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
Interquartile 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:
» Coefficient of Range
» Coefficient of Variation
» Coefficient of Standard Deviation
» Coefficient of Quartile Deviation
» Coefficient 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=HL= 7053= 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=LS
=357=28
Coefficient of range =LS x 100
L+S
= (357/35+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 150124 125129 130134 135139 140144
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.
HL 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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Interquartile Range (IQR)
The interquartile 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
Semiinterquartile 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
squareroot of the variance and is called standard deviation. Therefore, the standard
deviation, usually denoted by ¢ , is given by
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—— — —————— ———,
— U Shodeid Devicdion st a. dissete ma
and niinuows tsequency.
ls i. sis TS EVR
 ef rodlnd oy
(C20 I EY I) fe Bizeck
 Vala } nade
i N= = I; Four ——
Diarnete Yoous —F4eque
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xX £:  o—1c
= 5 3  q020 4
& 4  2030 B
J 2  3040 3
i é  4050 5
Fae 5  50 —60 10
1s 10 N Cluny Toateunin y= 30
TL e—
 Livert ot =m clus
OxXe ol Same =
i Twddugne –
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sido 9 soaglls nS
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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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er” 
—T1T Le _1 WS &i% =H)” _
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ht = Xi —A As pmeciined aean .
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Formulas for Standard Deviation
Population Standard 2 Xo
Deviation Formula 0g =
Sample Standard Deviation YN (XX)’
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 midpoint. 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 midpoint
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 Shortcut 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 —
=(12.5,1611)
=/[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®=(xXx)*
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 3040  4050 © 5060  6070  7080  8090  90100
Frequency 3 1 12 15 § 3 2
Co
Tsea  ridmint $ixi [ fix:®
fi C1 /
Rol40 2 25 05.  ~egx5
= iE 45 31s  14115
so60  12 55 460 % 6300
&70  15 “65 915 6717315
7040 % 15 600 46 000
Jogo  3 35 255  21615
90100 < 5185 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) — q@¥Aco00
S50
a 902 500
– – 1313
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Class  Frequency  Midpoint fx,  (x~%)* f(xx)*
(f) (x)
3040 3 35 105 729 2187
4050 7 45 315 280 2023
5060 12 55 660 49 588
6070 15 65 075 0 135
7080 8 75 600 169 1352
8090 3 85 255 529 1587
90100 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 3040 14050] 5060] 6070{ 7080] 8090{90100
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  Midpoint  v= i 0 ¥,2 Si ov, fiv2
I; XxX;
3040 3 35 —3 9 —9 27
4050 ra 45 — 2 4 — 14 28
5060 12 55 — 1 1 — 12 12
6070 15 65 8] 8] 0] [8]
TO80 8 75 1 1 8 8
8090 3 85 2 4 6 12
90100 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 shortcut method
Height  7075] 75808085]8590 9095 195100  100105{105110] 110115
1n cms
No. of 3 4 7 7 15 0 6 6 3
children
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1
Thank You
J)