that indicates the degree or extent to which the observations in a set spread around the central tendency.

Set A: 80, 82, 86, 89 and 93

86xSet B: 83, 85, 86, 87 and 89

Set C: 84, 85, 86, 87 and 88 86x

86x

The spread of the scores in each sets are different

Set A has the most spread of score, while Set C has the least

Computing for the degree of dispersion of the scores from the average will describe a set of distribution adequately

Sets of Data

Range

Easiest and simplest to determine, depends pair of extreme values

Unstable, easily fluctuates with the change of highest and lowest score

Most unreliable measure, does not give the dispersion or spread of the scores in between two extreme values

Characteristics

Range

The difference between the highest and th lowest score

Difference between the exact lower limit of the lowest score and the exact upper limit of the highest score

Exclusive Range

Inclusive Range

Difference between the highest and the lowest score

Interquartile Range (IR) & Semi-Interquartile Range or Quartile Deviation (QD)

distance from the first quartile to the third quartile

one half the distance from the first quartile to the third quartile

IR = Q3 – Q1

2.. 13 QQ

DQ

Illustrative Example:

Daily Allowances

No. Of Students

P80 – P8990 – 99

100 – 109110 – 119120 – 129130 – 139140 - 149

591012302014

Below is a distribution of the daily allowances of 100 freshman college students.

Table 1

Range

IR = Q3 – Q1

= 149.5 – 79.5

= P705

1424366686

100

= 134 – 110.33

= P 23.67

<cf

1

1

1

41

.1 QQ

B

Q Cf

CfnLQ

1012

2425 109.5

33.110P

3

3

3

43

.3 QQ

B

Q Cf

CfnLQ

1020

6675 129.5

00.134P

213 QQ

QD

2

33.110134

84.11P

Mean or Average Deviation

measures the average deviation of the values from the arithmetic mean, ignoring the algebraic sign of each deviation.

Ungrouped Data Grouped Data

n

xxfMD

k

iii

1

n

xxMD

n

ii

1

Where: xi = class mark fi = class frequency n = total frequency

The life expectancies (in months) of a hypothetical species of birds in captivity are : 20, 24, 32, 36, 40 and 46. Solve for the mean deviation.

Illustrative Example MD Ungrouped Data

xi

202432364046

xxi

n

xxMD

n

ii

1

Solution:

6

46

= 7.67mos.

139137

13

46198

n

xx

n

ii

1

6

198

= 33mos.

Find the mean deviation of the data in Table Y

Daily Allowances

fi

P 80 – P 8990 – 99

100 – 109110 – 119120 – 129130 – 139140 – 149

59

1012302014

Illustrative Example MD Grouped Data

ix xxi xxf ii

Table Y

1, 356.8

n

xxfMD

k

iii

1

100

0.360,1

60.13P

84.594.5

104.5114.5124.5134.5144.5

36.926.916.9

6.93.1

13.123.1

184.5242.1169.0

82.893.0

262.0323.4

Variance and Standard Deviation

similar to mean deviation since it is computed based on the difference of each value form the mean, with two expectations – the deviations are squared and the average of the deviations is found using n-1 as divisor instead of n. variance will be denoted by s2.

Formula Variance Ungrouped Data

1

2

12

n

xxs

n

ii

Variance

Standard Deviation

1

1

2

n

xxs

n

ii

Formula Variance Ungrouped Data

N

xn

ii

1

2

Sample mean is given Population mean is given

Square root of the variance is more frequently used of dispersion or variability.

Solve for the variance and the standard deviation of the life expectancy of six birds in captivity. The mean is 33 months.

202432364046

Illustrative Example:

ix xxi

478

2xxi 1

2

12

n

xxs

n

ii

5

4782 s

6.952 s

6.95s

moss 78.9

16981

19

49169

-13-9-137

13

Variance and Standard Deviation of Grouped Data

)1(1

2

1

2

nn

xfxfn

s

k

i

k

iiiii

1

1

2

2

n

xxfs

k

iii

)1(1

2

1

2

2

nn

xfxfn

s

k

i

k

iiiii

1

1

2

n

xxfs

k

iii

Where: xi = class mark fi = class frequency

n = total frequency

Compute the variance and the standard deviation of the data in Table Y. The mean is P 121.40.

Daily Allowance

6808.056512.492856.10571.32288.30

3432.207471.54

ix xxi if

Table Y

27939.00 100

1361.61723.61285.61

47.619.61

171.61533.61

-36.9-26.9-16.9-6.93.1

13.123.1

84.594.5

104.5114.5124.5134.5144.5

59

1012302014

P 80 – P 89 90 – 99100 – 109110 – 119120 – 129130 – 139140 – 149

2xxi 2xxf ii

Illustrative example:

Solution

1

1

2

2

n

XXifs

k

ii

99

27939

21.282s

s = P16.80

The Coding Formula for Variance and for Standard Deviation

ClassP80 – P8990 – 99100 – 109

110 – 119120 – 129130 – 139140 – 149

59

1012302014

2iiUfiiUfiUif

2

2

11

2

2

)1(c

nn

UfUfn

s

k

iii

k

iii

cnn

UfUfn

s

k

iii

k

iii

)1(

2

11

2

-3-2-1

-15-18-10

0304042

4536100

3080

126

123

0

100 69 327

2

2

11

2

2

)1(c

nn

UfUfn

s

k

iii

k

iii

22

2 10)99(100

)69()327(100

s

21.2822 Ps

21.282s

s = P 16.80

Relative Dispersion

amount of variability relative to an average

100.. X

SVC

10013

13

QQ

QQCQD

A. Coefficient of Variation expresses the standard deviation as a percentage of the mean.

B. Coefficient of Quartile Deviation makes use of the first and third quartiles.

100.. X

SVC

10013

13

QQ

QQCQD

Illustrative examples: For the data in Table Y

10040.121

80.16

%84.13

10033.110134

33.110134

%69.9

Standard Score

A student scored 43 in a long test in Chemistry wherein

the mean and the standard deviation of the scores were

38 and 6, respectively. In a long test in Physics, the

student scored 45. the mean and standard deviation in

Physics were respectively 42 and 10. in which long test

was the student’s relative standing higher?

s

xxZ

Illustrative example:

The Z scores of the student are:

6

3843 Z

10

4245 Z

In Chemistry,

In Physics,

The student’s relative performance was higher in

Chemistry than in Physics.

Solution:

83.0

30.0

SKEWNESS

curve trails off to the left, then the distribution is negatively skewed

s

xxSk

)~(3

Sk > 0 (skewed to the right)

Sk = 0 (symmetric distribution)

Sk < 0 (skewed to

the left)

frequency curve

curve trails off to the right, then the distribution is positively skewed

Illustrative Example

s

xxSk

)~(3

For a distribution whose mean is 50.5, median is 51.6 and the standard deviation is 4.2, the Pearsonian coefficient of skewness is:

Skewness

42

)6.515.50(3

79.0

Solution:

KURTOSIS

Symmetrical curves vary in shape because they do not have the same peakedness

41

4)(

ns

xxfKur

k

iii

for ungrouped data

41

4)(

ns

xxKur

k

ii

for grouped data

k > 3, the curve

is leptokurtic

k < 3, the curve is platykurtic

k = 3, the curve

is mesokurtic

Class fi

18 – 2627 – 3536 - 4445 – 5354 – 6263 – 7172 – 8081 – 8990 - 98

2356

10111283

4)( xxf ii

41

4)(

ns

xxKur

k

ii

xxi

2)4.329(60

137,797,15

43.2

Table XYZ

60

7.63x-41.7-32.7-23.7-14.7

5.73.3

12.321.330.3

6,047,4773,430,1431,577,478

280,16910,556

1,305247,664

1,646,6772,528,668

15,797,137

Illustrative Example

223140495867768594

xi