Measures of Dispersion - UnitedWorld School of Business Presentation - Unitedworld School of...

7/28/2019 Measures of Dispersion - UnitedWorld School of Business Presentation - Unitedworld School of Business

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Measures of DispersionAre measures of scatter ( spread) about anaveragei.e. extent to which individual items vary

Measures of DispersionAbsolute Measures measure value in same units ageRelative Measures - % or coefficient of absolutemeasures


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Measures of Dispersion

1. Range2. Inter-quartile range3. Quartile deviation4. Mean deviation5. Standard deviation


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Measures of Dispersion

Q1. Calculate range & co-efficient of range fromfollowing information480,562,570,322,435,497,675,732,375,482,791,820,275


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B. Quartiles1.Inter quartile range = Q3 Q12. Quartile dviationor semi inter quartile range = ( Q3 Q1)/2

a. In a normal distributionQ1 < Q2 < Q3Q2 = Mb. In a symmetrical distribution

Q2 + Quartile Dviation = Q3Q2 - Quartile Dviation = Q1


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Q1 = first quartile or lower quartile

Q2 = second / middle Quartile or medianQ3 = third quartile or upper quartile

Q3 Q1Coefficient of Quartile deviation = -----------

Q3 + Q1

Coefficient of Quartile DeviationDeviation by Quartiles =---------------------- x 100

Median


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Calculation of Quartile deviation undercontinuous series

1. If inclusive class intervals , convert toexclusive class intervals2. Size of class intervals should be equal

throughout distribution

3. L2 of first class interval should be equal toL1 of next class interval

4. If mid values are given , it is necessary todetermine class intervals

5. If it is open end type of frequencydistribution , coefficient of variation issuitable measure


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Calculation of Quartile deviationN+1Q1=size of (---------) th item of the series

4

3(N+1)Q3=size of ---------) th item of the series

4


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Q2. Calculate quartile deviation & its co-efficientfor the data given below

168

147 150 169 170 154 156 171 162 159 174 173 166 164 172


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Q3. Compute quartile deviation & its coefficientfor following data

X 10 12 14 16 18 20 22 24 28 30 34 36 38

F 3 6 10 15 20 24 30 22 18 14 10 6 6

Soln. calculate cumulative frequencycalculate Q1=N+1/4 &Q3=3(N+1)/4 th observation


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Procedure:Compute cumulative frequency

N

Find out Q1 & Q3 classes by m (Q1)=---------4

3N

& m(Q3)= ---------4


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After locating l1, l2 , f & c substitute values in

l2-l1 NQ1= l1+ --------- ( m-c) where m =---------f 4

N/4 - C

Q1 =l1+ --------- (l2-l1)f

l1= lower limit of quartile classl2 = upper limit of quartile class

f =frequency of quartile classc =cumulative frequency before quartile classM = quartile position


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After locating l1, l2 , f & c substitute values inl2-l1 3N

Q3= l1

+ --------- ( m-c) where m =---------f 4

3N/4 - CQ3 =l1+ --------- (l2-l1)

f


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Q4. Compute quartile deviation & its s

coefficient for marks of 215 student

Marks 0-

10

10-

20

20-

30

30-

40

40-

50

50-

60

60-

70

70-

80

80-

90

90-

100

Students 10 15 28 32 40 35 26 14 10 5

Soln. condition if class interval inclusive convert intoexclusive, class size equal

calculate cumulative frequencycalculate m (Q1)=N/4 &m(Q3)=3N/4 th observationinter quartile range = (Q3-Q1)quartile deviation = (Q3-Q1)/2


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X 10 12 14 16 18 20 22 24 28 30 34 36 38

F 3 6 10 15 20 24 30 22 18 14 10 6 6cf 3 9 19 34 54 78 10

8

130 148 162 172 178 184

Soln. calculate cumulative frequencycalculate Q1=N+1/4 &Q3=3(N+1)/4 th observation

inter quartile range = (Q3-Q1)

quartile deviation = (Q3-Q1)/2Q3-Q1

coefficient of quartile deiation = -----------------Q3+Q1


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l2-l1 N

Q1 = l1+--------------* (m-c) m= ------------m 4

l1- lower limit of Q1 class , l2= upper limit of Q1 classf = frequency of Q1 class , c= cumulative frequencybefore Q1 class


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l2-l1 3NQ3 = l1+--------------* (m-c) m= -------------

f 4l1- lower limit of Q3 class , l2= upper limit of Q3 classf = frequency of Q3 class , c= cumulative frequencybefore Q3 class


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Inter quartile range = (Q3-Q1)Quartile deviation = (Q3-Q1)/2

Q3-q1Coefficient of quartile deiation = -----------------

q3+q1


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Mean Deviation = sum of absolute deviationsfrom an average divided by total number of

itemsCoefficient of Mean Deviation = meanDeviation / Mean


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f(x-a)mod f dmod

Mean deviation = ------------- = --------------- fx N


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Q5A. calculate mean deviation & coefficient of

mean for the following two series

A

105 112 110 125 138 149 161 175 185 190

B 22 24 26 28 30 32 34 40 44 50


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Standard deviation of a series is the squareroot of the average of the squared deviationsfrom the mean ( Average Arithmatic mean)


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Standard deviation positive square root ofarithmetic mean of squares of deviations

dx2 fdx2 = (-------)= (--------)

N N

For frequencies of a value Coefficient of Standard deviation = ------------------

average

Coefficient of variation = --------------- x 100

average


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Q5Calculate standard deviation & coefficient ofvariation

X 65 67 68 68 69 71 72 72


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Q6.Calculate standard deviation & coefficient ofvariation

X 95 10

0

10

5

11

5

12

5

13

0

13

5

14

0

15

0

16

0

170

f 5 8 12 15 35 40 30 20 10 10 10


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Q6.Calculate standard deviation & coefficient ofvariation

X 95 100 105 115 125 130 135 140 150 160 170

f 5 8 12 15 35 40 30 20 10 10 10

dx=(x-

130)

-35 -30 -25 -15 -05 0 5 10 20 30 40


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Standard deviation positive square root ofarithmetic mean of squares of deviations dx dx2 dx

= (-------)2 = ------- - (------------) 2

N N N

fdx fdx2 fdx = (-------)2 = ------- - (------------) 2

f f f


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A 158 160 163 165 167 170 172 175 177 181B 163 158 167 170 160 180 170 175 172 175

By using standard deviation find out which seriesis more variable


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A 15

8

16

0

16

3

16

5

16

7

17

0

17

2

17

5

17

7

18

1

1688 168.8

B 16

3

15

8

16

7

17

0

16

0

18

0

17

0

17

5

17

2

17

5

1690 169

dxA -12 -10 -7 -5 -3 0 2 5 7 11 -12 (dxA)2


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A 15

8

16

0

16

3

16

5

16

7

17

0

17

2

17

5

17

7

18

1

1688 168.8

B 16

3

15

8

16

7

17

0

16

0

18

0

17

0

17

5

17

2

17

5

1690 169

dxA -12 -10 -7 -5 -3 0 2 5 7 11 -12 (dxA)2

dx2

A

14

4

10

0

49 25 9 0 4 25 49 12

1

526


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A 158 160 163 165 167 170 172 175 177 181 1688 168.8

B 163 158 167 170 160 180 170 175 172 175 1690 169

dxA -12 -10 -7 -5 -3 0 2 5 7 11 -12 (dxA)2

dx2A 144 100 49 25 9 0 4 25 49 121 526

dxB -7 -12 -3 0 -10 10 0 5 2 5 10 (dxB)2


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A 158 16

0

16

3

16

5

16

7

17

0

17

2

17

5

17

7

181 1688 168.8

B 163 15

8

16

7

17

0

16

0

18

0

17

0

17

5

17

2

175 1690 169

dx

A

-12 -10 -7 -5 -3 0 2 5 7 11 -12 (dxA)2

dx2

A

144 10

0

49 25 9 0 4 25 49 121 526

dx

B

-7 -12 -3 0 -10 10 0 5 2 5 10 (dxB)2

dx2

B

49 14

4

9 0 10

0

10

0

0 25 4 25 456


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dx dx2 dx 526 12 = (-------)2 = ------- - (------------) 2 = [ ----------- - ( ------)2

N N 1010

= [ 52.6- 1.2*1.2] = 52.6-1.44 = 51.46 =7.2Coefficient of variation = / x bar = 7.2*100/ 166.8 = 4.26%


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dx dx2 dx dx dx2 dx 456 10

= (-------)2 = ------- - (------------) 2 = [ ----------- - ( ------) 2

N N 1010

= [ 45.6- 1] = 44.6 = 6.7

Coefficient of variation = / x bar = 6.7*100 / 169 = 3.96%


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fdx fdx2 fdx

= (-------)2 x i = [ ------- - (------------) 2 ]xi f f f

i= class interval


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fdx fdx2 fdx

= (-------)2 x i = [ ------- - (------------) 2 ]xi f f f

i= class interval


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class 80-

84

75-

79

70-

74

65-

69

60-

64

55-

59

50-

54

45-

49

40-

44

35-

39

30-

34

25-

29

Mid v 82 77 72 67 62 57 52 47 42 37 32 27

frequenc

y

1 1 1 4 4 7 6 6 6 3 0 1

dx=x-52 30 25 20 15 10 5 0 -5 -10 -15 -20 -25

dx=(x-52/5)

6 5 4 3 2 1 0 -1 -2 -3 -4 -5

fdx 6 5 4 12 8 7 0 -6 -12 -9 0 -5dx2 36 25 16 9 4 1 0 1 4 9 16 25

fdx2 36 25 16 36 16 7 0 6 24 27 0 25

= i* A = 5* dx fdx2 fdx 218 10 = (-------)2 = ------- - (------------) 2 = [ --------- - ( ------) 2

N N N 40 40= [ 5.45- 0.25 ]= 5.20 = 2.32

= i* A = 5* = 5*2.32 = 11.60


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PropertiesStandard deviation is independent of change oforigin but not of scaleIf dx = x-A

x

=dx-A

If dx =------ x= i.di


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N112 + N222 + N332+ .Nnn212...n= --------------------------------------------

N1 + N2+ N3+.. Nn


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compute coefficient of variation & comment whichfactory profits are more consistent

Particular

s

Factory A Factory B

Averageprofits

19.7 21

Standard

deviation

6.5 8.64


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907/A Uvarshad,GandhinagarHighway,Ahmedabad 382422.

Ahmedabad Kolkata

Infinity Benchmark,10th Floor, Plot G1,Block EP & GP,Sector V, Salt-Lake,Kolkata 700091.

Mumbai

Goldline Business CentreLinkway Estate,Next to Chincholi FireBrigade, Malad (West),Mumbai 400 064.


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