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  • 1. 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 Σ 600 Σ 600 ξ 120 ξ 120 Example – 2 A series B series x d = (x - 0 ) x d = (x - 0) 100 -20 300 -20 110 -10 310 -10 120 0 320 0 130 10 330 10 140 20 340 20 Σ 600 1600 0 120 320 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 85
  • 2. 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 analysis The measures of dispersion are essential for studying the statistical tools. Requisites of a good measure of dispersion A 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 Dispersion The following are the important methods of studying variation. 1. Range 2. 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 Range Individual series ILLUSTRATION = 01 The net profit of a business concern in thousands of Rs is given below Year 1996 1997 1998 1999 2000 2001 2002 Profit 100 160 150 220 300 190 200 Find out Range and its co efficient. SOLUTION Largest item = 300 Smallest item = 100 ∴ Range = L – S = 300 – 100 = Rs 200 thousand 86
  • 3. Co efficient of Range = L – S = 300 –100 = 200 = 0.5 or 50% L+ S 300 +100 400 Discrete Series ILLUSTRATION = 02 Find out Range and its coefficient of the following data Size 3 5 7 9 11 13 Frequency 11 15 13 19 14 2 Solution: Largest Value = 13 Smallest Value = 3 Range = L – S = 13 – 3 = 10 Coefficient of Range = L – S L+S = 13 – 3 = 10 = 0.625 13 +3 16 Continuous Series ILLUSTRATION-3 Find out Range and its co efficient of the following frequency distribution. SBE 0-10 10-20 20-30 30-40 40-50 Frequency 01 03 12 06 03 Solution:- Largest Value = Upper limit of highest class interval L=50 Smallest Value= Lower limit of the lowest class Interval L=0 Range =L –S = 50 – 0 = 50 Co efficient of Range = L –S L+S = 50 – 0 = 50 = 1 or 100 % 50 +0 50 ILLUSTRATION =4 Calculate the Range and its co efficient from the following data. Size 0 –10 10 –20 20 –30 30 –40 40 –50 Frequency 01 03 12 06 03 Solution Convert inclusive class intervals into exclusive form. Class 0.5 –10.5 10.5 –20.5 20.5 –30.5 30.5 –40.5 40.5 –50.5 Frequency 3 7 20 13 6 Largest Value = 50. 5 L Smallest Value = 0.5 S Range = L – S = 50.5 – 0.5 = 50 Co efficient of Range = L – S = 50.5 – 0.5 = 50 = 0.98 L+S 50.5+ 0.5 51 Uses of Range: 1. Range is used in industries for the statistical quality control of the manufactured product by the construction of control chast. 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. 87
  • 4. 3. The meteorological department uses the Range for weather fore casts. Merits 1. It is simple to compute and understand. 2. It gives a rough, but quick answer 3. 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. Demerits 1. 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 RANGE Semi 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: - Co-efficient of quartile Deviation = Q3 – Q1 Q3 + Q1 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 series ILLUSTRATION =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 Marks arranged in ascending order Sl. No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Marks 15 20 20 21 22 22 24 25 28 28 29 30 32 33 35 Q1 = The size of N+1 th item Quartile 4 Deviation= Q3 – Q1 = The size of 15+1 16 4th item 2 4 4 = 30 –21 = 4.5 = 4th item is 21=Q1 2 th Q3 = The size of 3(N+1) item 4 Coefficient of = The size of 3(15+1) = 3X16 = 12th item Q.D = Q3 – Q1 = 30 –21 = 0.176 88
  • 5. 4 4 Q3 + Q1 30+21 = 12th item is 30 =Q3 Discrete series ILLUSTRATION=06 Find the quartile Deviation and its coefficient from the following. Age in years 15 16 17 18 19 20 21 No, of students 4 6 10 15 12 9 4 Solution Ages in years No. of Students Q1= The size of N+1th items Value x cf 4 f 15 4 4 = The size of 60+1 61 = 15.25 16 6 10 4 4 q1 Q1=17 17 10 20 Q3 = The size of 3(N+1)th items value 18 15 35 q3 4 19 12 47 = The size of 3(60+1) = 45.75 20 9 56 4 21 4 60 Q3 =19 N = 60 Quartile Deviation = Q3 – Q1 19 – 17 = 2 = 1 2 2 2 Coefficient of quartile Deviation = Q3 – Q1 = 19 – 17 = 2 = 0.0556 Q3 +Q1 19 + 17 36 Continuous Series ILLUSTRATION = 07 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 10 20 30 40 30 20 15 5 workers Solution Quartile Deviation = Q3 – Q1 2 Coefficient of Quartile Deviation = Q3 –Q1 Q3 + Q1 x F cf Q1= L1 + L 2-L1 (q1-c) where q1=N/4 = 170/4 = 42.5 120 –130 10 10 f 130 –140 20 30 q1 = 140+150 –140 (42.5-30) 140 –150 30 60 30 150 –160 40 100 = 140 + 10 x12.5 = Q1 =144.167 160 –170 30 130 30 170 –180 20 150 Q3 = L1 + L2 –L1(q1-c) where q3=3N 3 x 170 =127.5 180 –190 15 165 f 4 4 190 –200 5 170 =160 + 170 –160 (127.5 –100) = Q3 =169.167 170 30 Coeff. of Q .D = Q3 – Q1 = 169.167 – 144.167 = 25 = 0.079 Q3+Q1 169167 + 144.167 313.334 ILLUSTRATION =08 Calculate Quartile Deviation and its coefficient from the data given below. 89
  • 6. Mid Value 1 2 3 4 5 6 7 8 9 10 Frequency 2 9 11 14 20 24 20 16 5 3 Solution: Form Lower Classes and upper classes from the given mid value X f Cf Q1= L1+L2 – L1( q1-c) where q1 =N/4 = 140/4 = 35 0.5 – 1.5 2 2 f 1.5 –2.5 9 11 = 3.5 + 4.5 –3.5 (31-22) = 3.5 + 1 x9 = 4.14 2.5 –3.5 11 22 q1 14 14 3.5 –4.5 14 36 Q3 = L1+L2 –L1 (q3-c) where q3 = 3N = 3 x 124 = 93 4.5 –5.5 20 56 f 4 4 5.5 –6.5 24 80 q3 = 6.5 + 7.5 – 6.5( 93-80) = 6.5 +1 x 13 =7.15 6.5 –7.5 20 100 20 20 7.5 –8.5 16 116 Q.D = Q3 – Q1 = 7.15 – 4.14 =1.505 8.5 –9.5 5 121 2 2 9.5 –10.5 3 124 Coefficient of QD =Q3 – Q1 = 7.15 – 4.14 = 3.01= 0.266 124 Q3+Q1 7.15 +4.14 11.29 ILLUSTRATION:9 Calculate 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 &abv No. of worker 12 18 24 16 30 20 15 5 Solution:- Convert the given open end cum inclusive series into exclusive form X f Cf Q1=L1+L2 – L1(q1-c) Where q1= N/4 = 140/4 = 35 89.75 –99.75 12 12 f 99.75 –109.75 18 30 35 = 109.75 + 119.75 –109.75(35 –30) 109.75 –119.75 24 54 24 119.75 –129.75 16 70 = 109.75 + 10 x 5 = 109.75 +50 = 111.83 129.75 –139.75 30 100 105 24 24 139.75 –149.75 20 120 Q3= L1+L2 –L1(q3-c) where q3 = 3N = 3X140 = 105 149.75 –159.75 15 135 f 4 4 159.75 –169.75 5 140 = 139.75 + 149.75-139.75 (105 –100) 140 = 139.75 + 10 x 5 = 139.75+ 50 = 142.25 20 20 Q.D = Q3 –Q1 = 142.25 –111.83 = 15.21 2 2 Co eff of Q.D = Q3 –Q1 = 142.25 –111.83 = 30.42 = 0.119 Q3 +Q1 142.25 +111.83 254.08 ILLUSTRATION =10 Calculate Quartile Deviation and its coefficient from the data given Marks Below 50 55 60 65 70 75 80 85 90 95 100 No. of students 10 13 14 17 19 15 31 48 70 185 200 2 0 8 0 0 Convent the given cumulative frequency table into ordinary table. 90
  • 7. Solution x f cf Q1 = L1+ L2 –L1(q1-c) where q1= N/4 = 200/4 =50 45 –50 15 15 f 50 –55 16 31 = 60 + 65 –60(50 -48) 55 –60 17 48 22 = 60 +5 x 2 = 60 +10 = 60.45 60 –65 22 70 22 22 65 –70 32 102 Q3 = L1 + L2 – L1(q3-c) where q3 =3N =3x200 = 50 70 –75 28 130 f 4 4 75 –80 18 148 = 80+ 85 – 80(150 – 148) 80 –85 22 170 22 85 –90 15 185 = 80+ 5 x 2 = 80 + 10 = 80.45 90 –95 5 190 22 22 95 –100 10 200 Q.D = Q3 – Q1 = 80.450 – 60.45 = 10 2 2 200 Coefficient of Q. D = Q3 – Q1 = 80.45 – 60.45 = 0.141 Q3+Q1 80.45 + 60.45 ILLUUSTRATION =11 Calculate Quartile Deviation and its coefficient from the data given below. Wages Above 140 160 280 1200 1800 2000 2200 2400 2600 3000 0 0 0 In Rs. 100 95 85 72 61 54 43 30 16 10 Solution:- Convert the given more than cumulative frequency distribution in to an ordinary Table. X f cf Q1 =L1 + L2 –L1 (q1-c) where q1 = N/4 =100/4 = 25 5 f 1200 –1400 5 15 = 1600 + 1800 – 1600(25 –15) 1400 –1600 10 28 13 1600 –1800 13 39 = 1600 + 200 x 10 = 1600 + 2000 = 1753.846 1800 –2000 11 46 13 13 2000 –2200 07 57 Q3 = L1 + L2 – L1( q3 –c) where q3 = 3N = 3 x100 =75 2200 –2400 11 70 f 4 4 2400 –2600 13 84 = 2600 + 2800 – 2600(75 –70) 2600 –2800 14 90 14 2800 –3000 06 10 = 2600 + 200 x 5 = 2671.43 3000 –3200 10 0 14 100 Q.D = Q3 – Q1 = 2671.43 – 1753.846 = 458.79 2 2 Coefficient Q.D = Q3 – Q1 = 2671.43 – 1753.85 = 917.58 = 0.207 Q3 + Q1 2671.43 +1753.85 4425.28 Merits of Quartile Deviation 1. 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. Demerits 1. 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. 91
  • 8. 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 dispersion 4. What is Range? Explain its merits & demerits 5. 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 –24 No. 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 –70 Frequency 6 8 6 9 9 6 6 [Answer Q D =15.67, coefficient of Q D = 0.46 8. Calculate Quartile Deviation and its Coefficient from the following data. Size more than 20 30 40 50 60 70 Frequency 68 63 40 40 18 7 [Answer Q .D = 12.845, coefficient 0.27] 9. Calculate Quartile Deviation and its Coefficient from the following data Marks less than 3 10 20 40 50 60 70 80 90 0 No/ of persons 9 5 15 242 367 405 425 438 439 8 [Answer Q.D = 8.08, coeff = 0.21] 92
  • 9. 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 deviation Following 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 deviation Individual series – direct method Steps:- 1. Calculate arithmetic mean, 0 of the series 2. Obtain deviations from arithmetic mean dx = x – 0 3. square these deviations and final their sum ∑d2x 4. use formula for standard deviation σ = √ ∑ d2x or √∑ (x – 0 ) n n 0 = Arithmetic mean d2x = Square of the Deviation from 0 n = Numbers of items. ILLUSTRATION= 01 Find the standard Deviation of the monthly salaries of 10 persons given below. Persons A B C D E F G H I J Salaries in Rs 120 110 115 122 126 140 125 121 120 131 SOLUTION Salaries in Rs Squares Persons Deviations from mean (x-0) ax (x –123) x d2x A 120 -3 9 B 110 -13 169 C 115 -8 64 D 122 -1 1 E 126 +3 9 F 140 +17 289 G 125 +2 4 H 121 -2 4 I 120 -3 9 J 131 +8 64 ∑x 1230 ∑d0 622∑d2x 93
  • 10. 0 = ∑x = 1230 = 123 n 10 0 = 123 S. D = √∑d2x N = √622 10 = √62.2 σ = 7.89 ILLUSTRATION –02 Calculate the standard deviation from the following data. Variables x 10 15 20 12 8 4 15 16 25 35 SOLUTION x Deviation from Mean dx = x -0(x – 16) dx2 10 -6 36 15 -1 1 20 +4 16 12 -4 16 8 -8 64 4 -12 144 15 -1 1 16 0 0 25 +9 81 35 +19 361 160∑x 0 ∑d2x 720 Mean 0 = ∑x N = 160 = 16 10 Standard deviation = √∑d2x n = √720 10 = √72 S.D = 8. 485 SHORT – CUT METHOD OR DEVIATIONS TAKEN FROM ASSUMED MEAN This 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 ∑d2x 5. Use the formula σ = √∑d2x - (∑dx)2 x C n n ILLUSTRATION = 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 10 Marks 43 48 65 57 31 60 37 48 78 59 94
  • 11. Solution:- A= 31, C=1 dx = x - A C Marks x-31, Sl. No. d2x x dx 1 43 12 144 S.D =√Σd2x - Σdx 2 x C 2 48 17 289 n n 3 65 34 1156 = √6424 – 216 2 x 1 4 57 26 676 10 10 5 31 0 0 = √642.4 –(21.6)2 x 1 6 60 29 841 = √642.4 – 466.56 7 37 6 36 = √175.84 8 48 17 289 S.D = 13.26 9 78 47 2209 10 59 28 784 N=10 ∑dx 216 6424 ILLUSTRATION = 04 Calculate 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 Values - x (x –65) dx d2x S . D = √∑d2 x – ∑dx 2 xC 58 -7 49 n n 59 -6 36 60 -5 25 = √686 – -40 2 x 1 54 -11 121 10 10 65 0 0 = √68.6 – (-4)2 x 1 66 1 1 =√68.6 – 16 52 -13 169 =√52.6 75 10 100 S.D = 7.252 69 4 16 0 = A + ∑dx x C 52 -13 169 n n =10 -40 686 ∑d2x = 65 +(-40 x 1) = 61 0 10 DISCRETE SERIES Calculation of standard deviation DIRECTMETHOD OR ACTUAL MEAN METHOD Step:- 1. Calculate the mean of the series 95
  • 12. 2. Find deviations for various items from the mean dx = x –0 3. Square the deviations (d2) and multiply by the respective frequencies we get fd2x 4. Get the total product ∑fd2 and use the formula σ = √∑fd2x N= frequency total N ILLUSTRATION = 05 Calculate Standard Deviation from the following data. Marks 10 20 30 40 50 60 No. of Students 8 12 20 10 7 3 Solution x – 30.8 x f fx Dx dx2 dd2x 0 = Σfx = 1850 =30.83 10 8 80 -20.8 432.64 3461.12 N 60 20 12 240 -10.8 116.64 1399.68 S.D = √Σfd x 2 30 20 600 -0.8 0.64 12.8 N 40 10 400 9.2 84.64 846.40 = √10858.40 50 7 350 19.2 368.64 2580.48 60 60 3 180 29.2 852.64 2557.92 = √180.97 = 13.45 N= 60 1850Σfdx Σfd2x 10858.40 ILLUSTRATION = 06 No. of accidents 0 1 2 3 4 5 6 7 8 9 10 11 12 Persons involved 16 16 21 10 16 8 4 2 1 2 2 0 2 Solution (x – 3) No/of Accidents No/ of Persons fx dx d2x fd2 x 0 7 0 -3 9 144 1 16 16 -2 4 64 2 16 42 -1 1 21 3 21 30 0 0 0 4 10 64 1 1 16 5 16 40 2 4 32 6 8 24 3 9 36 7 4 14 4 16 32 8 2 8 5 25 25 9 1 18 6 36 72 10 2 20 7 49 98 11 2 0 8 64 0 12 0 24 9 81 162 2 702 N=100 ∑fx 300 ∑fd2x 0 = ∑fx n = 300 = 3 100 S.D = √∑fd2x N = √ 702 96
  • 13. 100 = √7.02 S.D = 2.649 SHORT – CUT METHOD OR ASSUMED MEAN Steps:- 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) C 3. Multiply the square deviations by the respective frequencies and get the Σfdx 4. Square the deviations d2x. 5. Multiply the squared deviations (d2x) by the respective frequencies and get ∑fd2x 6. Use the formula S.D = √∑d2x – ∑fdx 2 x C N N ILLUSTRATON = 07 Calculate standard deviation for the following data. Marks 5 6 7 8 9 10 11 12 13 Students 5 2 4 3 6 5 7 2 6 Solution A=9, C=1, dx = x –A C (x-9) S.D = √∑fd2x - ∑fdx 2 x C Marks x f Fdx d2x Fd2x dx N N 5 5 -4 -20 16 80 6 2 -3 -6 9 18 =√ 264 – 12 2 x1 7 4 -2 -8 4 16 40 40 8 3 -1 -3 1 3 9 6 0 0 0 0 = √ 6.6 – 0.9 10 5 1 5 1 5 = √ 6.51 11 7 2 14 4 28 S.D = 2.551 12 2 3 6 9 18 13 6 4 24 16 96 40 ∑fdx 12 ∑fd2x 264 ILLUSTRATION =08 Find standard deviation for the following data. Values 60 70 80 90 100 110 120 frequency 3 6 9 13 8 5 4 Solution (x –90) A= 90, C=10, dx = x – A fd2x = fdx x dx 10 c 97
  • 14. x f dx fdx fdx2 S.D = √∑fd2x – ∑fdx 2 xC 60 3 -3 -9 27 N N 70 6 -2 -12 24 80 9 -1 -9 9 = √124 – 0 2 x 10 90 13 0 0 0 48 48 100 8 1 8 8 =√2.583 x 10 110 5 2 10 20 = 1.6072 x 10 120 4 3 12 36 S.D = 16.072 N48 ∑fdx ∑fdx 0 124 CONTINUOUS SERIES Calculation of standard deviation In case of grouped series, find mid values and proceed as in case of discrete series. Direct Method ILLUSTRATION –09 Find arithmetic mean and standard deviation for the following data. Class 0-4 4-8 8-12 12-16 16-20 Frequency 2 3 10 3 2 Solution (x - 0) Class x f Mid value fm d d2 Fd2 0 = ∑fm = 200 = 10 0 –4 2 2 4 -8 64 128 N 20 4 –8 3 6 18 -4 16 48 8 –12 10 10 100 0 0 0 S.D = √∑fd2 = √352 12 –16 3 14 42 4 16 48 N 20 16 -20 2 18 36 8 64 128 S.D = 4.19 N= 20 200 0 352 SHORT – CUT METHOD OR ASSUMED MEAN METHOD In the continuous series the method of calculating standard deviation is almost the same as in a discrete frequency distribution. But one additional step here is obtained of mid value. The step deviation method is widely used . Formula for standard deviation calculation. S.D = √Σfd2x – Σfd2x x C N N ILLUSTRATION = 10 Calculate the standard deviation from the following data Class x 0 –10 10 –20 20 –30 30 –40 40 –50 50 –60 60 –70 Frequency 8 12 17 14 9 7 4 Solution = A =35, C=10, dx =(x –A)/c (x –35)/10 98
  • 15. X f Mid x dx fdx fd2x 0 = A + Σfdx x C 0 –10 8 5 -3 -24 72 N 10 –20 12 15 -2 -24 48 = 35 + -302 x 70 20 –30 17 25 -1 -17 17 71 30 –40 14 35 0 0 0 = 35 – 300 = 35 – 4.225 = 30.775 40 –50 9 45 1 9 9 71 50 –60 7 55 2 14 28 S.D =√Σfd2 – Σfdx x C 60 –70 4 65 3 12 36 N N 71 Σfd2x 210 = √210 – –30 2 x 10 71 71 = √2.957 – (-0.422)2 x 10 =√2.7785 x 10 =1.668 x 10 = 16.668 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 50-55 No of worker 170 110 80 45 40 30 25 Solution x – 37.5, A =37.5, C =5 , dx =(x – A)/c 5 mid Fd2x = fdx x dx Age x f dx fdx fd2x x 0 = A+ ∑fdx x C 20-25 170 22.5 -3 -510 1530 N 25-30 110 27.5 -2 -220 440 = 37.5 + –635 x 5 = 31.15 30-35 80 32.5 -1 -80 80 500 35-40 45 37.5 0 0 0 S.D = √∑fd2x – ∑fdx 2 x C 40-45 40 42.5 1 40 40 N N 45-50 30 47.5 2 60 120 =√2435 – –635 2 x 5 = √4.87 –1.613 x 5 50-55 25 52.5 3 75 225 500 500 N= 500 ∑fdx -635 2435 ∑fd2x = 1.804 x 5 = 9.020 ILLUSTRATION = 12 Calculate arithmetic mean and standard deviation from the following data. Wages more than 100 200 300 400 500 600 700 800 No 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 0 = A + ∑fdx xC 100 –200 45 150 -3 -135 405 N 200 –300 88 250 -2 -176 352 = 450 + -128 x 100 300 –400 146 350 -1 -146 146 660 400 –500 206 450 0 0 0 = 450 – 19.39 500 –600 79 550 1 79 79 0 = 430.61 600 –700 52 650 2 104 208 S.D = √∑fd2x – ∑fdx 2 x C 700 –800 30 750 3 90 270 N N 800 -900 14 850 4 56 224 = √ 1684 – –128 2 x 100 = √2.51 x 100 1684 660 660 N= 660 ∑fdx -128 = 1.584 x 100 =158 . 4 ∑fd x 2 IILUSTRATION 13 Calculate arithmetic mean and standard deviation. Age in less than years 10 20 30 40 50 60 70 80 99
  • 16. No of workers 15 30 53 75 100 110 115 125 Solution (x –35)/10, A = 35, C = 10, dx = (x – A)/C X f Mid x dx fdx fd2x 0 = ∑fdx x C = 35 + 2 x 10 = 35 + 20 =35.16 0-10 15 5 -3 -45 135 N 125 125 10-20 15 15 -2 -30 60 20-30 23 25 -1 -23 23 30-40 22 35 0 0 0 S.D = √ ∑fd2x – (∑fdx)2 x C = √ 488 – ( 02 )2 x 10 40-50 25 45 1 25 25 N N 125 125 50-60 10 55 2 20 40 = √3.904 – (0.016) x10 = √ 3.904 – 0.000256 2 60-70 5 65 3 15 45 = 1.975 x 10 70-80 10 75 4 40 160 = 19.75 N=125 ∑fdx 02 488=∑fd2x ILLUSTRATION 14 Calculate the arithmetic mean and standard deviation from the following data. Income under Under 100 10-104.9 105-109.9 110.114.9 115-119.9 120-124.9 125-129.9 No. of workers 20 40 30 20 50 15 5 Solution: - A = 112.45, C =5, dx =(x –A)/C X f Mid x dx fdx fd2x 0 = A+ ∑fdx X C 95-99.9 20 97.45 -3 -60 180 N 100-104.9 40 102.45 -2 -60 160 = 112.45 + -75 x 5 105-109.9 30 107.45 -1 -30 30 180 = 112.45 – 2 .08 = 110.37 110-114.9 20 112.45 0 0 0 S.D = √∑fd2x - ( ∑fdx)2 X C 115-119.9 50 117.45 1 50 50 N N 120-124.9 15 122.45 2 30 60 = √525– -75 2 x 5 = √2.9166 –(0.416)2 x 5 125-129.9 05 127.45 3 15 45 180 180 N = 180 -75 525 = √2.916 – 1.173 =√2.7436 x 5 = 1.656 x 5 = 8.28 COEFFICIENT OFVARIATION = CV The 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 = σ2 ∴ σ = √ variance Coefficient of standard deviation = σ 0 For better comparison purpose, this coefficient of standard deviation is multiplied by 100 gives the coefficient of variation Coefficient of variation = σ x 100 0 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 SERIES USING COEFFICIENT OF VARIATION ILLUSTRATION =15 The index numbers of prices of cotton and cool shares in a year are given below. 100
  • 17. Index no of prices of Index no, of prices Month cotton on shares, x of coal shares January 188 131 February 178 130 March 173 130 April 164 129 May 172 129 June 183 129 July 185 127 August 184 127 September 211 130 October 217 137 November 232 140 December 240 142 Compare the variations of the price of the two shares using coefficient of variation. Solution :- Calculation of coefficient of variation (x – 185) (y –127) X d2 y d2y dx dy 188 3 9 131 4 16 178 -7 49 130 3 9 173 -12 144 130 3 9 164 -21 441 129 2 4 172 -13 169 129 2 4 183 -2 4 129 2 4 185 0 0 127 0 0 184 -1 1 127 0 0 211 26 676 130 3 9 217 32 1024 137 10 100 232 47 2209 140 13 169 240 55 3025 142 15 225 107 7751 549 N=12 ∑dy=57 ∑dx ∑d x 2 ∑d2y 0 = A + ∑dx xC n = 185 + 107 x1 12 = 193.916 y = A + ∑dy x C n = 127 +57 x 1 12 = 127 + 4.75 = 131.75 S.D = √∑d2x - ( ∑dx )2 x C S.D = √∑d2y - ( ∑dy)2 X C n n n n = √7751 – (107)2 x 1 = √ 549 -(57)2 X 1 12 12 12 (12) = √645.92 – (8.916)2 = √45.75 – ( 4.75 )2 = √645.92 – 79.495 = √45.75 - 22.5625 = √566.425 0 = 23.799 = √ 23.1875 = 4.815 C.V= σ x 100 C.V = σ X 100 0 γ = 23.799 x100 = 12.27% = 4.815 x 100 = 3.65% 101
  • 18. 193.91 131.75 Hence cotton shares are more variable in price than the coal shares. ILLUSTRATION = 16 Following are the runs scored by two bats man A & B. Find a. Who is better scored run getter? b. Who is more consistent batsman? A 101 22 0 36 82 45 7 13 65 14 B 97 12 40 96 13 8 85 8 56 15 Solution Ax = 82, Ay = 13 A x – 82 B y – 13 d2x d2y X dx y dy 101 19 361 97 84 7056 22 -60 3600 12 -1 1 0 -82 6724 40 27 729 36 -46 2116 96 83 6889 82 0 0 13 0 0 45 -37 1369 8 -5 25 7 -75 5625 85 72 5184 13 -69 4761 8 -5 25 65 -17 289 56 43 1849 14 -68 4624 15 2 4 -435 29469 300 21762 ∑dx ∑d2x ∑dy ∑d2x 0 = A + ∑dx x C n = 82 + -435 x1 = 38.5 10 y = A + ∑dy x C n =13 + 300 x1 =43 10 x- series y-series S.D =√∑d x - (∑dx) x C 2 2 S.D = √∑d y -( ∑dy)2 xC 2 n n n n =√29469 – (-43.5)2 x 1 =√21762 - (300 )2 x 1 10 10 10 ( 10 ) =√ 2946.9 – 1892.25 = √2176.2 – (30)2 x 1 = √1054.65 = 32.475 =√2176.2 – 900 = √1276.2 = 35.723 C.V = σ x 100 C V = σ x 100 y 0 = 35.723 X 100 = 32.475 x 100 = 84 .35% 43 38.5 = 83.07% Conclusion:- 1. Batsman B is better run getter, because he has scored 430 runs compare to batsman A 385 runs 2. Batsman B is more consistent. COMPARISION OF TWO GROUPS OF DATA USING COEFFICEINT OF VARIATION DISCRETE SERIES ILLUSTRATION = 17 The goals scored by tow teams A & B in the football matches were as follows. Goals 0 1 2 3 4 102
  • 19. Matches A= 27 9 8 4 5 Teams B= 17 9 6 5 3 Find which team is more consistent. Solution A=2 Team –A Team-B Gools Team A, x –2 fAdx fAd2x dx fB fBdx fBd2x x fA dx 0 27 -2 -54 108 -2 17 -34 68 1 9 -1 -9 9 -1 9 -9 9 2 8 0 0 0 0 6 0 0 3 4 1 4 4 1 5 5 5 4 5 2 10 20 2 3 6 12 -49 141 -32 94 N1=53 N2= 40 ∑fAdx ∑fAd2x ∑fBdx ∑fBd2x S.D =√∑fAd2x - (∑fAdx )2 X C S.D =√∑fBd2x - (∑fBdx)2 x C N1 N1 N2 N2 =√141 - (-49) 2 x 1 = √ 94 - (-32)2 x1 53 ( 53) 40 40 = √2.66 – (-0.925)2 x 1 = √1.8052 = 1.3436 = √2.35 – (-0.8)2 x1 0A = A + ∑fadx x C = √ 1.71 = 1.308 N 0B= A + ∑fBdx x C = 2+ -49 x 1 N 53 = 2 + -32 x 1 = 1.2 = 2 –0.925 =1.075 40 C.V = σ X100 = 1.3436 x 100 C.V = σ x 100 = 1.308 x 100 = 109% 0A 1.075 0B 1.2 = 124.93% Team B is more consistent as it has less variation CONTINUOUSSERIES ILLUSTRATION =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.V Life in 000 hrs Company Company X A B 50-59 18 15 60-69 22 24 70-79 26 30 80-89 25 18 90-99 9 13 100 100 Solution Computation of coefficient of variation for both the series. x – 74.5 X Mid x fA fAdx fAd2x fB fBdx fBd2x 10 dx 103
  • 20. 50-59 54.5 -2 18 -36 72 15 -30 60 60-69 64.5 -1 22 -22 22 24 -24 24 70-79 74.5 0 26 0 0 30 0 0 80-89 84.5 1 25 25 25 18 18 18 90-99 94.5 2 9 18 36 13 26 52 100 -15 155 100 -10 154 N 1= N1 ∑fAdx ∑fAd2x N2 ∑fBdx ∑fBd2x S.D =√∑fAd2x - (∑fAdx )2 x C sssS.D =√∑fBd2x - (∑fBdx )2 x C N1 N1 N2 N2 =√155 - (-15) 2 x 10 = √ 154- (-10)2 x10 100 (100) 100 (100) = √1.55-( -0.15) 2 x 10 = √1.54 – (-0.1)2 x10 = √1.55-0.0225 = √ 1.54 – 0.01 x 10 = √1.5275 = 1.2359 x10 =12.359 = √1.53 x 10 = 1.2369 X 10 = 12.369 0A = A + ∑fAdx x C 0B= A + ∑fBdx x C N N =74.5 + (-15) x10 = 74.5 + (-10) X10 100 100 =74.5-1.5 = 73.0 = 74.5 -100 =73.5 C.V = σ X 100 = 12.359 X 100 = 16.93% 100 0A 73 C.V = σ X 100 = 12.369 x 100 = 16.83% 0B 73.5 Conclusion 1. 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 60-70 Frequency A 10 18 32 40 22 18 Frequency B 18 22 40 32 18 10 Solution Computation of coefficient of variation . Frequency A Frequency B Variable x-35 mid x fA fAdx fAd2x dx fA fBdx fBd2x X 10 dx 104
  • 21. 10-20 15 -2 10 -20 40 -2 18 -36 72 20-30 25 -1 18 -18 18 -1 22 -22 22 30-40 35 0 32 0 0 0 40 0 0 40-50 45 1 40 40 40 1 32 32 32 50-60 55 2 22 44 88 2 18 36 72 60-70 65 3 18 54 162 3 10 30 90 140 100 348 140 40 288 0A = A + ∑fAdx X C N1 =35 + 100 X 10 140 = 42.1429 0B = A + ∑fBdx X C N2 = 35 + 40 X 10 140 = 37.8571 A-Series B-Series S.DA =√∑fAd2x - (∑fAdx)2 x C S.DB =√∑fBdx - (∑fBdx )2 x C N1 N1 N2 N2 =√348 - (100) 2 X 10 =√288 - (40) 2 X 10 140 140 140 140 = √2.4857-( -0.7142) 2 x 10 = √2.05714 -(0.2857) 2 X 10 S.D= 14.055 = 14.055 C.V = σ X 100 = 14.055 x 100 C.V = σ X 100 = 14.055 X 100 0A 42.1429 0B 37.8571 = 33.35% = 37.127% Conclusion:- Series B is more Variable ILLUSTRATIONS:-20 Two 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 40-45 No/ of Tyres: x: 1 22 64 10 3 y: 3 21 74 1 1 Solution: computation of coefficient of variation, Brand A Brand B x-32.5 Life in month x mid x fAdx fdx fd2x f fdx fd2x 5 20-25 22.5 -2 01 -2 4 3 -6 +12 25-30 27.5 -1 22 -22 22 21 -21 21 30-35 32.5 0 64 0 0 74 0 0 35-40 37.5 1 10 10 10 1 1 1 40-45 42.5 2 3 6 12 1 2 4 100 -8 48 100 -24 38 01 = 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 = 31.3 105
  • 22. 100 S.D =√∑fd2x - (∑fdx )2 x C S.D =√∑fd2x - (∑fdx)2 X C N1 N1 N2 N2 =√ 48 - (-8) 2 x 5 =√ 38 - (-24) 2 X 5 100 100 100 100 = 3.44 = 2.839 C.V = σ X 100 = 3.44 x 100 C.V = σ X 100 = 2.839 X 100 0 32.1 0 31.30 = 10.72% = 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 follows Factory Average Standard Deviation No. of workers A 345 5 476 B 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? Solution 1. Calculation of total weekly wage payment Total wages paid by factory A= Rs345 x 476 = 164220 Total wages paid by factory B= Rs 285 x 524 = 14 9340 Therefore, factory A pays out larger amount as weekly wages 2. Calculation of coefficient of variation Factory A ------- CV = σ x 100 = 5 x 100 = 1.449% 0 345 Factory B--------- CV = σ x 100 = 4.5 x 100 = 1.578% 0 285 Factory B has greater variability in individual wages since CV of factory B is greater than CV of factory A. ILLUSTRA TION=22 Particulars regarding income of two villages are given below. Particulars Village A Village B Average income 1750 1860 Variance 100 81 State in which village is the variation in income greater ? Solution Calculation of coefficient of variation 1. village . A------CV = σ x 100 =10 x 100 = 0.57% SD =√Variance =√100 = 10 0 1750 2. village. B------ CV = σ x 100 = 9 x 100 = 0.483% S.D √Variance = √81 = 9 0 1860 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? Solution x-Series x-Series C.V = σ = x 100 CV = σ X 100 0 Y 106
  • 23. 58 = 21.2 x 100 69 = 15.6 0 y 0 = 2120 y = 1560 58 69 0 = 36.55 y = 22.6 ILLUSTRAAATION = 24 Coefficient of variation of two series are 75% and 90% and their arithmetic means are 20 & 20 respectively find their standard deviations. x-series y-series CV = σ x 100 CV = σ x 100 0 y 58 = σ x 100 90 = σ x 100 20 20 75 =100σ 90 =100σ 20 20 σ = 75 x 20 = 15 σ = 90 x 20 = 18 100 100 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 up 3. 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 comparison 5. 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 3. What is coefficient of variation? What purpose does it serve. PRACTICAL PROBLEMS 1. 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 90-100 No of workers 12 9 8 5 7 9 [Answers: 0 = 67-6, SD = 18.31] 2. Calculate the standard deviation from the following data. Monthy 78-82 73-77 68-72 63-67 58-62 53-57 48-52 43-47 38-42 33-37 28-32 Exp. No. of 3 6 7 12 17 13 9 7 4 2 1 worker [Answer SD = 11] 107
  • 24. 3. Find mean standard deviation and coefficient of variation from the following data. Age in under year 10 20 30 40 50 60 70 80 No. of persons 15 30 53 75 100 110 115 125 [Answers; 0 = 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 90 No. of students 100 90 80 65 50 20 15 10 5 2 [ Answers: 0 = 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 49 Share B price 108 107 105 105 106 107 104 103 104 101 [ Answers : 0 A = 53, A= 2.646, CV = 4.99 % ] 0B = 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 60 M. 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 consistent batsman? Scores 53 54 55 56 57 58 59 60 Total No/of innings A 2 0 0 4 3 5 3 3 20 B 1 2 3 6 3 3 2 0 20 [ Answers 0A= 57.4, A = 1.9 6 CV = 3.4% 0B = 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 Number of bags x Company A Company B under-100 50 100 100-104.9 150 75 105-109.9 120 125 110-114.9 80 65 115-119.9 130 135 120-124.9 70 140 125-129.9 150 60 130 & above 50 100 Total 800 800 (Ans: 0A = 114.64, °0A=10.57, CVA= 9.22%) (0B=115.75, °0B=11.095, CVB= 9.64%) 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 Number of lamps manufactured X Company A Company –B 108
  • 25. Up to 2000 150 100 2000-3999 150 175 4000-5999 120 125 6000-7999 80 65 8000-9999 130 135 10000-11999 170 140 1200 & above 200 60 1000 800 [Answers 0A = 7399.5, °0A = 4308.13, 0B = 6549.5, °0 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 0 1 = 33.3 0 2 = 20] 109