Measures of Variability

OBJECTIVES•To understand the different measures of

variability•To determine the range, variance, quartile

deviation, mean deviation and standard deviation for ungrouped and grouped data

Measures of dispersion (variability or spread)

consider the extent to which the observations vary

MEASURES OF VARIATION

RANGE QUARTILE DEVIATION MEAN DEVIATION VARIANCE STANDARD DEVIATION

1. Range, R The difference in value between the

highest-valued data, H, and the lowest-valued data, L

R = H – L Example: 3, 3, 5, 6, 8

R = H – L = 8 – 3 = 5

2. Quartile Deviation, QD

or semi-interquartile range obtained by getting one half the

difference between the third and the first quartiles

2

QQD 13 Q

SOLVE FOR Q1 and Q3

w4Q1if

cfN

ll

w43

Q3if

cfN

ll

classes)simply (or interval class of sizeor width class w

class quantile theof f

class quantile thebefore

frequency cumulative than less cf

i

frequency

Problem

The examination scores of 50 students in a statistics class resulted to the following values:

Q3 = 75.43 Q1 = 54.24 Determine the value of the quartile

deviation or semi-interquartile range.

Solution

10.60 2

24.5443.75

213

QQQD

Problem Compute the value of the semi-inter quartile range or quartile deviation

The performance ratings of 100 faculty members of a certain college are presented in a frequency distribution as follows:

Class interval or Classes f <cf

71-74 3 3 75-78 10 13 79-82 13 26 1st quartile

class

83-86 18 44 87-90 25 69 91-94 19 88 3rd quartile

class

95-98 12 100

Solution (Grouped data)

19.82

413

13255.784

2

1

11

13

Q

wf

cfN

llQ

QQQD

Q

76.91

419

69755.90

43

3

33

Q

wf

cfN

llQQ

Solution cont’d…

78.42

19.8276.91

QD

QD

Substitute

3. Mean Deviation, MD

– based on all items in a distribution

For ungrouped data For grouped data

size samplen

ondistributi theofmean x

valuesindividual the x

x

frequencyf

deviationmean MD n

xfMD

n

xMD

i

ii

xx

where

4. Variance, s2

- most commonly used measure of variability

- the square of standard deviation

For ungrouped data

size samplen

score ax

mean the x

mean thefrom score a ofdeviation the

nsobservatio ofset a of variance s

sor

2

2

2

2

2

2

xxx

wherenn

xx

n

xs

i

i

Note:

The greater the variability of the observations in a data set, the greater the variance. If there is no variability of the observations, that is, if all are equal and hence, all are equal to the mean then s2 = 0

For grouped data

size samplen

ndistriutio in the valuesindividualx

mean the x

mean thefrom score a ofdeviation the

nsobservatio ofset a of variance s

sor

2

2

2

2

2

2

xxx

wheren

n

fxfx

n

fxs

i

i

5. Standard Deviation, s

- the positive square root of the variance

2ss

Problem: Find the (a) range, (b) quartile deviation, © mean deviation, (d) variance and (e) standard deviation Student Score 1 50 2 48 lowest value 3 72 4 67 5 71 6 65 7 73 highest value 8 62 9 64 10 60

(a) Range, R

R = H – L R = 73 – 48 = 25

(b) Quartile Deviation, QD Arrangement in ascending order

48 50 60 62 64 65 67 71 72 73

Using method 3 for finding Qn (ungrouped data)

Q1 is located at n/4 = 10/4 = 2.5 Q1 = (50+60)/2 = 55

Q3 is located at 3n/4 = 3(10)/4 = 7.5

Q3 = (67+71)/2 =69

QD cont’d…

72

5569

213

QD

QQQD

© Mean Deviation, MD

56.610

6.65

n

xMD i

First, solve for the meanUngrouped data

20.63

10

48506062646567717273

x

x

Data for mean deviation, MD

Score, x xi = x- xi2

73 9.8 96.04 72 8.8 77.44 71 7.8 60.84 67 3.8 14.44 65 1.8 3.24 64 0.8 0.64 62 -1.2 1.44 60 -3.2 10.24 50 -13.2 74.24 48 -15.2 231.04 TOTAL xi = 65.6 669.60

x

(d) Variance, s2

96.66106.669

2

2 n

xs i

(e) Standard Deviation, s

18.896.66

96.662

s

s

Problem: The following are marks obtained by a group of 40 students on an English examination

Classes f <cf

95-99 2 40

90-94 2 38

85f-89 4 36

80-84 6 32

75-79 5 26

70-74 4 21

65-69 5 17

60-64 2 12

55-59 2 10

50-54 4 8

45-49 1 4

40-44 2 3

35-39 1 1

Find the following:

a. range b. quartile deviation c. mean deviation d. variance e. standard deviation

Solution

a. Range, R = H – L = 99 – 35 = 64

b. Quartile Deviation, QD

83.8256

26305.79

43

Q where

2

33

13

wf

cfn

ll

QQQD

Q

Solve for Q1

50.5952

8105.54

4Q1

1

w

f

cfn

llQ

Substitute

11.67 2

59.50-82.83

213

QQQD

c. Mean Deviation, MD

9.1240

516

table theRefer to

7140

2840

n

xfMD

xxx

x

i

i

Data for mean deviation, MD

Class interval

x f fx |xi| f|xi|

95-99 97 2 194 26 52

90-94 92 2 184 21 42

85-89 87 4 348 16 64

80-84 82 6 492 11 66

75-79 77 5 385 6 30

70-74 72 4 288 1 4

65-69 67 5 335 4 20

60-64 62 2 124 9 18

55-59 57 2 114 14 28

50-54 52 4 208 19 76

45-49 47 1 47 24 24

40-44 42 2 84 29 58

35-39 37 1 37 34 34

Total 40 2840 516

d. Variance, s2

241.5 40

96602

2

n

fxs i

Data for the variance, s2

Class interval

x f fx xi fxi2

95-99 97 2 194 26 1352

90-94 92 2 184 21 882

85-89 87 4 348 16 1024

80-84 82 6 492 11 726

75-79 77 5 385 6 180

70-74 72 4 288 1 4

65-69 67 5 335 -4 80

60-64 62 2 124 -9 162

55-59 57 2 114 -14 392

50-54 52 4 208 -19 1444

45-49 47 1 47 -24 576

40-44 42 2 84 -29 1682

35-39 37 1 37 -34 1156

Total 40 9660

e. Standard Deviation, s

54.155.241

s variance, theofroot square positive

theis s, deviation, standard 2

s

The

New Topic…

Objectives

To know the measures of skewness and kurtosis

To find the Pearsonian coefficient of skewness

Measures of Skewness

summarize the extent to which the observations are symmetrically distributed

Skewness

the degree to which a distribution departs from symmetry about its mean value

or refers to asymmetry (or "tapering") in the distribution of sample data

Positive skew the right tail is longer the mass of the distribution is

concentrated on the left of the figure

has a few relatively high values the distribution is said to be right-

skewed mean > median > mode the skewness is greater than zero

Negative skew

the left tail is longer the mass of the distribution is

concentrated on the right of the figure

has a few relatively low values the distribution is said to be left-

skewed mean < median < mode the skewness is lower than zero

No skew

the distribution is symmetric like the bell-shaped normal curve

mean = median = mode

xxx ˆ~

OR…

Exercise

Pearsonian coefficient of skewness

modex̂

median x~

mean x

skewness of

t coefficien S

~3

ˆ

k

Pearsonianwheres

xxSor

s

xxS kk

Skewness based on quartiles

quartile 3Q

quartile 2Q

quartile 1Q

3

2

1

13

1223

rd

nd

st

where

QQ

QQQQSk

Interpretation

If skewness is positive, the data are positively skewed or skewed right, meaning that the right tail of the distribution is longer than the left. If skewness is negative, the data are negatively skewed or skewed left, meaning that the left tail is longer.

Interpretation cont’d…

If skewness = 0, the data are perfectly symmetrical. But a skewness of exactly zero is quite unlikely for real-world data, so how can you interpret the skewness number? In the classic Principles of Statistics (1965), M.G. Bulmer suggests this rule of thumb:

Interpretation cont’d…

If skewness is less than −1 or greater than +1, the distribution is highly skewed.

If skewness is between −1 and −½ or between +½ and +1, the distribution is moderately skewed.

Interpretation cont’d…

If skewness is between −½ and +½, the distribution is approximately symmetric.

Example: With a skewness of −0.1082, the

sample data are approximately symmetric.

Problem

Find the Pearsonian coefficient of skewness of the set of data shown in the following table:

Scores of ten students in a mathematics ability testStudent Score 1 50 2 48 3 72 4 67 5 71 6 65 7 73 8 62 9 64 10 60

Computed values

Refer to the previous computations

76.118.8

)682.63(3

)~(3

682

7165~

18.8

20.63

k

k

Ss

xxS

Mdnx

s

x

Interpretation

Negative sign means the tail extends to the left the mean is less than the mode by

176% considered a substantial departure

from symmetry

Problem

Find the Pearsonian coefficient of skewness for the following set of data:

83.82)5(12

25.79ˆ

)(ˆ

54.15

71

21

1

x

wdd

dllMox

s

x

Class

intervalx f fx |xi| f|xi|

95-99 97 2 194 26 52

90-94 92 2 184 21 42

85-89 87 4 348 16 64

80-84 82 6 492 11 66

75-79 77 5 385 6 30

70-74 72 4 288 1 4

65-69 67 5 335 4 20

60-64 62 2 124 9 18

55-59 57 2 114 14 28

50-54 52 4 208 19 76

45-49 47 1 47 24 24

40-44 42 2 84 29 58

35-39 37 1 37 34 34

Total 40 2840 516

761.054.15

83.8271ˆ

83.82)5(12

25.79ˆ

)(ˆ21

1

k

k

Ss

xxS

x

wdd

dllMox

Interpretation

Negative (-) computed value means the mean is less than the mode by

76.1% considered quite negligible departure

from symmetry given set of data is more or less

evenly distributed

Problem

Find the Pearsonian coefficient of skewness for the distribution whose

5 s deviation, standard

and 18.6 x̂ , mode

20.5 x ,

mean

Solution

38.05

6.185.20ˆ

k

k

Ss

xxS

Interpretation

Positive sign indicates the tail of the distribution extends to the

right Computed value means

the mean is greater than the mode by 38%

considered negligible skewness

Measures of Kurtosis

Kurtosis - the degree of peakedness (or flatness) of a distribution

2

1

4

444

4

sdeviation standard s

1m '

measure kurtosis edStandardiz

n

xxiwhere

s

mm

n

ii

Types of Kurtosis

Mesokurtic distribution a normal distribution, neither too

peaked nor too flat its kurtosis (Ku) is equal to 3

Leptokurtic distribution

has a higher peak than the normal distribution

with narrow humps and heavier tails

its kurtosis (Ku) is higher than 3

Platykurtic distribution

has a lower peak than a normal distribution

flat distributions with values evenly distributed about the center with broad humps and short tails

its kurtosis (Ku) is less than 3