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Measures of dispersion PDF / PPT

Measures of dispersion

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Dispersion

» Definition: “Dispersion (also called variability,
scatter or spread) is the extent to which a
distribution is stretched or squeezed.”

» Common examples of measures of statistical
dispersion are the variance, standard deviation
and interquartile range.

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Types of Measures of Dispersion

There are two main types of dispersion methods in statistics which
are:

«Absolute Measure of Dispersion
Relative Measure of Dispersion

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Common measures of Dispersion
Absolute dispersion

» An absolute measure of dispersion contains the same unit as the original
data set. Absolute dispersion method expresses the variations in terms of
the average of deviations of observations like standard or means
deviations. It includes range, standard deviation, quartile deviation, etc

Range

Inter-quartile range (IQR)
Variance

Standard deviation

Mean deviation

Standard error of the mean
Lorenz Curve

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Relative Measure of Dispersion

» The relative measures of depression are used to compare the distribution of
two or more data sets. This measure compares values without units.
Common relative dispersion methods include:

» Co-efficient of Range

» Co-efficient of Variation

» Co-efficient of Standard Deviation
» Co-efficient of Quartile Deviation

» Co-efficient of Mean Deviation

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The common coefficients of dispersion are:

C.D. In Terms of Coefficient of dispersion

Range C.D. = (Xmax — Xmin) ¥ (Xmax
+ Xmin)

Quartile Deviation C.D.=(Q3 —-Q1) 7 (Q3 +
Q1)

Standard Deviation C.D. =S.D. ¥ Mean

(S.D.)

Mean Deviation C.D. = Mean

deviation/Average

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Range

®» [he range of a data set is the difference between the largest
and smallest data values.

» Example:
®» Range of the sample:53, 55, 70, 58, 64, 57, 53, 69, 57, 68, 53

»/ Where, R= Range H = Highest value in the observation= 70
L = Lowest value in the observation= 53
®» Range, R=H-L= 70-53= 7

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RANGE

Example: Calculate the range and coefficient of range of the given dataset —

13, 20,7, 15, 29, 35
Solution:
R=L-S
=35-7=28
Coefficient of range =L-S x 100
L+S
= (35-7/35+7)x 100 = 66.66%

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Example:

Find out the Range and coefficient of range of following data.
59,46,30,23,27,40,52,35,29

ans. 36, 0.44

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Find range of coefficient

Calculate the coefficient of range for the following data:

Heights 150-124 125-129 130-134 135-139 140-144

IN cm.

No, of

6 9 10 12 8

students
Answer
First let us find the limits of heights (limits of class intervals) and their mid points.
Height Height limits Midpoint of heightsNo. of students

120 —124119.5 — 124.5122 i}
125 — 129124.5 — 129.5127 . 9
130 — 134129.5 — 134.5132 10
135 — 139134.5 — 139.5137 12
140 — 144139.5 — 144.5142 8

Upper limit of the highest height, H — 144.5, lower limit of the lowest height, L = 119.5.
H-L 144.5—119.5 25
H+L 1445+ 119.5 264

Coefficient of range = = 0.095.

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Advantages & disadvantages of the Range

» Advantages:

®» FqQsy to compute.

» Easy to understand.
» Scores exist in the data set.

=» Disadvantages:

» Value depends only on two scores.

» |nfluenced by sample size.

» Very sensitive to outliers.

® |nsensitive to the distribution of scores within the two extremes.
» [For example: 1,2,2,3,4,6,7 vs. 1,1,1,1,1,1,7 both have Range, R=6

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Inter-quartile Range (IQR)

The inter-quartile range of a data set is the difference between the

third quartile(75%) and the first quartile (25%).

It is the range for the middle 50% of the data.

It overcomes the sensitivity to extreme data values.
IQR= Q3 – Q1

Where, Q1 = First quartile Q3 = Third quartile

Semi-inter-quartile range is also known as quartile deviation,

QD =IQR/2=Q3 – Q1 /2

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smallest
value
(minimum)

median
SO0th

25th percentile 75th
percentile percentile

First Quartile Third Quartile
(Q,) (Qs)

Nr pr—

Interguartile Range = Q,- Q,

largest
value
(maximum)

100th
percentile

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Advantages of IQR
It is not affected by extreme values as in the case of range.

It is useful in estimating dispersion in grouped data with open ended
class.

Disadvantages of IQR

IQR as a measure of dispersion is most reliable only with symmetrical
data series. Unfortunately, in social sciences most of data
distributions are generally asymmetrical in nature. So, its use in social
sciences is usually limited to data which are moderately skewed.

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MEAN DEVIATION

It is the average of the absolute values of the
deviation from the mean (or median or mode).

Mean deviation or MD or § — > | =N

where, MD= mean deviation;
x = deviation from actual mean
| | = not considering sign (+ve or -ve )
Deviation, x = X — X

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= MERITS AND DEMERITS OF MEAN DEVIATION

®» Mean deviation is easy to calculate but since mean
deviation has less mathematical value , it is rarely
applied for biological statistical analysis. It is also not

meaningful because negative sign of deviations Is
ignored.

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STANDARD DEVIATION

® [t may be defined as “the square root of the airthmetic

mean of the squares of deviations from the airthmetic
mean.”

» [Sample Standard Deviation, s=Vs2

» [Population Standard Deviation, o= Vo?

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Variance

Ivy
Thus, we can take “Ll =)” asa quantity which leads to proper measure

of dispersion. This number, 1.¢., mean of the squares of the deviations from mean 1s

called the variance and 1s denoted by – (read as sigma square). Therefore, the

variance

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15.5.1 Standard Deviation In the calculation of variance, we find that the units of
individual observations x and the unit of their mean y are different from that of variance,

since variance involves the sum of squares of (xy). For this reason, the proper

measure of dispersion about the mean of a set of observations is expressed as positive
square-root of the variance and is called standard deviation. Therefore, the standard

deviation, usually denoted by ¢ , is given by

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i = Khushs 00
TT = =X = (Goo +23 FO + THO + Z 30¢ 300)
= y r—— rm

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“Bale Loo 206 LH4A HE
i \fisha —n AF0 isi ~~ Fe EEE
: 1 130 226
a 4 %0 XY – 123¢
Klaas 300 b= 94 6&3
p + X y . = (x—21) = 1
Li: 394 10420
Re I shh wd BOF, (206°
490 — 234 = 36 (F6)S_
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T= d= Cnn =
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—— — —————— ———,

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ls i. sis TS EVR
| ef rodlnd oy
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i é | 40-50 5
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1s 10 N Cluny Toateunin y= 30
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sido 9 soaglls nS
a 2

Reduce ©.5 Awova Lowes Limit st aud
1A 0.5 do pps lived ot cul cluac.

Foxe lusive clus §

– 0. 5 —-1. pe |
9. SY —»19.5
1.5% = 25

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= es es
“B andogd__Jeviediov: Toweala ox Gesonped ——
=] Pode = ill ce J
er” |
—T1T Le -_1 WS &i% =H)” _
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h= width of clus intesvels i
ht = Xi —A As pmeciined aean .

hh
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sn) |

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ature d “meu,

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Formulas for Standard Deviation

Population Standard 2 Xo
Deviation Formula 0g =

Sample Standard Deviation YN (X-X)’
Formula 7

Notations for Standard Deviation

* 0 = Standard Deviation
* Xj = Terms Given in the Data
e X = Mean

e n = Total number of Terms .
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15.5.2 Standard deviation of a discrete frequency distribution Let the given discrete

frequency distribution be
XoooX, XX…

where N= y fi
i=l

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15.5.3 Standard deviation of a continuous frequency distribution The given
continuous frequency distribution can be represented as a discrete frequency distribution
by replacing each class by its mid-point. Then, the standard deviation is calculated by
the technique adopted in the case of a discrete frequency distribution.

[f there is a frequency distribution of n classes each class defined by its mid-point
x, with frequency f, the standard deviation will be obtained by the formula

0 = Ic Lis 9k \

where ¥ is the mean of the distribution and N = y fi.

i=l

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Standard Deviation Formula Based on
Discrete Frequency Distribution

For discrete frequency distribution of the type:

X: X1, X2, X3, … Xp and

f: fq, fs, fs, —- fn

The formula for standard deviation becomes:

0 =| % Ti file — 2)”

Here, N is given as:

N=”%1f;

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Standard Deviation Formula for Grouped
Data

There is another standard deviation formula which is
derived from the variance. This formula is given as:

0 = [Si fiat — (T fi)

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Another formula for standard deviation :

2
pa NY f= Yoh
i=l i=l

xX, —A
h

where fi is the width of class intervals and y, =

mean.

and A is the assumed

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MERITS AND DEMERITS OF STD.
DEVIATION

Std. Dev. summarizes the deviation of a large distribution from mean in one
figure used as a unit of variation.

It indicates whether the variation of difference of a individual from the
mean is real or by chance.

Std. Dev. helps in finding the suitable size of sample for valid conclusions.
It helps in calculating the Standard error.
DEMERITS-

It gives weightage to only extreme values. The process of squaring
deviations and then taking square root involves lengthy calculations.

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CALCULATION OF STANDARD DEVIATION- DISCERETE SERIES OR
GROUPED DATA

Three Methods ©. | | |
a) Actual Mean Method or Direct Method

b) Assumed Mean Method or Short-cut Method :

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Example : During a survey, 6 students were asked how many hours per day
they study on an average? Their answers were as follows: 2, 6, 5, 3, 2, 3.
Evaluate the sample standard deviation.

Solution:

Find the mean of the data:

(1615131248) _ 5 o
Step 2: Construct the table:
Xq Xp -X (x1 – %)?
2 -1.5 2.25
4] 2.5 6.25
EH] 1.5 2.25
3 -0.5 0.25
2 -1.5 2.25
3 -0.5 0.25
=13.5

Step 3: Mow, use the Standard Deviation formula

Sox — xX)

Te— 1

Sample Standard Deviation = 5 —

=-(12.5,16-11)
=-/[2.71

=1.643 .
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Example : Find the mean respiration rate per minute and its standard deviation when in 4
cases the rate was found to be : 16, 13, 17 and 22.

Standard deviation = o=|

* Solution:

(xi — 0)? _ i az

n mn

_ ¥x 16+13+17+22 68
Here Mean=x = = = =17
n 4 4
(x) d=x—Xx d®=(x-Xx)*
16 1 1
13 a 16
17 0 0
22 5 25
Dx = 68 > d? = 42

Standard deviation = = [RED – JF a_ iE =3.2
mn mn 4

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Example 11 Find the standard deviation for the following data :

cl 38 3823
fl 7 [0] of 6
E__
oe wc. TN Eon if =hixi)-
Ne = = _]
= 3 £: Fix LEDEY —]
3 IE 5 21 g3 —
< 10 <€o 84H0 p=
13 15 195 1253S =(=%ix1) —
1€ 10 15 3240 | = (e14)~
23 06 13% 31F4L | = 23694
|ZF=4% |Zhxi= 614 phx 236521
on: [4g (9652) — AF6996 |
YY
= 1 [ <6m00 _
21g a) |
= 2935
44% _

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Solution : In the given data.
fiz = 21, 80, 195, 180, 138
Fiz? = 63, 640, 2535, 3240, 3174
Now, standard deviation 1s given bw,

o = ~/N fix? — (> fime)”
Here. N = > fi = 48
>» fiw: = 614

> fiz = 9652
So, putting these values,

1 2
aT \/48(9652) — 614

1
= a = ——+/463296 — 3T6998

1 203.76
eo — — ./B6300 — ———’%
T= 4s 48

oo — 6.12

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Example 9 Find the variance and standard deviation for the following data:

i

4

8

11

17

20

24

32

i;

3

5

9

5

4

3

1

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Ladle 12.0

Yi Ji fix: x, -X (x, =x) f(x; =x)”
4 3 12 —-10 100 300
8 5 40 —6 36 180
11 9 99 —3 9 81
17 5 83 3 9 45
20 4 80 6 36 144
24 3 72 10 100 300
32 1 32 18 324 324
30 420 1374
1 7
N=30, Y fx; =420, } f(x, —%)” =1374
i=1 i=1
7
Therefore T= Lf 1 420 =14
N 30
: 1 & 5
Hence variance (og?) = ~N Lim x)
_L 1374=4538
S30 CET
and Standard deviation (¢)=+/45.8 = 6.77

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Example 10 Calculate the mean, variance and standard deviation for the following
distribution:

Class 30-40 | 40-50 © 50-60 | 60-70 | 70-80 | 80-90 | 90-100
Frequency 3 1 12 15 § 3 2

Co
Tsea | ridmint $ixi [ fix:®
fi C1 /

Ro-l40 2 25 05. | ~egx5
= iE 45 31s | 14115
so-60 | 12 55 460 % 6300
&-70 | 15 “65 915 6717315
70-40 % 15 600 46 000
Jogo | 3 35 255 | 21615
90-100 < 51-85 i90 T%06 50

1 5° : 31006 | 202.250
Tia) 5 2100) = Fico ‘ 5

oc = 1 IN TESGS — (ZhxR

| AA

i

A = 1 S50 (202350) — [email protected]¥Aco00
S50
a 902 500
– – 1313

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Class | Frequency | Mid-point| fx, | (x~%)*| f(x-x)*
(f) (x)
30-40 3 35 105 729 2187
40-50 7 45 315 280 2023
50-60 12 55 660 49 588
60-70 15 65 075 0 135
70-80 8 75 600 169 1352
80-90 3 85 255 529 1587
90-100 2 05 190 1089 2178
50 3100 10050
1 3100 i
Mean x = — = = 62
Thus N Yin = i
: a) 1 y fix, — x)?
Variance (o ) =N = i Xj
1
= —x10050 =201
50
and Standard deviation (¢) =+/201=14.18

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Examples 12 Calculate mean, variance and standard deviation for the following

distribution.
Classes 30-40 140-50] 50-60] 60-70{ 70-80] 80-90{90-100
Frequency 3 7112115] 8 3 2

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Solution Let the assumed mean A = 65. Here 2 = 10
‘We obtain the following Table 15.11 from the given data :

Table 15.11

.— 65
Class |Frequency | Mid-point | v= i 0 ¥,2 Si ov, fiv2
I; XxX;
30-40 3 35 —3 9 —9 27
40-50 ra 45 — 2 4 — 14 28
50-60 12 55 — 1 1 — 12 12
60-70 15 65 8] 8] 0] [8]
TO-80 8 75 1 1 8 8
80-90 3 85 2 4 6 12
90-100 2 a5 3 9 6 18
N=50 — 15 105
Therefore em A 2S yes 15 166
50 S50
– . h2 pe 2
Variance 2 = ~2 NY fivim — (= Ji yi )
10
_ oy? > 5 [50>105— 15) |
(50)
1
— — [5250 — 225]=201
25
and standard deviation (o)= A201 = 14.18

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=» Example.

Find the variance and standard deviation of the following data by using
assume mean method.

6,8, 10,12, 14, 16, 18, 20, 22, 24

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» Solution:

+ – = x; —14 Deviations from mean =F)
‘ 2 x—x) o
6 =i —9 81
8 3 7 49
10 a] -5 25
12 1 -3 9
14 0 ~1 1
16 1 1 1
18 2 3 9
20 3 5 25
22 4 7 49
24 5 9 381
= 330
n
ya
. _— 5
Therefore Mean X = assumed mean + = xh = 4+ x2=13
} 1 10 5 1
and Variance (og?) = ~)(x-%) = 197330 =33
i=1

Thus Standard deviation (o ) = </33 = 5.74

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Find the mean, variance and standard deviation using short-cut method

Height | 70-75] 75-80|80-85]85-90| 90-95 195-100 | 100-105{105-110] 110-115
1n cms

No. of 3 4 7 7 15 0 6 6 3
children

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1

Thank You

J)

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