Dispersion Presentation - Unitedworld School of Business

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

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Measures of Dispersion

Are measures of scatter ( spread) about an

average

i.e. extent to which individual items vary

Absolute Measures – measure value in same

units – ageRelative Measures - % or coefficient of absolute

measures

1. Range

2. Inter-quartile range

3. Quartile deviation4. Mean deviation

5. Standard deviation

A. 1. Range = Xmax – X min = L-S

Xmax – X min L-S2. Coefficient of Range = ---------------- = ------

Xmax + X min L+S

Q1. Calculate range & co-efficient of range fromfollowing information

480,562,570,322,435,497,675,732,375,482,791,8

20,275

B. Quartiles

1.Inter quartile range = Q3 – Q1

2. Quartile deviation

or semi inter quartile range = ( Q3 – Q1)/2

a. In a normal distributionQ1 < Q2 < Q3

Q2 = M

b.In a symmetrical distributionQ2 + Quartile Deviation = Q3

Q2 - Quartile Deviation = Q1

Q1 = first quartile or lower quartile

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

Q3 – Q1

Coefficient of Quartile deviation = -----------Q3 + Q1

Coefficient of Quartile Deviation

Deviation by Quartiles =---------------------- x 100

Median

Calculation of Quartile deviation under continuous

series1. If inclusive class intervals , convert to exclusive

class intervals

2. Size of class intervals should be equal

throughout distribution3. L2 of first class interval should be equal to L1 of

next class interval

4. If mid values are given , it is necessary to

determine class intervals

5. If it is open end type of frequency distribution ,

coefficient of variation is suitable measure

Calculation of Quartile deviation

Q1=size of (---------) th item of the series

3(N+1)

Q3=size of ---------) th item of the series

Q2. Calculate quartile deviation & its co-efficient for the

data given below

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

Q3. Compute quartile deviation & its

coefficient for following dataX 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

ProcedureCompute cumulative frequency

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

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

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

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

N/4 - C

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

l1= lower limit of quartile class

= upper limit of quartile class

f =frequency of quartile class

c =cumulative frequency before quartile class

M = quartile position

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

l2-l1 3N

Q3= l1+ --------- ( m-c) where m =---------

3N/4 - C

Q3 =l1+ --------- (l2-l1)

Q4. Compute quartile deviation & its coefficient

for marks of 215 students

Marks 0-

Students 10 15 28 32 40 35 26 14 10 5

Soln. condition if class interval inclusive convert into

exclusive, class size equal

calculate cumulative frequencycalculate m (Q1)=N/4 &m(Q3)=3N/4 th observation

inter quartile range = (Q3-Q1)

quartile deviation = (Q3-Q1)/2

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

cf 3 9 19 34 54 78 10

130 148 162 172 178 184

Soln. calculate cumulative frequency

calculate Q1=N+1/4 &Q3=3(N+1)/4 th observation

inter quartile range = (Q3-Q1)

quartile deviation = (Q3-Q1)/2

coefficient of quartile deiation = -----------------

l2-l1 NQ1 = l1+--------------* (m-c) m= ------------

l1- lower limit of Q1 class , l2= upper limit of Q1

f = frequency of Q1 class , c= cumulative

frequency before Q1 class

l2-l1 3N

Q3 = l1+--------------* (m-c) m= -------------

f 4l1- lower limit of Q3 class , l2= upper limit of Q3 class

f = frequency of Q3 class , c= cumulative frequency

before Q3 class

Inter quartile range = (Q3-Q1)

Quartile deviation = (Q3-Q1)/2

Coefficient of quartile deiation = -----------------

Mean Deviation = sum of absolute deviations

from an average divided by total number of

Coefficient of Mean Deviation = mean

Deviation / Mean

Σ f(x-a)mod Σ f dmod

Mean deviation = ------------- = ---------------

Σ fx N

Q5A. calculate mean deviation & coefficient of mean for

the following two series

105 112 110 125 138 149 161 175 185 190

B 22 24 26 28 30 32 34 40 44 50

Standard deviation of a series is the square

root of the average of the squared deviations

from the mean ( Average – Arithmatic mean)

Standard deviation σ – positive square root of

arithmetic mean of squares of deviations Σ dx2 Σ fdx2

σ = √ (-------) = (--------)

For frequencies of a valueσ

Coefficient of Standard deviation = ------------------

average

σCoefficient of variation = --------------- x 100

average

Q5Calculate standard deviation & coefficient

of variation

X 65 67 68 68 69 71 72 72

Q6.Calculate standard deviation & coefficient of

variation

X 95 10

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

Q6.Calculate standard deviation & coefficient of

variation

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-

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

arithmetic mean of squares of deviations Σ dx Σ dx2 Σ dx

σ = √ (-------)2 = √ ------- - (------------) 2

Σ fdx Σ fdx2 Σ fdxσ = √ (-------)2 = √ ------- - (------------) 2

Σ f Σ f Σ f

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

By using standard deviation find out whichseries is more variable

1688 168.8

1690 169

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

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

144 100 49 25 9 0 4 25 49 121 526

A 158 16

177 181 1688 168.8

B 163 15

172 175 1690 169

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

dx2A 144 10

49 25 9 0 4 25 49 121 526

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

A 158 16

181 1688 168.8

B 163 15

175 1690 169

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

144 10

49 25 9 0 4 25 49 121 526

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

9 0 10

0 25 4 25 456

Σ dx Σ dx2 Σ dx 526 12 σ = √ (-------)2 = √ ------- - (------------) 2 =√ [ ----------- - ( ------)

N N 10 10

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

Σ dx Σ dx2 Σ dx

Σ dx Σ dx2 Σ dx 456 10

σ = √ (-------)2 = √ ------- - (------------) 2 =√ [ ----------- - ( ------) 2

N N 10

=√ [ 45.6- 1] =√ 44.6 = 6.7

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

Σ fd’x Σ fd’x2 Σ fd’xσ = √ (-------)2 x i = [√ ------- - (------------) 2 ]xi

Σ f Σ f Σ f i= class interval

Σ fd’x Σ fd’x2 Σ fd’x

σ = √ (-------)2 x i = [√ ------- - (------------) 2

]xiΣ f Σ f Σ f

i= class interval

class 80-

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

frequency 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

d’x=(x-52/5) 6 5 4 3 2 1 0 -1 -2 -3 -4 -5

fd’x 6 5 4 12 8 7 0 -6 -12 -9 0 -5

d’x2 36 25 16 9 4 1 0 1 4 9 16 25

fd’x2 36 25 16 36 16 7 0 6 24 27 0 25

σ = i* σA = 5* σΣ dx Σ fd’x2 Σ fd’x 218 10

σ = √ (-------)2 = √ ------- - (------------)2 =√ [ --------- - ( ------)

N N N 40 40

=√ [ 5.45- 0.25 ] =√ 5.20 = 2.32

σ = i* σA = 5* σ = 5*2.32 = 11.60

arithmetic mean of squares of deviations

Properties

Standard deviation σ is independent of change of

origin but not of scaleIf dx = x-A σx=σd

If d’x =------ σx= i.σdi

arithmetic mean of squares of deviations

N1σ12 + N2σ2

2 + N3σ32 + …….Nnσn

σ12...n = √ --------------------------------------------

N1 + N2+ N3 +……….. Nn

compute coefficient of variation & comment which factory

profits are more consistent

Particular

Factory A Factory B

Average

profits

19.7 21

Standarddeviation

6.5 8.64

Campus Overview

907/A Uvarshad,

Gandhinagar

Highway, Ahmedabad – 382422.

Ahmedabad Kolkata

Infinity Benchmark,

10th Floor, Plot G1,

Block EP & GP,

Sector V, Salt-Lake,

Kolkata – 700091.

Mumbai

Goldline Business Centre

Linkway Estate,

Next to Chincholi Fire

Brigade, Malad (West),

Mumbai – 400 064.

Thank You

Dispersion Presentation - Unitedworld School of Business

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