UNIT –5 MEASURES OF DISPERSION Measures of central tendencies indicates the central Tendency of a frequency distribution in the form of an average. An average is a single significant value which is used to describe a distribution. These averages tell something about the general level of magnitude of the distribution but they fail to show anything further about the distribution. An average does not tell the full story and it is hardly fully representative of a mass, unless we know the manner in which the individual items scatter around it. A further description of the series is necessary to gauge the worthness of average. Thus, it may be such that in several series the average may be the same, but variables may highly differ in magnitudes and therefore, the central tendency calculated from such variables may not be the most typical representative in many cases. To know the extent of spread about these averages or the variations of items, it is necessary to observe the following examples. Example1 A series B series 100 50 110 150 120 100 130 100 140 200 600 600 120 120 Example – 2 A series B series x d = (x - ) x d = (x - ) 100 -20 300 -20 110 -10 310 -10 120 0 320 0 130 10 330 10 140 20 340 20 600 1600 120 320 In the example-1 though the formation of series is different, but their averages are same. However, in the example-2 though their means are different, the deviations of individual items from ‘x’ are same. Therefore, it is clear that one should not hurriedly conclude that the series A & B 85
Transcript
UNIT –5MEASURES OF DISPERSION
Measures of central tendencies indicates the central Tendency of a frequency distribution in the form of an average. An average is a single significant value which is used to describe a distribution. These averages tell something about the general level of magnitude of the distribution but they fail to show anything further about the distribution. An average does not tell the full story and it is hardly fully representative of a mass, unless we know the manner in which the individual items scatter around it. A further description of the series is necessary to gauge the worthness of average.
Thus, it may be such that in several series the average may be the same, but variables may highly differ in magnitudes and therefore, the central tendency calculated from such variables may not be the most typical representative in many cases.
To know the extent of spread about these averages or the variations of items, it is necessary to observe the following examples.
x d = (x - ) x d = (x - )100 -20 300 -20110 -10 310 -10120 0 320 0130 10 330 10140 20 340 20 600 1600 120 320
In the example-1 though the formation of series is different, but their averages are same. However, in the example-2 though their means are different, the deviations of individual items from ‘x’ are same. Therefore, it is clear that one should not hurriedly conclude that the series A & B are same as their x are same or the series are not same as their x are different. That means averages them selves are not sufficient indicators of all the characteristics of a given data therefore they must further be subjected to other statistical analysis. Such further step in statistical analysis are the measures of dispersion.
MEANING OF MEASURES OF DISPERSION:- Dispersion is an important measure sought for describing the character of variability of data. Dispersion finds out how individual values fall apart on an average from the representative value. The average is derived from the actual values, but dispersion is known by averaging the deviations from representative value.
Definition: Let us go through some definitions
“ Dispersion is the measure of the variation of the items”A. L. Bowley
“ The degree to which numerical data tend to spread about an average value is called the variation or dispersion of the data”-Spiegel
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Above given definitions focuses an variation. In order to understand the actual amount of variation must present in a given set of value, the size of variation must be measured and expressed in terms of numbers. This is known as measure of dispersion.
Objectives of measures of dispersion The following are the main purpose of measuring dispersion
1. To test the reliability of an average.The variation measure is the only means to test the representative character of an Average. If the scatter is large, average is less reliable. On the other hand if the scatter it small the average is a typical value.
2. To serve as a basis for control of variability.Measures dispersion are indispensable to determine the nature and find the causes of variation. When these are known, it is easy to control the variation itself.
3. To compare two or more series with regard to their variability.The degree of uniformity or the consistency of data can be found out through study of measure of dispersion. when comparing two series, as regards the reliability of the averages, due considerations may be given to dispersion ,which is a good basis for comparison.
4. To facilitate as a basis for further statistical analysisThe measures of dispersion are essential for studying the statistical tools.
Requisites of a good measure of dispersionA good measure of dispersion should have the following properties.
1. It should be simple to understand and rigidly defined.2. It should be easy to compare.3. It should be based an all items.4. It should be free from sampling fluctuations.5. It should be capable of further algebraic treatment.6. It should remain un affected by extreme items.
Methods of Measuring DispersionThe following are the important methods of studying variation.
1. Range2. Inter-quartile Range & Quartile Deviation.3. Mean Deviation (Not included in your syllabus)4. Standard Deviation.
Range:The range is the simplest measure of dispersion. It is a rough measure of dispersion. Its measure
depends upon the extreme items and not on all the items. It does not tell us anything about the distribution of values in the series relative to a typical value
Thus Range = Largest Value – Smallest value R= L –S
Co efficient of Range = L –S L+S
To Compare the series, the relative measure of dispersion is used.
Computation of RangeIndividual series
ILLUSTRATION = 01The net profit of a business concern in thousands of Rs is given below
Range = L – S = 50.5 – 0.5 = 50Co efficient of Range = L – S = 50.5 – 0.5 = 50 = 0.98
L+S 50.5+ 0.5 51Uses of Range:
1. Range is used in industries for the statistical quality control of the manufactured product by the construction of control chast.
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2. Range is useful in studying the variations in the prices of stick, shares and other commodities that are sensitive to price changes from one period to another period.
3. The meteorological department uses the Range for weather fore casts.Merits1. It is simple to compute and understand.2. It gives a rough, but quick answer3. When items are limited as in the case of sample lots for quality control pourpose, these methods are quite
handy.4. It is rigidly defined.Demerits1. It is not reliable, because it is affected by the extreme items.2. It cannot be applied to open end cases.3. Range is too indefinite to be used as a practical measure of dispersion.
QUARTILE DEVIATION OR SEMI-INTER QUARTILE RANGESemi inter quartile range or quartile deviation is defined as half the distance between the third and the
first quartiles.Symbolically:-
Semi-inter quartile Range Or =Q2 – Q1
Quartile Deviation 2
It means, the items below the lower quartile and the items above the upper quartile are not at all included in the computation. Thus we are considering only the middle half portion of the distribution. The range so obtained is divided by two as we are considering only half of the data.
The quartile deviation gives the average amount by which the two quartiles differ from median in an asymmetrical distribution. It is a measure of partition rather than a measure of dispersion. The smallest the value of Q.D, the minimum is the dispersion of middle half of the distribution around the median. However it provides no indication of the degree of dispersion lying beyond the limits of the two quartiles.
Quartile deviation is an absolute measure of dispersion. The relative measure of dispersion, known as co-efficient of quartile deviation, is calculated as follows: -
Quartile deviation is an improved measure over the range, as it is not calculated from extreme items, but on quartiles.
Computation of Quartile Deviation & its Coefficient Individual seriesILLUSTRATION =5
15 students of a class obtained the following marks in statistics. Calculate the quartile Deviation and its coefficient.Marks:( 15, 20, 20, 21, 22, 22, 24, 25, 28, 28, 29, 30, 32, 33, 35.)Solution
Calculate quartile deviation and its coefficient from the following data.Wages in Rs 120 –130 130 –140 140 –150 150 –160 160 –170 170 –180 180 –190 190 –200
No.of workers
10 20 30 40 30 20 15 5
SolutionQuartile Deviation = Q3 – Q1
2 Coefficient of Quartile Deviation = Q3 –Q1
Q3 + Q1x F cf Q1= L1 + L 2-L1 (q1-c) where q1=N/4 = 170/4 = 42.5
ILLUSTRATION:9Calculate quartile Deviation and its coefficient from the data given below.Wages in Rs. 100 100-109.5 110-119.5 120-129.5 130-139.5 140-149.5 150-159.5 160 &abvNo. of worker 12 18 24 16 30 20 15 5
Solution:-Convert the given open end cum inclusive series into exclusive formX f Cf Q1=L1+L2 – L1(q1-c) Where q1= N/4 = 140/4 = 35
ILLUSTRATION =10 Calculate Quartile Deviation and its coefficient from the data given
Marks Below 50 55 60 65 70 75 80 85 90 95 100No. of students 15 31 48 70 102 130 148 170 185 190 200Convent the given cumulative frequency table into ordinary table.
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Solution
x f cf Q1 = L1+ L2 –L1(q1-c) where q1= N/4 = 200/4 =50 f = 60 + 65 –60(50 -48) 22 = 60 +5 x 2 = 60 +10 = 60.45 22 22 Q3 = L1 + L2 – L1(q3-c) where q3 =3N =3x200 = 50 f 4 4= 80+ 85 – 80(150 – 148) 22= 80+ 5 x 2 = 80 + 10 = 80.45 22 22Q.D = Q3 – Q1 = 80.450 – 60.45 = 10 2 2Coefficient of Q. D = Q3 – Q1 = 80.45 – 60.45 = 0.141 Q3+Q1 80.45 + 60.45
Merits of Quartile Deviation1. It is easy to calculate and simple to understand.2. It is rigidly defined.3. It gives an idea about the variation of central 50% of the observation.4. It is not unduly affected by extreme values.
Demerits1. It is not based on all the observations.2. It is not capable of further of algebraic treatment.3. It is much affected by fluctuations.4. It is necessary to arrange the data in ascending order.
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TERMINAL QUESTIONS (5, 10 & 15 Marks)1. Define the term Dispersion. State the objectives of measurements of dispersion.2. What is meant by dispersion? Explain different methods measuring dispersion.3. Briefly, explain the relative merits and demerits of the various measures of dispersion4. What is Range? Explain its merits & demerits5. What is quartile Deviation? State it merits & demerits 6. Find out quartile Deviation and its coefficient from the following data.
Weight in kg 10 –12 12 –14 14 –16 16 –18 18 –20 20 –22 22 –24No. of Boxes 6 9 20 25 24 15 5
[Answer Q.D = 2.09, coefficient of Q.D = 0 .12]7. Calculate Quartile Deviation and its coefficient from the following data.Class 1 –10 11 –20 21 –30 31 –40 41 –50 51 –60 61 –70Frequency 6 8 6 9 9 6 6[Answer Q D =15.67, coefficient of Q D = 0.468. Calculate Quartile Deviation and its Coefficient from the following data.Size more than 20 30 40 50 60 70Frequency 68 63 40 40 18 7 [Answer Q .D = 12.845, coefficient 0.27]9. Calculate Quartile Deviation and its Coefficient from the following dataMarks less than 10 20 30 40 50 60 70 80 90No/ of persons 5 15 98 242 367 405 425 438 439 [Answer Q.D = 8.08, coeff = 0.21]
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STANDARD DEVATION
Concept of standard Deviation was introduced by Karl Pearson in 1893. It is the most important measure of dispersion and is widely used in many statistical formula. Standard Deviation is also called Root – mean square Deviation. The reason is that it is the square root of the means of the squared deviation from the arithmetic mean. In this method the draw back of ignoring the algebraic sign in mean Deviation is overcome by taking the square of deviation, there by making all the deviations as positive.
Therefore, it is defined as positive square root of the arithmetic mean of the squares of the deviations of the given observations from the arithmetic mean. The standard deviations is denoted by the greek letter (sigma) the square of standard deviation is known as variance.
Features of standard deviationFollowing are the feature of standard deviation 1. The standard deviation measures the absolute dispersion of a distribution. The greater the dispersion greater
is the standard deviation.2. A small standard deviations means a high degree of uniformity of the observation as well as homogeneity. A
large standard deviation the opposite.3. If we have two or more comparable series with nearly identical means, it is the distribution with the smallest
standard deviation that the most representative mean.4. The standard deviation has the same unit as the original variables have.
Computation of standard deviationIndividual series – direct methodSteps:- 1. Calculate arithmetic mean, of the series
2. Obtain deviations from arithmetic mean dx = x –
3. square these deviations and final their sum d2x 4. use formula for standard deviation
= d 2 x or (x – ) n n
= Arithmetic mean d2x = Square of the Deviation from n = Numbers of items.
ILLUSTRATION= 01Find the standard Deviation of the monthly salaries of 10 persons given below.
Persons A B C D E F G H I JSalaries in Rs 120 110 115 122 126 140 125 121 120 131
160x 0 d2x 720Mean = x N = 160 = 16 10Standard deviation = d 2 x n = 720 10 = 72 S.D = 8. 485
SHORT – CUT METHOD OR DEVIATIONS TAKEN FROM ASSUMED MEANThis method is adopted when the arithmetic average is a fractional value. Taking deviations from
fractional value would be a very difficult and tedious task. To save time and labour we apply shore cut method.Steps:- 1. Assume any one of the item in the series as an average = (A) 2. Find out the deviations from the assumed mean dx = x – A C 3. Find out the total of the deviations dx
4. Square the deviations and add up the squares of deviations i.e d2x5. Use the formula = d 2 x - ( dx) 2 x C
n nILLUSTRATION = 03
The below given table gives the marks obtained by 10 students in statistics examination. Calculate standard deviation.Sl. No 1 2 3 4 5 6 7 8 9 10Marks 43 48 65 57 31 60 37 48 78 59
ILLUSTRATION = 04Calculate the standard deviation from the following data.
Values (x) 58 59 60 54 65 66 52 75 69 52
Solution A = 65, C = 1, dx =(x –A)/C
DISCRETE SERIESCalculation of standard deviation
DIRECTMETHOD OR ACTUAL MEAN METHODStep:-
1. Calculate the mean of the series 2. Find deviations for various items from the mean dx = x –3. Square the deviations (d2) and multiply by the respective frequencies we get fd2x4. Get the total product fd2 and use the formula
= fd 2 x N= frequency total N
ILLUSTRATION = 05Calculate Standard Deviation from the following data.
Marks 10 20 30 40 50 60No. of Students 8 12 20 10 7 3
Solution x – 30.8x f fx Dx dx2 dd2x = fx = 1850 =30.83
1. Assume any one of the item in the series as an average ( A )2. Find out the deviations from assumed mean after considering common factor if any dx =(x –A)
C3. Multiply the square deviations by the respective frequencies and get the fdx4. Square the deviations d2x.5. Multiply the squared deviations (d2x) by the respective frequencies and get fd2x6. Use the formula S.D = d 2 x – fdx 2 x C
N N
ILLUSTRATON = 07Calculate standard deviation for the following data.
X f Mid x dx fdx fd2x = A + fdx x C N = 35 + -302 x 70 71 = 35 – 300 = 35 – 4.225 = 30.775 71S.D = fd 2 – fdx x C N N = 210 – –30 2 x 10 71 71 = 2.957 – (-0.422)2 x 10 =2.7785 x 10 =1.668 x 10 = 16.668
0 –1010 –2020 –3030 –4040 –5050 –6060 –70
8121714974
5152535455565
-3-2-10123
-24-24-17091412
724817092836
71 fd2x 210
ILLUSTRATION = 11 The following data relate to the age of a group of workers calculate the arithmetic Mean and standard
deviation.Age in years 20-25 25-30 30-35 35-40 40-45 45-50 50-55No of worker 170 110 80 45 40 30 25 Solution x – 37.5, A =37.5, C =5 , dx =(x – A)/c
5
Age x fmid
xdx fdx fd2x
Fd2x = fdx x dx = A+ fdx x C N = 37.5 + –635 x 5 = 31.15 500S.D = fd 2 x – fdx 2 x C N N
ILLUSTRATION = 12Calculate arithmetic mean and standard deviation from the following data.
Wages more than 100 200 300 400 500 600 700 800No of workers 660 615 527 381 175 96 44 14
[K U V BBM 2002]Solution ( x – A)/100, A =450, C= 100, dx,=(x – A)/C
x f Mid x dx fdx fd2x = A + fdx xC N = 450 + -128 x 100 660 = 450 – 19.39 = 430.61S.D = fd 2 x – fdx 2 x C N N = 1684 – –128 2 x 100 = 2.51 x 100 660 660 = 1.584 x 100 =158 . 4
IILUSTRATION 13 Calculate arithmetic mean and standard deviation.Age in less than years 10 20 30 40 50 60 70 80No of workers 15 30 53 75 100 110 115 125
Solution (x –35)/10, A = 35, C = 10, dx = (x – A)/CX f Mid x dx fdx fd2x = fdx x C = 35 + 2 x 10 = 35 + 20 =35.16
N 125 125
S.D = fd 2 x – ( fdx) 2 x C = 488 – ( 02 )2 x 10 N N 125 125 = 3.904 – (0.016)2 x10 = 3.904 – 0.000256 = 1.975 x 10 = 19.75
0-1010-2020-3030-4040-5050-6060-7070-80
151523222510510
515253545556575
-3-2-101234
-45-30-23025201540
13560230254045160
N=125 fdx 02 488=fd2x
ILLUSTRATION 14Calculate the arithmetic mean and standard deviation from the following data.
Income under Under 100 10-104.9 105-109.9 110.114.9 115-119.9 120-124.9 125-129.9No. of workers 20 40 30 20 50 15 5
Solution: - A = 112.45, C =5, dx =(x –A)/CX f Mid x dx fdx fd2x = A+ fdx X C
N = 112.45 + -75 x 5 180 = 112.45 – 2 .08 = 110.37S.D = fd 2 x - ( fdx) 2 X C N N = 525– -75 2 x 5 = 2.9166 –(0.416)2 x 5 180 180 = 2.916 – 1.173 =2.7436 x 5 = 1.656 x 5 = 8.28
COEFFICIENT OFVARIATION = CVThe standard deviation is an absolute measure of dispersion. It is expressed in terms of units in which
the original figures are collected and stated. The standard deviation of heights of students cannot be compared with the standard deviation of weights of students, as both are expressed in different units. Therefore, the standard deviation must be converted into a relative measure of dispersion for the purpose of comparison. The relative measure is known as the coefficient of variation.
Variance:- square of standard deviation is called variance symbolically Variance = 2
= variance Coefficient of standard deviation =
For better comparison purpose, this coefficient of standard deviation is multiplied by 100 gives the coefficient of variation
Coefficient of variation = x 100
Prof. Karl Pearson suggests this measure of coefficient of variation as the most commonly used measure of relative variation. The series for which the c v is greater, indicates that the series is more variable or less uniform if the coefficient of variation is less, it indicates that the series is less variable or more stable or more consistent
COMPARISON OFTWO SERIESUSING COEFFICIENT OF VARIATION
ILLUSTRATION =15The index numbers of prices of cotton and cool shares in a year are given below.
= A + dx x C n = 82 + -435 x1 = 38.5 10 y = A + dy x C n=13 + 300 x1 =43 10
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x- series S.D = d 2 x - ( dx) 2 x C n n =29469 – (-43.5)2 x 1 10 10 = 2946.9 – 1892.25 = 1054.65 = 32.475 C V = x 100 = 32.475 x 100 = 84 .35% 38.5
y-seriesS.D = d 2 y -( dy) 2 xC n n =21762 - (300 )2 x 1 10 ( 10 ) = 2176.2 – (30)2 x 1 =2176.2 – 900 = 1276.2 = 35.723C.V = x 100 y = 35.723 X 100 43 = 83.07%
Conclusion:-1. Batsman B is better run getter, because he has scored 430 runs compare to batsman A 385 runs2. Batsman B is more consistent.
COMPARISION OF TWO GROUPS OF DATAUSING COEFFICEINT OF VARIATION
DISCRETE SERIESILLUSTRATION = 17
The goals scored by tow teams A & B in the football matches were as follows.Goals 0 1 2 3 4Matches A= 27 9 8 4 5Teams B= 17 9 6 5 3Find which team is more consistent.
= 2.66 – (-0.925)2 x 1 = 1.8052 = 1.3436 A = A + fadx x C N = 2+ -49 x 1 53 = 2 –0.925 =1.075 C.V = X100 = 1.3436 x 100 A 1.075 = 124.93%
S.D = f Bd 2 x - ( f Bdx)2 x C N2 N2
= 94 - (-32)2 x140 40
= 2.35 – (-0.8)2 x1 = 1.71 = 1.308B= A + fBdx x C N = 2 + -32 x 1 = 1.2 40C.V = x 100 = 1.308 x 100 = 109% B 1.2
Team B is more consistent as it has less variation
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CONTINUOUSSERIESILLUSTRATION =18
Following data gives life of electric bulb manufactured by two companies. Calculate. A. which of the two makes has a higher average life?B. If prices of both the bulbs is same, which company’s bulb would you prefer to buy and why? Use
C.VLife in 000 hrs
XCompany
ACompany
B50-5960-6970-7980-8990-99
182226259
1524301813
100 100Solution
Computation of coefficient of variation for both the series.
X Mid xx – 74.510 dx
fA fAdx fAd2x fB fBdx fBd2x
50-5960-6970-7980-8990-99
54.564.574.584.594.5
-2-1012
182226259
-36-2202518
722202536
1524301813
-30-2401826
602401852
N1=100 N1
-15fAdx
155fAd2x
100N2
-10fBdx
154fBd2x
S.D = f Ad 2 x - ( f Adx )2 x C N1 N1
=155 - (-15) 2 x 10 100 (100) = 1.55-( -0.15) 2 x 10 = 1.55-0.0225 = 1.5275 = 1.2359 x10 =12.359 A = A + f Adx x C N =74.5 + (-15) x10 100 =74.5-1.5 = 73.0C.V = X 100 = 12.359 X 100 = 16.93% A 73
sssS.D = f Bd 2 x - ( f Bdx )2 x C N2 N2
= 154- (-10)2 x10 100 (100) = 1.54 – (-0.1)2 x10 = 1.54 – 0.01 x 10 = 1.53 x 10 = 1.2369 X 10 = 12.369B= A + f Bdx x C N = 74.5 + (-10) X10 100 = 74.5 -100 =73.5 100C.V = X 100 = 12.369 x 100 = 16.83% B 73.5
Conclusion1. B company bulbs have higher average life i.e. 73.5 hours.2. B company bulbs are more uniform and consistent in giving life, hence buyer would prefer to buy B
companies bulbs.
ILLUSTRATION =19
From the data given below state which of the two series is more variable. Use coefficient of variation. Variable 10-20 20-30 30-40 40-50 50-60 60-70Frequency A 10 18 32 40 22 18 Frequency B 18 22 40 32 18 10
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SolutionComputation of coefficient of variation .
Frequency A Frequency BVariable
Xmid x
x-3510 dx
fA fAdx fAd2x dx fA fBdx fBd2x
10-2020-3030-4040-5050-6060-70
152535455565
-2-10123
101832402218
-20-180404454
401804088162
-2-10123
182240321810
-36-220323630
72220327290
140 100 348 140 40 288
A = A + fAdx X C N1
=35 + 100 X 10 140 = 42.1429 B = A + fBdx X C N2
= 35 + 40 X 10 140 = 37.8571 A-Series B-Series S.DA = f Ad 2 x - ( f Adx)2 x C N1 N1
=348 - (100) 2 X 10 140 140 = 2.4857-( -0.7142) 2 x 10 S.D= 14.055 C.V = X 100 = 14.055 x 100 A 42.1429 = 33.35%
S.DB = f Bdx - ( f Bdx )2 x C N2 N2
=288 - (40) 2 X 10 140 140 = 2.05714 -(0.2857) 2 X 10 = 14.055 C.V = X 100 = 14.055 X 100 B 37.8571 = 37.127%
Conclusion:- Series B is more Variable
ILLUSTRATIONS:-20Two brands of types are tested for their life and the following results were obtained. State which brand
of Tyres are more consistent?Life in months 20-25 25-30 30-35 35-40 40-45No/ of Tyres: x: 1 22 64 10 3
y: 3 21 74 1 1 Solution: computation of coefficient of variation,
Brand A Brand B
Life in month x mid xx-32.5
5fAdx fdx fd2x f fdx fd2x
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20-2525-3030-3535-4040-45
22.527.532.537.542.5
-2-1012
012264103
-2-220106
42201012
3217411
-6-21012
+1221014
100 - 8 48 100 -24 38
1 = A + fdx X C N1
=352.5+ -8 X 5 100 = 32.5-0.4 =32.1 y = A + f dx X C N2
= 32.5+ - 24 X 5 = 32.5 –1.20 = 31.3 100
S.D = fd 2 x - ( fdx )2 x C N1 N1
= 48 - (-8) 2 x 5 100 100 = 3.44 C.V = X 100 = 3.44 x 100 32.1 = 10.72%
S.D = fd 2 x - ( fdx )2 X C N2 N2
= 38 - (-24) 2 X 5 100 100 = 2.839 C.V = X 100 = 2.839 X 100 31.30 = 9.07%
‘Y’ Brand Tyres are more consistent than brand x Tyres.
ILLUSTRATION:21:- In two factories A& B engaged in the industrial area the average weekly wages in Rs and the standard
deviations are as followsFactory Average Standard Deviation No. of workers
A 345 5 476B 285 4.5 524
Find 1) Which factory A or B pays out a larger amount as weekly wages?2) Which factory A or B has greater variability in individual wages?
Solution1. Calculation of total weekly wage payment
Total wages paid by factory A= Rs345 x 476 = 164220Total wages paid by factory B= Rs 285 x 524 = 14 9340Therefore, factory A pays out larger amount as weekly wages
2. Calculation of coefficient of variation Factory A ------- CV = x 100 = 5 x 100 = 1.449% 345 Factory B--------- CV = x 100 = 4.5 x 100 = 1.578% 285 Factory B has greater variability in individual wages since CV of factory B is greater than CV of factory A.
ILLUSTRA TION=22Particulars regarding income of two villages are given below.
Particulars Village A Village BAverage income 1750 1860Variance 100 81 State in which village is the variation in income greater ?Solution
Calculation of coefficient of variation
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1. village . A------CV = x 100 =10 x 100 = 0.57% 1750 2. village. B------ CV = x 100 = 9 x 100 = 0.483% 1860
SD =Variance =100 = 10
S.D Variance = 81 = 9
Conclusion:- village A has greater variation than village B.
ILLUSTRATION=23.Coefficient of variation of two series are 58% and 69%. Their standard Deviations are 21.2 and 15.6
what are their arithmetic means?
Solutionx-Series x-Series
C.V = = x 100 58 = 21.2 x 100 = 2120 58 = 36.55
CV = X 100 Y69 = 15.6 yy = 1560 69 y = 22.6
ILLUSTRAAATION = 24Coefficient of variation of two series are 75% and 90% and their arithmetic means are 20 & 20
respectively find their standard deviations.x-series y-seriesCV = x 100 58 = x 100 20 75 =100 20 = 75 x 20 = 15 100
CV = x 100 y90 = x 100 20 90 =100 20 = 90 x 20 = 18 100
USE OF STANDARD DEVIATIONS:Standard Deviation is the best measure of dispersion. It is widely used statistics because it possesses
most of the characteristics of an ideal measure of dispersion. It is widely used in sampling theory and biologist. It is used in coefficient correlation and in the study of symmetrical frequency distribution.
Merits of standard Deviation: 1. It is rigidly defined and its value is always definite and based on all the observations.2. As it is based on arithmetic mean, it has all the merits of arithmetic mean.3. It is the most important and widely used measure of dispersion.4. It is possible for further algebraic treatment.5. It is less affected by the fluctuations of sampling and hence stable.6. It is the basis for measuring the coefficient of correlation sampling and statistical inferences 7. The coefficient of variation is considered to be the most appropriate method for comparing the variability of
two or more distributions and this is based on mean and standard deviation.Demerits:1. It is not easy to understand and it is difficult to calculate.2. It gives more weight to extreme values, because the values are squared up3. It is affected by the value o f every item in the series.4. As it is an absolute measure of variability, it cannot be used for the purpose of comparison5. It ha not found favour with the economists & businessmen.THEORETICAL QUESTIONS(5, 10 & 15 Marks)1. Why is that standard deviation is considered to the most popular measure of dispersion ?2. What is standard deviation? State their merits & demerits
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3. What is coefficient of variation? What purpose does it serve.PRACTICAL PROBLEMS1. Calculate mean and standard deviation f r o m the following data.Wages in rs 40-50 50-60 60-70 70-80 80-90 90-100No of workers 12 9 8 5 7 9 [Answers: = 67-6, SD = 18.31]
2. Calculate the standard deviation from the following data.Monthy
[Answer SD = 11]3. Find mean standard deviation and coefficient of variation from the following data.Age in under year 10 20 30 40 50 60 70 80No. of persons 15 30 53 75 100 110 115 125 [Answers; = 35.16, SD = 19.76, CV = 56.2% ]4. Find Mean, standard deviation and coefficient of variation from the following data .Marks more than 0 10 20 30 40 50 60 70 80 90No. of students 100 90 80 65 50 20 15 10 5 2 [ Answers: = 38.7, SD = 21.3, CV=55 %] 5. From the prices pf shares of company ‘A’ and company ‘B’ given below, state which is more stable in value?Share A price 55 54 52 53 56 58 52 50 51 49Share B price 108 107 105 105 106 107 104 103 104 101 [ Answers : A = 53, A= 2.646, CV = 4.99 % ] B = 105, B=2 CV = 1.90% ( B company share prices are more stable)6. From the following table of marks obtained by 10 students, find the coefficient of variation and determine
the marks of which subject are more variable.Statistics 25 50 45 30 70 42 36 38 34 60M. Accounting 10 70 50 20 95 55 42 60 48 80 [Answers CV for stat=30.49%, CV for A/C= 45.9% ] (Variation is greater in the marks of M. Accounting)7. The scores of two bats men A & B for 20innings are as under which of the two may be regarded as the more
B 1 2 3 6 3 3 2 0 20 [ Answers A= 57.4, A = 1.9 6 CV = 3.4% B = 56.2, B= 1.6 CV=2.86 %] (Batsman B can be considered as more consistent)8. Samples of polythene Bags from two manufacturers A & B are tested by a prospective buyer for bursting
pressure with the following results.Bursting pressure in lbs
xNumber of bags
Company A Company Bunder-100 50 100100-104.9 150 75105-109.9 120 125110-114.9 80 65115-119.9 130 135120-124.9 70 140125-129.9 150 60
130 & above 50 100Total 800 800
(Ans: A = 114.64, A=10.57, CVA= 9.22%)(B=115.75, B=11.095, CVB= 9.64%)
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9. The life of two types of lamps is given below Find 1. Which of the two makes has a higher average life?2. If prices are same for both, which type would you prefer to buy and why (use CV)
Life in hours
XNumber of lamps manufactured
Company A Company –BUp to 20002000-39994000-59996000-79998000-9999
10000-119991200 & above
15015012080130170200
1001751256513514060
1000 800 [Answers A = 7399.5, A = 4308.13, B = 6549.5, B = 3820.68]10. Coefficients of variation of two series are 60% & 80% respectively. Their standard Deviation are 20 and 16
respectively what are their means?[ Answer 1 = 33.3 2 = 20]
