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Statistics is the science of collection, classification, tabulation, analysis and
interpretation of numerical data. Measures of central tendency give a birds eye view
of the huge mass of statistical data. Measures of dispersion give an idea about thehomogeneity or hetrogeneity of the distribution.
You are already familiar with
* Collection of data and preparing frequency distribution table for a given set of scores.
* Computation of the measures of central tendency namely arithmetic mean, median
and mode for both ungrouped and grouped data.
* Determination of the measures of dispersion such as the range, quartile deviation
and mean deviation for given distribution.
This chapter deals with the measures of dispersion namely the standard deviation
and co-efficient of variation.
After learning this chapter you will be able to :
* Compute the measure standard deviation for both ungrouped and grouped data.
* Determine the relative measure co-efficient of variation and understand the variability.
* Interpret the data based on these statistical measures.Dispersion refers to the variability in the size of items of the distribution. The degree
to which numerical data tend to spread about an average value, is the variation or
dispersion of the data.
The measures of dispersion, which are in common use are : (i) Range (ii) Quartile
Deviation (iii) Mean Deviation and (iv) Standard Deviation.
We have already discussed about the measures of dispersion such as range, quartile
deviation and mean deviation. Besides these measures of dispersion, Standard Deviation
is an ideal measure of dispersion.
1. Standard Deviation :
Statistical data is the set of observations of characteristics of individuals as age,
height, weight, income, marks scored etc., These characteristics are generally called
variables. The values of variables may be close to the arithmetical average or scattered
away from the average. The values vary from the mean and the measure of such variation
is called the Variance of the distribution.
The square root of Variance is the Standard Deviation of the distribution.
3 STATISTICS
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2. Standard Deviation for an Ungrouped Data
Model 1 : Calculation of variance and standard Deviation in a series of individual
observations.
X be the variable given by X : x1, x
2 x
3 .... x
n
N : The number of items
X : The Arithmetic Mean
D : Deviation from the Arithmetic Mean = (X X )
The Variance of the distribution is given by Variance =2
DN
1
Symbolically Variance is denoted by22
D
N
1 = . As the Standard Deviation
is the square root of Variance of the distribution.
S.D = Variance =
S.D =N
D2
The Standard Deviation is conventionally represented by the Greekletter Sigma :
Standard Deviation is the square root of the arithmetic average of
the squares of the deviations from the mean.
Working Rule :
When individual observations are given.1) The arithmetic mean is computed.
2) The deviations of the individual scores from the arithmetic mean are obtained.
3) The squares of the deviations are calculated.
4) The sum of the squared deviations is divided by the number of observations.
5) This value is the Variance.
6) The positive square root of Variance is the standard deviation.
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Worked Examples :
1) Calculate the Variance and Standard Deviation of the following 10 scores.
14, 16, 21, 9, 16, 17, 14, 12, 11 and 20. Also interpret the results.
Table of calculation
Scores Deviation Square of
X (D = X X ) the deviation D2
14 -1 1
16 +1 1
21 +6 36
9 -6 36
16 +1 1
17 +2 4
14 -1 112 -3 9
11 -4 16
20 +5 25
X =150 2D = 130
Solution : The number of observations = N = 10
(i) Find the arithmetic mean by using
Mean =N
XX =
=10
150Arithmetic Mean = 15
The average score = 15
(ii) Find the variance by using
Variance = 2DN
1
2 =
10
1 (130) = 13 Variance = 13
(iii) Find the standard deviation by using
S.D = Variance
= 13 = 3.6 S.D. of the scores = 3.6
On an average, the individual scores deviate from the arithmetic mean by 3.6
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2) The runs scored by a batsman in eight innings are given as 35, 42, 23, 34, 39,
36, 32 and 31 Find (i) the average score and (ii) the standard deviation of scores.
How do you interpret the results?
Solution : The number of individual observations = N = 8
Table of calculation
Scores Deviation Square of
X (D = X X ) the deviation D2
35 +1 1
42 +8 64
23 11 121
34 0 0
39 +5 25
36 +2 4
32 -2 431 -3 9
X = 272 2D = 228
i) Arithmetic MeanN
XX
==
=8
272= 34 Arithmetic Mean = 34
ii) S.DN
D
2=
=8
228 = 5.28 = 5.34 S.D. = 5.34
The average of runs scored by the batsman is 34 and on an average, the individual
scores deviate from the average score by 5.34 runs.
Alternate method
This method is used when the arithmetic mean ( X ) is not an integer.
Step 1 : Assume one of the scores as an assumed mean (A)
Step 2 : Find out the deviations from the assumed mean, D = (X-A)
Step 3 : Find out the sum of the deviations, DStep 4 : Square the deviations and find out the sum of the squares of the deviations,
2D
Step 5 : By using the formula, determine the standard deviation =
22
N
D
N
D
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3) The marks obtained by 10 students in an examination are given as 43, 48, 55,
57, 42, 50, 47, 48, 58 and 50. Find (i) the Mean and (ii) the Standard deviation.
Interpret the results
Solution : The number of individual observations = N = 10. Let the assumed mean
A = 50
Table of calculationScores Deviation Square of
X (D = X X ) the deviation D2
43 -7 49
48 -2 4
55 +5 25
57 +7 49
42 -8 64
50 0 047 -3 9
48 -2 4
58 +8 64
50 0 0
X = 498 D = -2 2D = 268
i) Arithmetic MeanN
XX
==
=10
498= 49.8 Arithmetic Mean = 49.8
The average of marks scored by 10 students is 49.8
ii) S.D
22
N
D
N
D
==
=
2
10
2
10
268
= 2)2.0(8.26 = 04.08.26
= 76.26 = 5.173 S.D = 5.173
The individual scores differ from the average score by 5.173 marks.
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Model 2 : Calculation of Standard Deviation for a grouped data:
X be the variable and f be the frequency, then the distribution is given by
X x1
x2
x3
........... xn-1
xn
f f1
f2
f3
........... f n-1
fn
N = f = The sum of the frequencies
X = The Arithmetic Mean =N
fX
D = Deviation from the arithmetic mean = (X X )
Variance of the distribution is given byN
fD2
Variance =N
fD2
2 =
Standard deviation of the distribution is given by
S.D. =N
fD
2=
Working Rule
To calculate the Variance and Standard Deviation
Step 1: Determine the arithmetic mean X , by using the formulaN
XX
= or
N
fXX
=
Step 2: Calculate the values of deviations, using D = (XX ) for all the values of X.
Step 3: Square the values of D and obtain D2.
Step 4: Multiply the corresponding values of f and D2 and then determine 2fd
Step 5: CalculateN
fD2
2 = This gives the Variance of the distribution.
Step 6: Take the positive square root of the VarianceN
fD
2=
This gives the standard deviation of the distribution.
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4) Calculate (i) the arithmetic mean (ii) the standard deviation of the given distribution.
X 10 15 20 25 30 35
f 3 8 5 9 4 1
Table of calculation
Score frequency Deviation Square of f.D2
X f fX D = (X X ) deviation D2
10 3 30 -11 121 363
15 8 120 -6 36 288
20 5 100 -1 1 5
25 9 225 +4 16 144
30 4 120 +9 81 324
35 1 35 +14 196 196
f =30 f =630 f D2=1320
Solution : The sum of the frequencies = N = f = 30
i) Arithmetic Mean =N
fXX
=
=30
630 = 21
Arithmetic Mean = 21
The average score = 21
ii) S.DN
fD
2==
=30
1320
= 44
= 6.63
S.D. = 6.63
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5) Calculate (i) Arithmetic Mean (ii) Variance and (iii) Standard deviation of the following
distribution
X 2.5 3.5 4.5 5.5 6.5
f 4 3 5 10 3
Table of calculation
Score frequency Deviation Square of f.D2
X f fX D = (X X ) deviation D2
2.5 4 10.0 -2.2 4.84 19.36
3.5 3 10.5 -1.2 1.44 4.32
4.5 5 22.5 -0.2 0.04 0.20
5.5 10 55.0 +0.8 0.64 6.406.5 3 19.5 +1.8 3.24 9.72
f =25 f = 117.5 f D2=40.0
Solution : The sum of the frequencies = N = f = 25
i) Arithmetic MeanN
XX
==
=25
5.117 = 4.5
Arithmetic Mean = 4.7
ii) Variance =N
fD2
2 =
=2540 = 1.6
Variance = 1.6
iii) S.D = Variance
= 6.1 = 1.26
S.D = 1.26
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Alternate method :
Step 1: Take any one of the values of X as an assumed average (A)
Step 2: Find out the deviations from the assumed average D = (X-A)
Step 3: Multiply the deviations by their respective frequencies and obtain the sum, f D
Step 4: Square the deviations (D2)
Step 5: Multiply the squared deviations by their respective frequencies and obtain fD2
Step 6: Compute Arithmetic Mean and Standard Deviation by using the formula,
Arithmetic Mean = X = A +N
fD
Standard Deviation
22
N
fD
N
fD
==
6) Compute : (i) Arithmetic Mean and (ii) Standard Deviation for the following
distribution and interpret the results
Marks 35 40 45 50 55
Number of students 2 4 8 5 1
Solution : Choose the assumed mean A = 45
Table of calculation
Score frequency Deviation Square of f.D f.D2
X f D = (X X ) deviation D2
35 2 -10 100 -20 20040 4 -5 25 -20 100
45 8 0 0 0 0
50 5 +5 25 25 125
55 1 +10 100 10 100
N = 20 fD = -5 fD2 = 525
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i) MeanN
fDAX +==
= 45 +20
)5(
= 45 0.25 Arithmetic Mean = 44.75
Average Marks is 44.75
ii) S.D
22
N
fD
N
fD
=
2
20
5
20
525
= 0625.025.26 =
1875.26= = 5.117 S.D. = 5.117
The average marks of 20 students is 44.75. The scores of the students deviate
from the Mean score by about 5 marks.
3. Calculation of Standard Deviation for a grouped data
Standard deviation for a grouped data can also be determined as similar to
ungrouped data.
Know this :
1) The average of the class intervals are represented by their middle points.
2) The Deviations from the arithmetic mean are obtained.
3) The squares of these deviations are multiplied by the respective frequencies.
4) The total of these products is divided by the total of the frequencies and
the result is the Variance.
5) The positive square root of the Variance is the Standard Deviation of thedistribution.
7) Calculate the Arithmetic Mean and Standard Deviation for the given frequency
distribution.
Class Interval frequency
1-5 2
6-10 3
11-15 4
16-20 1
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Solution :
Step 1: Find the mid-points for each class intervals (X)
Step 2: Find the product f.X for each class interval.
Class Frequency Mid-point Deviation
Interval f X fX from Mean D2 f.D2
D = (X X )
1-5 2 3 6 -7 49 98
6-10 3 8 24 -2 4 12
11-15 4 13 52 3 9 36
16-20 1 18 18 8 64 64
N = 10 fX = 100 fD2 = 210
Step 3: Find arithmetic mean, by using the formula
Arithmetic MeanN
fxX
==
=10
100Arithmetic Mean = 10
Step 4: Find the deviation D from the arithmetic mean X , for each class interval.
i.e., D = X X
Step 5: Square the deviations and obtain D2.
Step 6: Find the product, fD2 for each class interval and find fD2
Step 7: Compute Standard Deviation by using
S.D.N
D.f
2==
10
210=
21= = 4.6 S.D. = 4.6
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8) The distribution of the ages (in years) of 20 persons in a locality, are recorded
as follows :
Age (in years) 30-34 35-39 40-44 45-49 50-54
Number of Persons 2 5 6 5 2
Calculate (i) the average age (ii) the variance and (iii) standard deviation of the distribution
of ages.
Solution : Table of calculation
Age Frequency Mid-point Deviation
(in yrs) f X fX from Mean D2 f.D2
C.I. D = (X- X )
30-34 2 32 64 -10 100 200
35-39 5 37 185 -5 25 12540-44 6 42 252 0 0 0
45-49 5 47 235 +5 25 125
50-54 2 52 104 +10 100 200
20 840 650
Observations : N = f = Sum of the frequencies = 20
fX = 840, f.D2 = 650
i) Arithmetic Mean =N
fxX
=
=20
840 = 42 Arithmetic Mean = 42
Average age (in years) is 42
ii) Variance =NfD
22 =
=20
650 = 32.5 Variance = 32.5
iii) S.D. = Variance
= 5.32 = 5.7 S.D = 5.7
Standard deviation of the distribution of ages (in years) is 5.7
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Step - Deviation Method
1) Choose one of the middle values of the class intervals (X) as an assumed
mean (A).
2) Find the step-deviations from the assumed mean as d =i
AX , where i is the
size of the class interval.3) Multiply these step-deviations by their respective frequencies and obtain the sum
of the products, f.d.
4) Square the step-deviations, (d2)
5) Multiply the squares of the step deviations by their respective frequencies and obtain
the sum of the products, f.d2
6) Compute the arithmetic mean and standard deviation, by using
Arithmetic Mean ixNfd
AX
+==
Standard Deviation ixN
fd
N
fd
22
==
9) The marks obtained by 60 students in a test are given as follows:
Marks 5-15 15-25 25-35 35-45 45-55 55-65
Number of students 8 12 20 10 7 3
Calculate (i) the arithmetic mean and (ii) standard deviation of the distribution.
Solution :
Marks Mid point Number Step d2 f.d f.d2
X of students deviation
f d=
i
AX
5-15 10 8 -2 4 -16 32
15-25 20 12 -1 1 -12 12
25-35 30 20 0 0 0 0
35-45 40 10 +1 1 10 10
45-55 50 7 +2 4 14 28
55-65 60 3 +3 9 9 27
N = 6 0 fd = 5 2fd =109
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Choose the assumed mean A = 30
Observations : Total frequency N = 60Size of the class interval i = 10
f.d = 5 and f.d2 = 109
i) Arithmetic Mean ixNfdAX
+==
10x60
530
+=
6
530+= = 30 + 0.83
= 30.83 Arithmetic Mean = 30.83
ii) S.D ixN
fd
N
fd
22
==
10x60
5
60
1092
=
( ) ( ) 10x08.0817.1 2=
10x0064.0817.1 = 10x81.1=
= 13.45 x 10 = 13.45 S.D. = 13.45
The Average of marks scored by 60 students is 30.83
The Standard Deviation of the distribution of marks is 13.45
4. Co-efficient of variation
The co-efficient of variation is a relative measure of dispersion. It is based on
the arithmetic mean and standard deviation of a frequency distribution.
The co-efficient of variation (CV) is given by C.V =
M.A
D.Sx 100
In symbols, C.V =
X
x 100
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Remember :
* Co-efficient of variation is a relative measure of dispersion.
* It is generally expressed as a percentage.
* Mean ( X ) and S.D. ( ) determine the co-efficient of variation.
* It is independent of units.
* Consistency or variability is determined by the co-efficient of variation.10) The total runs scored by two cricket players Arun and Bharath in 15 matches are
1050 and 900 with standard deviations 4.2 and 3.0 respectively. Who is better
run getter? Who is more consistent?
Solution : Number of matches played = 15
Average score of Arun =15
1050 = 70
Average score of Bharath =15
900 = 60
Player Mean S.D. C.V. = 100xX
Arun 70 4.2 0.6100x
70
2.4=
Bharath 60 3.0 0.5100x60
0.3=
The average score of Arun is greater than the average score of Bharath. HenceArun is a better run getter.
The co-efficient of variation of Bharath is less than the co-efficient of variationof Arun. Hence Bharath is more consistent.
11) The performance of 20 students in an examination in English, Mathematics andScience is given below.
Subject Mean Score S.D. Score
X
English 56 5.75
Mathematics 73 6.25
Science 62 6.0
Using this data, determine in which subject their performance is more consistent?
X
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(i) English : Given X = 56 and = 5.75
C.V 100xX
=
100x56
75.5
=
C.V. in English = 10.27
(ii) Mathematics : Given X = 73 and = 6.25
C.V 100xX
=
100x7325.6
=
C.V. in Mathematics = 8.56
(iii) Science : Given X = 62 and = 6.0
C.V 100xX
=
100x62
0.6
=
C.V. in science = 9.68
On comparison, the co-efficient of variation in mathematics is the least.
Hence the performance in Mathematics is more consistent.
12) The marks obtained by two candidates Asha and Bhanu in 5 tests are given below.
Test No. 1 2 3 4 5
Asha 58 65 58 64 55
Bhanu 66 60 60 76 68
(i) Who is a better scorer? and (ii) Who is more consistent?
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Solution :
Table of calculation
Asha Bhanu
Scores Deviation Deviation
X from Mean D2 X from Mean D2
D = (X X ) D = (X X )
58 -2 4 66 0 0
65 +5 25 60 -6 36
58 -2 4 60 -6 36
64 +4 16 76 +10 100
55 -5 25 68 +2 4
300 74 330 176
(i) Asha :
The number of tests N = 5
Arithmetic MeanN
XX
==
5
300=
= 60 Arithmetic Mean = 60
Standard DeviationN
D
2==
5
74= 5.14=
= 3.85 S.D = 3.85
Co-efficient of variation 100xX
CV
==
100x60
85.3
=
= 6.42 C.V. of Asha = 6.42
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(ii) Bhanu : The number of observations N = 5
Arithmetic MeanN
XX
==
5
330= Arithmetic Mean = 66
Standard DeviationN
D
2==
5
176= 2.35= S.D. = 5.93
Co-efficient of variation 100xX
CV
=
100x66
93.5
= C.V. of Bhanu = 8.98
The average scores of Bhanu is higher than the average score of Asha. HenceBhanu is a better scorer.
The co-efficient of variation of Asha is less than the co-efficient of variation ofBhanu. Hence Asha is more consistent.
13) Calculate the Standard Deviation and Co-efficient of variation for the followingdistribution.
Marks (X) 10 20 30 40 50
Number of students (f) 4 3 6 5 2
Solution : The sum of the frequencies = N = 20
Table of calculation
Scores frequency Deviation Square of
X f f.X. from Mean the Deviation f.D2
D = (X X ) D2
10 4 40 -19 361 1444
20 3 60 -9 81 243
30 6 180 +1 1 6
40 5 200 +11 121 605
50 2 100 +21 441 882
N = 20 fX = 580 fD2 = 3180
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i) Arithmetic MeanN
fXX
==
20
580= = 29 Arithmetic Mean = 29
ii) S.D.N
D.f
2==
20
3180=
159= = 12.61 S.D. = 12.61
iii) C.V 100xM.AD.S
=
100x29
61.12
=
29
1261= = 43.48 C.V. = 43.48
14) The following frequency distribution shows the daily wages earned by 15 workers.
Wages (in Rs.) 30-40 40-50 50-60 60-70 70-80
No. of workers 2 3 5 3 2
Find i) Arithmetic Mean ii) Standard Deviation and iii) Co-efficient of variation
Solution :
Table of calculation :
Wages frequency Mid point Deviation Square
in Rs. f X fX from Mean of the f.D2
C.I D = (X X ) deviation
D2
30-40 2 35 70 -20 400 800
40-50 3 45 135 -10 100 300
50-60 5 55 275 0 0 0
60-70 3 65 195 +10 100 300
70-80 2 75 150 +20 400 800
15 fx = 8252
fd =2200
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Observations: The sum of the frequencies = N = 15
f.X = 825 f.D2 = 2200
i) Arithmetic MeanN
XX
==
=15
825
= 55 Arithmetic Mean = 55
The average wages (in Rs) is 55
ii) S.DN
D.f
2==
15
2200
= 67.146= = 12.11 S.D = 12.11
The standard deviation of the distribution of wages (in Rs) is 12.11
iii) C.V 100xM.A
D.S
=
100x55
11.12
=
=
55
1211 = 22.02 C.V = 22.02
Exercise : 3
1) Calculate the Standard Deviation from the following set of 10 observations.
8, 9, 15, 23, 5, 11, 19, 8, 10 and 122) The scores of a batsman in 8 innings are given as 48, 40, 36, 35, 46, 42, 36
and 37
Find (i) Average score (ii) Variance and (iii) Standard Deviation of scores.
3) Calculate Mean and standard deviation for the following distribution.
X 5 15 25 35 45
f 5 8 15 16 6
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4) The marks scored by 60 students in a mathematics test are given below.
Marks (X) 10 20 30 40 50 60
No. of students (f) 8 12 20 10 7 3
Find the Variance and Standard Deviation of the marks.
5) Calculate (i) Arithmetic Mean and (ii) Standard Deviation for the following frequency
distribution.
Class Frequency
Interval
20-25 8
25-30 3
30-35 15
35-40 12
40-45 8
45-50 4
N 50
6) The daily wages of 40 workers of a factory are given in the following table.
Wages in Rs. 30-34 34-38 38-42 42-46 46-50 50-54
No. of workers 4 7 9 11 6 3
Calculate (i) Mean (ii) Variance and (iii) Standard Deviation of wages and also
interpret the findings.
7) Batsman A gets an average of 64 runs per innings with standard deviation of 18
runs, while batsman B gets an average score of 43 runs with standard deviation
of 9 runs in an equal number of innings. Discuss the efficiency and consistency of
both the batsmen.
8) In two factories A and B, located in the same industrial area, the average weekly
wages in rupees and the Standard Deviations are as follows :
Factory Average of S.D. of
wages in Rs. wages in Rs.
A 34.5 6.21
B 28.5 4.56
Determine which factory has greater variability in individual wages?
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9) The Mean and Standard Deviation of the heights and weights of 20 persons are
given below :
Characteristic X Height (in cms) 175 3.5
Weight (in Kgs) 70 2.1
In which characteristic are they more variable?
10) Calculate (i) Arithmetic Mean (ii) Standard Deviation and (iii) Co-efficient of variation
for the following frequency distribution.
Class Interval Frequency
30-35 5
35-40 10
40-45 1645-50 15
50-55 4
Total 50
11) The runs scored by two batsmen A and B in six innings are given as follows :
Batsman A 48 50 54 46 48 54
Batsman B 46 44 43 46 45 46
Determine (i) Who is a better run getter? (ii) Who is a consistent player?
12) Marks obtained in a test by X standard students of two sections A and B are
given below :
Marks No. of students No. of students
in in
Section A Section B
25-30 5 5
30-35 10 12
35-40 25 20
40-45 8 8
45-50 2 5
Determine : (i) Which sections performance is better? and (ii) Which sections performance
is more variable?