BUSINESS STATICS

Putting the known values, we have Jl.f:asures ofDisp~rsion 3l=A+--4X0 10 NOTES 100 Self-Instructional A1lterial 91 or A=31 +4=35 A or assumed mean is the midpoint corresponding to the class having X value 0. As the class interval is of 10 and the variable under study is a continuous one, the class for which X= 0 will be 35-5 to 35 + 5, i.e., 30 to 40. A class next lower than this is 30-10 to 30, i.e., 20 to 30. Similarly other classes can be calculated. So all the class intervals are: 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Problem 3: The mean of 50 readings of a variable was 7.43 and their S.D. was 0.28. The following ten additional readings become available: 6.80, 7.81, 7.58, 7.70, 8.05, 6.98, 7.78, 7.85, 7.21 and 7.40. If these are included with original 50 readings, find (1) The mean, (iJ) The standard deviation of the whole set of 60 readings. Solution: Mean of 50 readings = 7.43 Mean of 10 additi.ona1readi.ngs = ~X N· 6.80 + 7.81 + 7.58 + 7.70 + 8.05 + 6.98 + 7.78 + 7.85 + 7.21 + 7.40 10 = 7.516 Meanof 60 readm. gs= 7.43x50+7.516x10 = 371.5+75.16 = 7.44 50+10 60 Standard Deviation, 0.28 = ~X2 - (7.43)2 50 ~xz 0.0784 = - - - 55.2 50 ~x- = (0.0784 + 55.2)50 = 2764.165 Sum of square of 10 additional readings, 46.24 + 61.00 + 57.46 + 59.29 + 64.80 + 48.72 + 60.53 + 61.62 + 51.99 + 54.76 = 566.55 Sum of the square of 60 readings= 2764.165 + 566.55 = 3330.71 :. S.D. of 60 readings = 3330.71 - (7.44)2 60 = ~55.52- 55.35 = ../0.71 = 0.41 Problem 4: The first of two subgroups has 100 items with mean 15 and S.D. 3. If the whole group has 250 items with mean 15.6 and S.D. .JB.44 , find the S.D. of the second group. Solution: Combined A.M. = X= ~ Xj + Ni Xi ~+Ni 15 _6 = (100 X15) + (150 XXi) 250 3900 = 1500 + 150 Xi X2 = 16 Therefore the A.M. of the second group is 16.

!Jeasures ofDispersion Combined S.D., NOTES 100 x 9 + 100 (15 -15.6)2 + 150cr~ + 150 (16 -15.6) 2 92 Self-Instructional Material .J13.44 = 250 13.44 = 900 + 36 + 150~ + 24 or 3360 = 960 + 150cr2 250 2 150crJ = 240, crJ = 16, o-2 = 4 Therefore the S.D. of the second group is 4. Problem 5: You are given the two variables Aand BUsing quartile deviations state which of the two is more dispersed? A B Mdpoint Frequency Mdpoint Frequency 15 15 100 340 20 33 150 492 25 56 200 890 30 103 250 1420 35 40 300 620 40 32 350 360 45 10 400 187 450 140 Solution: To compare the variability comparison of coefficient of quartile deviations is required. Coefficient of quartile deviation is, ~ - <4 ~+Q. MlriableA f Cumulative Mdpoint MlriableB f Cumulative Mdpoint Oass Interval Frequency Frequenc 100 Oass Interval 15 12.5-17.5 15 15 150 340 340 20 17.5-22.5 33 48 200 75-125 492 832 25 22.5-27.5 56 104 250 125-175 890 1722 30 27.5-32.5 103 207 300 175-225 1420 3142 35 32.5-37.5 40 247 350 225-275 620 3762 40 37.5-42.5 32 279 400 275-325 360 4122 45 42.5-47.5 10 289 450 325-375 187 4309 375-425 140 4449 425-475 Q1 has 4N,.1.e., 4289 or 72.25 Q1 has 4N'.I.e., -4424-9 or 1112.25 . below . 1tems 1t. items below it. :. It lies in the group 22.5-27.5 .. It lies in the group 175-225 Q = 22.5 + 72.25 - 48 X 5 Q = 175 + 1112.25-832 x50 I 56 I 890 = 24.67 = 190.7 Q3 has 3: or 216.75 items below it. Q3 has 3: items below it. :. Q3 lies in the group 32.5-37.5 Q3 lies in the group 275-325 Q = 32.5+ 216.75-207 x5 Q = 275+ 3336.75-3142 x50 3 40 3 620 = 33.72 = 290.7

Coefficient of Q.D. Coefficient of Q.D. .Measures ofDispersion = 33.72-24.67 = 0.15 = 290.7 - 190.7 = 0 21 NOTES 33.72 + 26.67 290.7 + 190.7 . Self-Instructional Material. 93 \\s coefficient of quartile deviation for B is higher, it is more variable. ?roblem 6: From the data given about four subgroups, calculate the average and the ;tandard deviation of the whole group. Subgroup NJ. of.Men An>rage «age Standard Deviation (f) «age (f) A 50 B 100 61.0 8 120 70.0 9 c 30 80.5 10 83.0 II D 300 Solution: .Men An>rage NX. 0' NJ )(_)( I'(X. - )(cj N ~ c Sub- 50 61 3050 8 3200 -13 8450 group 100 70 7000 9 8100 -4 1600 120 80.5 9660 10 12000 6.5 5070 A 30 83 2490 II 3630 9.0 2430 B 300 26930 17550 c D 22200 Combm. ed Mean - = LNX = 22200 = ~ 74 (X) -- -- c l\\ 300 (Combined Standard Deviation)2 L + L X- Xc)= - N-a-2 N( 2 .=..._____;__ _.::..:...__ LN LN = 26930 + 17550 = 44480 = 148.27 300 300 300 cr = .JI48.27 = ~ 12.18 Problem 7: For a certain group of wage-earners, the median and quartile wages per week were~ 44.3, ~ 43.0 and~ 45.9 respectively. Wages for the group ranged between~ 40 and ~50. 10 per cent of the group had under~ 42 per week, 13 per cent had~ 47 and over and 6 per cent ~ 48 and over. Put these data into the form of a frequency distribution, and hence obtain an estimate of the mean wage and the standard deviation. Solution: Assuming that the group has I00 workers the frequency distribution will take the following shape. Earnings NJ.of Midvalue !{X) d fd rei f «age- (X) earners (f) 40-42 10 41.00 410 -3.50 -35 122.50 15 42.50 637.50 -2.00 -30 60.00 42-43 - 25 43.65 1091.25 -21.25 18.06 25 45.10 1127.50 0.85 43-44.3 12 46.45 557.40 +0.60 15.00 9.00 44.3-45.9 7 47.50 332.50 23.40 43.63 45.9-47 6 49.00 294.00 1.95 21.00 63.00 47-48 3.00 27.00 121.50 48-50 L f= 100 4450.15 4.50 437.69

~ures ofDispersion .X = \"£/X = 4450.15 = 't 44.50 NOTES N 100 94 Self-Instructional Material 4.12 SUMMARY In this unit, you have learned about the various measures of dispersion. The measures of dispersion are important property ofa distribution and needs to be measured by an appropriate statistics. It describes absolute and relative measures of dispersion. In its absolute form it states the actual amount by which the value of an item on an average deviates from a measure of central tendency. Absolute measures are expressed in concrete units, i.e., units in terms of which the data have been expressed, e.g., rupees, centimetres, kilograms, etc., and are used to describe frequency distribution. A relative measure ofdispersion is a quotient obtained by dividing the absolute measures by a quantity in respect to which absolute deviation has been computed. It is as such a pure number and is usually expressed in a percentage form. Relative measures are used for making comparisons between two or more distributions. . In engineering problems too, the variability is an important concern. The amount of variability in dimensions ofnominally identical components is critical in determining whether or not the components of a mass-produced item will be really interchangeable. Thus, the measures of dispersion are useful in determining how representative the average is as a description of the data, in comparing two or more series with regard to their scatter, in designing a production control system which is based on the premise that if a process is under control, the variability it produces is the same over a period of time. You have learned the meaning and significance of range. You have also learned how to calculate quartile deviation, mean deviation and standard deviation. 4.13 ANSWERS TO 'CHECK YOUR PROGRESS' 1. Absolute measure of dispersion states the actual amount by which an item on an average deviates from a measure of central tendency. 2. Relative measure ofdispersion is a quotient computed by dividing the absolute measures by a quantity in respect to which absolute deviation has been computed. 3. The range of a set of numbers is the difference between the maximum and minimum values. It indicates the limits within which the values fall. 4. If quartile deviation is to be used for comparing the variability of any two series, it is necessary to convert the absolute measure to a coefficient ofquartile deviation. To do this the absolute measure is divided by the average size of the two quartile. Symbolically, Coefficient of Quartile Deviation = ~ - q ~+q 5. Mean deviation also called average deviation, of a frequency distribution is the mean of the absolute values of the deviation from some measure of central tendency. In other words, mean deviation is the arithmetic average of the variations (deviations) of the individual items of the series from a measure of their central tendency. 6. (a) It is easy to understand. (b) As compared to standard deviation (disc'ussed later), its computation is simple. (c) As compared to standard deviation, it is less affected by extreme values. (d) Since it is based on all values in the distribution, it is better than range or quartile deviation.

7. Standard deviation, cr (sigma) is defined as the square root of the mean of the squares ~asures ofDispersion of the deviations of individual items from their arithmetic mean. NOTES 8. 2.14 9. 4.09 Self-Instructional Miterial 95 10. (a) 7.46 (b) 5.4 approx 4.14 QUESTIONS AND EXERCISES Short-Answer Questions 1. What is range? How is it measured? 2. Write the definition and formula of quartile deviation. 3. How will you calculate the mean deviation of a given data? 4. Explain standard deviation. Why is it used in statistical evaluation of data? 5. Calculate standard deviation for the series 1, 2, 3, 5, 7. 6. For a group of 50 male workers the mean and standard deviation of their weekly wages are ~ 63 and ~ 9 respectively. For a group of 40 female workers these are ~ 54 and ~ 6 respectively. Find the standard deviation of the combined group of 90 workers. 7. (a) Mean and standard deviations of two distributions of 100 and 150 items are 50, 5 and 40, 6 respectively. Find the mean and standard deviations of all the 250 items taken together. (b) Mean and standard deviations of 100 items are found by a student as 9 and 5. If at the time of calculations two items are wrongly taken as 40 and 50 instead of 60 and 30, find the correct mean and standard deviations. Long-Answer Questions 1. Calculate mean deviation and its coefficient about median, arithmetic mean and mode for the following figures, and show that the mean deviation about the median is least. 103, 50, 68, 110, 108, 105, 174, 103, 150,200,225, 350, 103 2. Compute mean deviations of the two series and point out which is more variable. Ahnth IndexNJ. IndexNJ. Ahnth IndexNJ. Index NJ. Calcutta Delhi Calcutta Delhi 1970 April 93 107 1970 October 97 107 1970 May 97 108 1970 November 97 105 1970June 95 102 1970 December 92 101 1970 July 95 102 1971 January 93 100 1970 August 95 102 1971 February 89 97 104 96 1970 September 95 1971 March 89 3. Calculate (a) Median coefficient of dispersion and (b) Mean coefficient of dispersion from the following data: Size of Items 14 16 18 20 22 24 26 Frequency 24 5 32 14 4. Compute the mean deviation from the median and from the mean for the following distribution of the scores of 50 college students. Also complete the class interval. Scores 140- 150- 160- 170- 180- 190-200 Frequency 4 6 10 18 9 3

Af::asures ofDispersion 5. Find the mean deviation about the mean of the following data of ages of married men in a certain town. NOTES Ages 15-24 25-34 35-44 45-54 55-64 65-74 No. of Men 33 264 303 214 128 58 6. Calculate the mean deviation from the following data. What light does it throw on the social conditions of the community? Difference in age between husband and wife: Difference in Years: 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 Frequency: 449 705 507 281 109 52 16 4 7. The following figures give the income of 10 persons in rupees. Find the standard deviation. 114, 115, 123, 120, 110, 130, 119, 118, 116, 115 8. Calculate the mean and standard deviation ofthe following values ofthe world's annual gold output millions of pound (in for 20 different years) 94, 95, 96, 93, 87, 79, 73, 69, 68, 67, 78, 82, 83, 89, 95, 103, 108, 117, 130,97. Also calculate the percentage of cases lying outside the mean at distances ±a, ±2a, ±3a where a denotes standard deviation. 9. Calculate standard deviation from the following data: Size of Item 6 7 8 9 10 11 12 Frequency 3 6 9 13 8 5 4 10. Calculate the arithmetic mean and the standard deviation from the following data: Class Interval 5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45 Frequency 6 5 15 10 5 4 32 11. Calculate the mean and the standard deviations from the following data: Age Group Below 20-25 25-30 30-35 35-40 40-45 45-50 50-55 55 and above No. of Employees 26 44 60 101 109 846 66 10 12. Calculate mean and standard deviation from the following data: Age Under 10 20 30 40 50 60 70 80 No. of Persons Dying 15 30 53 75 100 110 115 125 13. The marks obtained by the students of class A and Bare given below: Marks 5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45 Class A 1 10 20 8 6 3 1 Class B 5 6 15 10 5 4 2 2 Calculate mean, median, mode and standard deviation for the distributions. Explain your results regarding composition of the class in respect to intelligence. 14. Explain clearly the ideas implied in using arbitrary working origin and scale for the calculation of the arithmetic mean and standard deviation of frequency distribution. The values of arithmetic mean and standard deviation of the following frequency distribution of a continuous variable derived from analysis are ~ 135.33 and~ 9.6 respectively. Find the upper and lower limits of the various classes: X -4 -3 -2 -1 0 +1 +2 +3 f 2 5 8 18 22 13 8 4 96 Self-Instructional Material

15. (a) Mean of 100 items is 50 and their standard deviation is 4. Find the swn and swn of .Measures ofDispersion squares of all the items. NOTES (b) The mean and the standard deviation ofa sample of 100 observations were calculated as 40 and 5.1 respectively by a student, who took by mistake 50 instead of 40 for one observation. Calculate the correct mean and standard deviation. 16. The following data gives the arithmetic averages and standard deviations of the three subgroups. Calculate the arithmetic average and standard deviation of the whole group. Subgroup NJ. of.Men Average S.D. of Wlge(() Wlge(() A 50 61.0 8.0 B 100 70.0 9.0 c 120 80.5 10.0 17. For a group containing 100 observations, the arithmetic mean and standard deviation are 8 and ..!fD.5. For 50 observations selected from these 100 observations the mean and the standard deviation are 10 and 2 respectively. Find the arithmetic mean and the standard deviation of the other half. 18. A group has cr = 10, N = 60, cr2 = 4. A subgroup of this has Xj = 11, 1\\{ = 40, crf = 2.25. Find the mean and standard deviation of the other subgroup. 19. Two cricketers scored the following runs in the several innings. Find who is a better run-getter and who is more consistent player. A 42, 17, 83, 59, 72, 76, 64, 45, 40, 32 B 28, 70, 31, 0, 59, 108, 82, 14, 3, 95 20. The following are some of the particulars of the distribution of weights of boys and girls in a class. Boys Girls Nwnber 100 50 Mean weight (kg) 60 45 Variance 94 (a) Find standard deviation of the combined data. (b) Which of the two distributions is more variable? 4.15 FURTHER READING Kothari, C.R. 1984. Quantitative 1ixhniques, 3rd Edition. New Delhi: Vikas Publishing House Pvt. Ltd. Chandan, J.S. 1998. Statistics for Business and Economics. New Delhi: Vikas Publishing House Pvt. Ltd. Chandan, J.S., Jagjit Singh and K.K. Khanna. 1995. Business Statistics, 2nd Edition. New Delhi: Vikas Publishing House Pvt. Ltd. Self-Instructional MJ.terial 97



UNIT 5 VARIANCE, MOMENTS, lilriance, !vf:Jnrnts, Skewness SKEWNESS AND KURTOSIS and Kurtosis Structure NOTES 5.0 Introduction Self-Instructional Jl.-hterial 99 5.1 Unit Objectives 5.2 Variance and Coefficient of Variation 5.3 Lorenz Curve 5.4 Moments 5.5 Moments about the Mean 5.6 Skewness 5.6.1 Karl Pearson's Measure of Skewness 5.6.2 Bowley's (Quartile) Measure of Skewness 5.6.3 Kelly's (Percentile) Measure of Skewness 5.7 Kurtosis 5.8 Solved Problems 5.9 Summary 5.10 Answers to 'Check Your Progress' 5.11 Questions and Exercises 5.12 Further Reading 5.0 INTRODUCTION In this unit, you will learn about variance, moment, skewness and kurtosis used to measure distribution and deviation. A whole series of measures is known as moments which when properly interpreted gives a wealth of information about the 'shape' of the distribution. It will be seen that the arithmetic mean (X) and the standard deviation (cr) are the first two members of the series. The symmetry and the peakness of the distribution can also be obtained from the higher members of this series. Though the concept of moments is a highly mathematical one, an elementary introduction which brings out the physical significance is discussed in this unit. Moments about the mean, skewness and kurtosis are also discussed in this unit. 5.1 UNIT OBJECTIVES After going through this unit, you will be able to: • Calculate variance and coefficient of variation • Measure deviation from the average using 'Lorenz Curve' • Define the concept of moments and their significance • Calculate the moment about the mean • Describe when frequency distribution is called skewed • Evaluate Karl Pearson's measure of skewness • Differentiate between Bowley's and Kelly's measure of skewness • Explain kurtosis and convexity of curves

l!:lriance, !vbments, Skewness 5.2 VARIANCE AND COEFFICIENT OF VARIATION. and Kurtosis The square of standard deviation, namely d, is termed as variance and is more often NOTES specified than the standard deviation. Clearly, it has the same properties as standard deviation. 100 Self-Instructional Material As is clear, the standard deviation cr or its square, the variance, cannot be very useful in comparing two series where either the units are different or the mean values are different. Thus, a cr of 5 on an examination where the mean score is 30 has an altogether different meaning than on an examination where the mean score is 90. Clearly, the variability in the second examination is much less. To take care ofthis problem, we define and use a coefficient ofvariation, V, V = cr x I00 expressed as percentage. :X Example 5.1: The following are the scores of two batsmen A and B in a series of innings: A 12 115 6 73 7 19 119 36 84 29 B 47 12 76 42 4 51 37 48 13 0 Who is the better run-getter? Who is more consistent? Solution: In order to decide as to which of the two batsmen, A and B, is the better run- getter, we should find their batting averages. The one whose average is higher will be considered as a better batsman. To determine the consistency in batting we should determine the coefficient of variation. The less this coefficient the more consistent will be the player. A B Scores X ~ Scores X ~ X -38 1,444 X 14 196 +65 4,225 -21 441 12 -44 1,936 47 43 1,849 115 +23 12 -43 529 76 9 81 6 -31 1,849 42 -29 841 73 +69 -4 324 7 -14 961 51 18 19 +34 4,761 37 4 16 119 -21 48 15 225 36 196 -20 400 84 1,156 13 -33 1,089 29 441 0 5,462 I;x =500 17,498 I;x =330 Batsman A: Batsman B: :X = 500 =50 :X = 330 = 33 10 10 cr = ~17,498 = 4 1.83 ~cr = 5•462 = 23.37 10 10 v = 23.37 X 100 v = 41.83 X }00 33 50 = 70.8 per cent = 83.66 per cent

A is a better batsman since his average is 50 as compared to 33 of B. But B is more 1-flriance, ~ments, Skewness consistent since the variation in his case is 70.8 as compared to 83.66 of A and Kurtosis Example 5.2: The following table gives the age distribution of students admitted to a NOTES college in the years 1914 and 1918. Find which of the two groups is more variable in age. Self-Instructional Material 101 Age Number ofStudents in 15 1914 1918 16 I6 3 34 17 8 22 18 12 35 19 14 20 20 13 7 21 5 19 22 23 23 3 24 25 26 27 Solution: A5Slnned 114ean-21 A5sumed 114ean-19 Age 1914 1918 f x' IX' IX-2 f x' IX fxa 15 0 -6 0 0 I --4 --4 16 16 I -5 -5 25 6 -3 -18 54 17 3 --4 -12 48 34 -2 --68 136 18 8 -3 -24 72 22 -I -22 22 19 12 -2 -24 48 -112 20 14 -I -14 14 35 0 0 0 20 I 20 20 -79 7 2 14 28 19 3 57 171 21 13 0 0 0 3 4 12 48 22 5 I 5 5 147 +103 495 23 2 2 4 8 24 3 3 9 27 -9 25 I 4 4 16 26 0 5 0 0 27 I 6 6 36 63 +28 299 -51 1914 Group: z2 ra L -[L(:') 299 -(~)2 63 63

lariance, MJnrnts, Skewness = .J4.476- 0.655 = \"'4.091 and Kurtosis = 2.02 NOTES x = 21 + ( ~1 ) = 21 - 8 = 20.2 Check Your Progress v = 2.02 X 100 20.2 1. Coefficients of variation of two series are 58% and =202=10 69%. Their standard 20.2 deviations are 21.2 and 15.6. What are their 1918 Group: arithmetic mean? cr 495 ( - 9 ) 2 = .J3.3673- 0.0037 2. When can coefficient of variations be greater than 147 147 100%? What can you say about the items ofthe given = .J3.3636 = 1.834 data in such a case? X = 19 +C~) 102 Self-Instructional Material = 19-0.06 = 18.94 V = }. 834 X 100 18.94 = 9.68 The coefficient of variation of the 1914 group is 10 and that of the 1918 group 9.68. This means that the 1914 group is more variable, but only barely so. Example 5.3: You are supplied the following data about the height ofboys and girls studying in a college. Boys Girls Number 72 38 Average height (inches) 68 61 Variance of distribution 94 You are required to find out: (a) In which sex, boys or girls, is there greater variability in individual heights. (b) Common average height in boys and girls. (c) Standard deviation of the height of boys and girls taken together. (d) Combined variability (C.V.). Solution: (a) C.V. of boys' height=~ x 100 = J9 x 100 = 4.41% x1 68 C.V. of girls' height= ~ x 100 = J4 x 100 = 3.28% x2 61 Thus there is a greater variability in the height of boys than that of the girls. (b) Combined height of boys and girls is given as, 72 X 68 + 38 X 61 7214 = 65.58 m. ches approx. ------ = -- 72 + 38 110

(c) The combined standard deviation may be calculated by applying the following lilriance, Nfoments, Skewness formula: and Knrtosis NOTES 72x9+38x4 72(65.58-68)2 +38(65.58-61)2 ----------+--~----~----~------~ 72 + 38 72 + 38 2018.794 = 18.35 110 a 12 4.28 inches (d) Combined variability= ~ x 100 = 4·28 x 100 = 6.53 \"X 65.58 Example 5.4: In a co-educational college boys and girls formed separate groups on the :oundation day when everyone had to put in physical labour. Compute standard deviation :or boys and girls separately and for the combined group. Did the separation by sex make !ach workgroup more homogeneous? Minutes ofLabour Given by NJ. ofGirls NJ. ofBoys Each Individual 20 120 60 60 100 55 100 200 50 450 355 45 450 350 40 300 500 35 250 350 30 100 20 25 Solution: Minutes X NJ. of .,x rI.x2 NJ.of f2X f~/2 ofLabour = -X-4-5 Girls Boys 5 Given by lj '2 Each Indivi- dua/(X) 60 +3 20 60 180 120 360 1080 55 +2 60 120 240 100 200 400 50 +I 100 100 100 200 200 200 45 0 450 0 0 355 0 0 40 -1 450 -450 450 350 -350 350 35 -2 300 -600 1200 500 -1000 2000 30 -3 250 -750 2250 350 -1050 3150 25 -4 100 -400 1600 20 -80 320 Total 1730 -1920 6020 1995 -1720 7500 Self-Instructional Material 103

lariance, JVbnx:nts, Skewness Girls: X- I = 45 - 11793200 x 5 = 45 - 5.55 = 39.45 and Kurtosis Boys: cr 1 = -60-20 - (--19-20)2 X5=.J3.4798 -1.2317 X5 = 1.5 X 5 = 7.5 NOTES 1730 1730 104 Self-Instructional Material V. = .22_x 100 = 19.0% I 39.45 X- 2 = 45 - 11979205 X5 = 45 - 4.31 = 40.69 a2 = 750 (-1720) 2 X5 = ~(3.7594- 0.7434) X5 1995- 1995 = 1.7366 X5 = 8.68 V:: = 8·68 X100 2 40.69 = 21.34% c b. d M X 1730 X 39.45 + 1995 X40.69 om me ean, 12 = 1730 + 1995 = 149424 .78 = 40.11 3725 d1 = 0.66 and ~ = -0.58 cr2 = 1730 (7.5) 2 + 1995 (8.68) 2 + 1730 (0.66) 2 + 1995 (-0.58) 2 12 1730 + 1995 = 66.86 cr 12 = 8.18 C.V. = 8·18 X 100 = 20.38% 40.11 Therefore, there doesn't seem to be any evidence that each workgroup was more homogeneous than the total population. Example 5.5: The values of the arithmetic mean and the standard deviation ofthe following frequency distribution of a continuous variable derived from short-cut method are 135.3 lbs and 9.6 lbs respectively. X -4 -3 -2 -1 0 2 3 Total Frequency 2 5 8 18 22 13 8 4 80 Determine the actual class interval. Solution: Calculation of standard deviation: X -4 -3 -2 -1 0 2 3 Total Frequency (I} 2 5 8 18 22 13 8 4 80 /(X) -8 -15 -16 -18 0 13 16 12 -16 f(X) 32 45 32 18 0 13 32 36 208 I -(IStandard Deviation = i x ~X) ~X)r .. Putting the known values, we have, 9.6 = jx 208 -( -16r = j X .../2.6- 0.04 80 80

or 9.6 = jx .)2.56 = jx 1.6 1-ilriance, MJnrnts, Skewness and Kllrtosis i= 9.6 =6 NOTES 1.6 Self-Instructional Material 105 Arithmetic Mean = A+ 'if( X) x i n :. Putting the known values, we have 135.3 = A+ -l6 X 6 =A- 1.2 80 or A= 135.3 + 1.2 = 136.5 A or assumed mean is the midpoint corresponding to the class having X value 0. As the class interval is of 6 and the variable under studying is a continuous one, the class for which X= 0 will be 136.5 -3 to 136.5 + 3, i.e., 133.5-139.5. A class next lower than this is 133.5-6 or 133.5, i.e., 127.5 to 133.5. Similarly other classes can be calculated. So all the class intervals are: 109.5-115.5 115.5-121.5 121.5-127.5 127.5-133.5 133.5-139.5 139.5-145.5 145.5-151.5 151.5-157.5 Determining Overall Performance If for a number of candidates each writes papers in several subjects, then the total marks obtained by candidates will not be a correct basis for determining their merit. This is also because the marks in different subjects are likely to have different amounts of spread and means. The extent of the spread of marks in a paper introduces 'weights' which will be to the advantage of those getting high marks in subjects where the spread is great, and to the disadvantage ofthose gaining high marks in subjects where range is small. A method to for the errors so introduced consists of converting the marks into 'standard scores' defined as X- X . This measures deviation of marks of a student from the mean a of that subject in the units of standard deviation and are termed as z-scores as well. This corrects both for variations of X and a in different subjects. The .cscores of a student in different subjects are then added to give a true measure of relative performance. Example 5.6: Candidate Marks in -- Economics Commerce Total 159 A 84 75 159 B 74 85 Average for Economics is 60 with standard deviation 13 Average for Commerce is 50 with standard deviation 11 Whose performance is better, A's orB's? Solution: z-Scores A Economics 84 - 60 = 1.85} 13 Collllrerce = 2.27 4.1 2 75-50 11 B. Economics 74 1-3 60 = 1.08} 85-50 4•26 Collllrerce = 3.28 11 Since Bs z-score is higher therefore his performance is better.

lflriance, M:Jments, Skewness 5.3 LORENZ CURVE and Kurtosis The Lorenz curve is a graphic method of measuring deviations from the average. It was devised by Max 0. Lorenz for measuring the inequalities in the distribution of wealth. But NOTES it can be applied with equal advantage for comparing the distribution of profits amongst different groups of business and such other things. It is a cwnulative percentage curve. In it the percentage of items are combined with the percentages of such other things as wealth, profits or tum-over, etc. In drawing a Lorenz curve the following steps are necessary: (a) The various groups of each variable should be reduced to percentages. Thus, if it is desired to show the distribution of income amongst the various groups of population of a country, the various groups of population should be reduced in the form of percentages of total population; so also the incomes derived by these groups in terms of the total income of the country. (b) The two sets of the percentages obtained by step 1 should then be used to determine the cumulated and cumulative percentages. (c) The cumulative percentages of these two variables should then be plotted along the axis of Yand axis of X The scale along the axis of Ybegins from zero at the point of intersection and goes upward up to l 00, while the scale along the axis of Xbegins with 100 at the point of intersection and goes up to zero towards the right. (d) The points 100, 100 along the axis of Yand the points 0, 0 along the axis of X should be joined by a straight line. The line so obtained is called the line ofequal distribution, and serves as the basis for the determination of the extent to which the actual distribution deviates from the ideal distribution given by this line. (e) The actual data map now be plotted on this graph in the ordinary manner and the plotted points may be connected by means of a curve. The farther the curve obtained under step 5 is from the line of equal distribution, the greater is the deviation. Example 5.7: The following table gives the population and earnings of two towns A and B. Represent the data graphically so as to bring out the inequality of the distribution of earnings. Town A TownB Persons Earning (!Alily) Process Earning (!Alily) 100 75 50 80 100 100 70 120 100 150 30 60 100 225 25 140 100 325 100 200 100 375 45 200 100 450 30 140 100 600 80 460 100 850 20 120 100 1850 50 480 lOOO 5000 500 2000 106 Self-Instructional Material

Solution: 1-ariance, MJnrnts, Skewness and Kurtosis Population and Daily Earnings ofTwo Towns A and B NOTES Town A TownE Persons Earnings Persons Earnings Cumu- Cumu- Cumu- Cumu- Cumu- Cumu- Cumu- Cumu- lative lative lative lative lative lative lative lat/ve Total Percentage Total Percentage Total Percentage Total Percentage 100 10 75 1.5 50 10 80 4 200 20 175 3.5 120 24 200 10 300 30 325 6.5 150 30 260 13 400 40 550 11 175 35 400 20 500 50 875 17.5 275 55 600 30 600 60 1250 25 320 64 800 40 700 70 1700 34 350 70 940 47 800 80 2300 46 430 86 1400 70 900 90 3150 63 450 90 1520 76 1000 100 5000 100 500 100 2000 100 I~ \"\"i\\,\\, LINE OF £QUALDIS11UBtmON 7\\ !i '.\\ ~~ \"\"\"i \\. \",\\ .B \"\"\"~ so ..!!! I \"\"\" \"\"\"40 '\\ ·\",'-.'....., \"\"'20 ~~ 1'. ~ \"\"'\"\"1 ' ~ .::- ~ ~ 100 90 tiO 70 60 so 40 30 20 10 0 I'ER.CIDITAOB OF POPULATION 5.4 MOMENTS The term moment is obtained from mechanics where the 'moment of a force' describes the tendency or capacity of a force to turn a pivoted lever (Fig. 5.1). Force Fj has a counter clockwise moment Fjx1 and force Fz has a clockwise moment FzlS· Force .Fj Force Fz PIVOT Fig. 5.1 Self..Instructional Material 107

lluiance, AfJments, Skewness and Kurtosis NOTES 0 Fig. 5.2 For a frequency distribution we imagine that at a distance x from the origin 0, a force equal to the frequency (f) associated with x acts and thus the moment about 0 is equal to f(x). Taking the contributions of the whole distribution, the moment then is \"'i:.f(x). This then, is, the moment on the lever produced if the whole distribution was sitting on a lever pivoted at the origin. To correct for the number of items involved (since we are interested in specifying only the 'shape' of the distribution) we divide by \"f.f, i.e., the total number of items. This 'first' moment about the origin then is v11 (nu one prime). l: f(x) (5.1) v11 = r : f Just like in mechanics, we can define higher order moments as well by multiplying f by higher powers of x: Thus I l: fil- v3l _ l :l &: f and Y, = -- , soon. 2 - l: f In genera1, v I = -l: I-X (5 .2) r l: f Instead of taking the moments about the origin we may also take them about any other point x0 (equivalent to pivoting the lever bar about that point). Then, l: f(x-x0 l (5.3) vr= l: f Thus, v; is a special case of vr with x0 = 0. The series of moments with x0 = x, i.e., moments about the mean have special significance and are denoted by J.l (mu). l: f(x- x)r (5.4) J.lr= l:f 5.5 MOMENTS ABOUT THE MEAN The moments of a frequency distribution about the mean :X have special significance. This is explained in detail. The zeroth moment J.lo is by equation (5.4) l: f(x-x)0 J.lo = l: f But since any number raised to power zero is one, it is clear that, l:f J.lo = l: f = 1 For all distributions. Similarly, But, by definition of the mean :X, the algebraic sum of the deviations, i.e., \"f. f(x- :X) is zero, so that JJ1 = 0. For all distributions. 108 Self-Instructional Material

Next J.lz =I: f(x- x)2/I: f is by definition the variance of the distribution. Thus, Vtlriance, JJ:Jments, Skewness and Kurtosis J.lo = 1 NOTES J.ll = 0 (5.5) J.lz = if For all distributions. There is yet another point which can be deduced about the moments. We have already seen, while discussing the properties of the arithmetic mean that the sum of deviations below the mean equals the sum of deviations above the mean. This shows that the negative and positive deviations cancel out. This would be so in all distributions whether symmetrical or asymmetrical, when the deviations are raised to the first power. When the deviations are raised to any even power, their signs will all be positive and will no longer cancel out. When the deviations are raised to any odd power (other than 1) and the sum of the negative deviations equals the sum ofthe positive deviations, the distribution is symmetrical. Thus in symmetrical distribution only, J.L 3 = 0 J.ls = 0 J.L7 = 0, etc. For this reason we can use these moments as measures ofasymmetry. Example 5.8: Find the first four moments about the mean from the following data: Class Interval 0-10 10-20 20-30 30--40 Frequency I 3 4 2 Solution: x = - -x - 2 5 Size X f 10 IX fj(2 fj(3 fj(4 0-10 5 I -2 -2 4 -8 +16 10--20 15 3 -I -3 3 -3 +3 20--30 25 4 0 0 0 0 0 30--40 35 2 I 2 2 2 +2 10 -3 9 -9 21 Moments about Arbitrary Mean v= -3 X 10 =-3 v. = _120_ x 1o2 = 90 I 10 2 V. = - 9 X 103 = -900 V= 31_0!. X 104 = 21000 3 10 4 5.6 SKEWNESS When a frequency distribution is not symmetrical it is said to be asymmetrical or skewed. The nature of symmetry and the various types of asymmetry are illustrated in the given example. The following table shows the heights of the students of a college: Class A BcD Interval ffff 56.5-58.5 58.5-60.5 5304 60.5-62.5 25 5 4 8 62.5-64.5 15 20 40 20 64.5-66.5 10 44 24 24 66.5-68.5 15 20 20 40 68.5-70.5 25 5 8 4 534 0 Self-Instructional Material 109

lflriance, MJments, Sla:wness N 100 100 100 100 and Kurtosis 63.5 63.5 63.5 63.5 M:an(~) 63.5 63.5 63 64 NOTES ~ian(Mi) - 63.5 61.9 65.1 MJde(MJ) The histograms and the corresponding curves are drawn in Figs. 5.3 and 5.4. A l,f --A I\"\\. , 1-- x!oMd I\\ /\\ 56.5 I r--1 B r+- 70.5 ;-- f-- r r---r- Il-l BI 56f5 .:.>!'f_- M.rp -- ~d 7 .5 J\\ '- / ~ Fig. 5.3 A glance at the data of each of the four classes given above makes a very interesting study. The shape of the curves, histograms and placement of equal items at equal distances on either side of the median clearly show that distributions A and Bare symmetrical. If we fold these curves, or histograms on the ordinate at the mean, the two halves of the curve or histograms will coincide. In distribution B, all the three measures of central tendency are identical. In A, which is a bimodal distribution, mean and median have the same value. Distributions Cand Dare asymmetrical. This is evident from the shape ofthe histograms and curves, and also from the fact that items at equal distances from the median are not equal in number. The three measures of central tendency for each of these distributions are of different sizes. A point of difference between the asymmetry of distribution Cand that of Dshould be carefully noted. In distribution G where the mean (63.5) is greater than the median (63) and the mode (61.9), the curve is pulled more to the right. In distribution Dwhere mean (63.5) is lesser than the median (64) and mode (65.1) the curve is pulled more to the left. In other words, we may say that if the extreme variations in a given distribution are towards higher values they give the curve a longer tail to the right and this pulls the median and mean in that direction from the mode. If, however, extreme variations are towards lower values, the longer tail is to the left and the median and mean are pulled to the left of the mode. It could also be shown that in a symmetrical distribution the lower and upper quartiles are equidistant from the median, so also are corresponding pairs of deciles and percentiles. This means that in a asynnnetrical distribution the distance of the upper and lower quartiles from median is unequal. 110 Self-Instructional Material

c~ I c 1\\ liiriance. l'v.foment~, Skwness I and Kurtosis I \\ I \\ NOTES I 1\\. :I TI I - Self-Instructional Material 111 I II \"J !\"-.. I II I II I II I II ht- r-- I II I II I II 56.5 I i I L ... 7 05 I I MEAN I • I I ~--- MEDIAN I~------ MO'DE D t-- Dn I r-lr- I I 56.5 I I I I I ll II -v I II \\ II r--.. II II 1-l .lII 70j5 I L...,MODE '- • MJ:DIA~ \"'MEAN Fig. 5.4 From the above discussion, we can summarize the tests for the presence of skewness as follows: 1. When the graph of the distribution does not show a symmetrical curve. 2. When the three measures of central tendency differ from one another. 3. When the sum of the positive deviations from the median are not equal to the negative deviations from the same value. 4. When the distances from the median to the quartiles are unequal. 5. When corresponding pairs ofdeciles or percentiles are not equidistant from the median. Measures of Skewness On the basis of the above tests, the following measures of skewness have been developed: 1. Relationship between three measures of central tendency--commonly known as theKarl Pearson's measure of skewness. 2. Quartile measure of skewness-known as Bowley's measure of skewness. 3. Percentile measure of skewness-also called the Kelly's measure of skewness. 4. Measures of skewness based on moments. All these measures tell us both the direction and the extent of the skewness. 5.6.1 Karl Pearson's Measure of Skewness It has been shown earlier that in a perfectly symmetrical distribution, the three measures of central tendency, viz., mean, median and mode will coincide. As the distribution departs from symmetry these three values are pulled apart, the difference between the mean and mode being the greatest. Karl Pearson has suggested the use of this difference in measuring skewness. Thus Absolute Skewness = Mean -Mode. (+) or (-) signs obtained by this formula would exhibit the direction of the skewness. If it is positive, the extreme variation in the given distribution is towards higher values. If it is negative, it shows that extreme variations are towards lower values. Pearsonian Coefficient of Skewness The difference between mean and mode, as explained in the preceding paragraph, is an absolute measure of skewness. An absolute measure cannot be used for making valid comparison between the skewness in two or more distributions for the following reasons: (1) The same size of skewness has different significance in distributions with small variation.

lflriance, !1-f:Jments, SkeWIIess and in distributions with large variation, in the two series, and (il) The unit of measurement and Kurtosis in the two series may be different. NOTES To make this measure a suitable device for comparing skewness, it is necessary to eliminate from it the disturbing influence of 'variation' and 'units of measurements'. Such 112 Self-Instructional Material elimination is accomplished by dividing the difference between mean and mode by the standard deviation. The resultant coefficient is called Pearsonian Coeflicient ofSkewness. Thus, the formula of Pearsonian Coefficient of Skewness is, Coefficient ofSkewness = - -Me-an--M-od-e - Standard Deviation Since, as we have already seen, in moderately skewed distributions that, Mode = Mean- 3 (Mean- Median) We may remove the mode from the formula by substituting the above in the formula for skewness, as follows: Coefficient of Skewness = Mean- [Mean- 3(Mean- Median)] Standard Deviation Mean- Mean+ 3(Mean- Median) Standard Deviation 3(Mean- Median) a The removal of the mode and substituting median in its place becomes necessary because mode cannot always be easily located and is so much affected by grouping errors that it becomes unreliable. Example 5.9: Find the skewness from the following data: Height 58 59 60 61 62 63 64 65 (in inches) 10 18 30 42 35 28 16 8 N.Jmber ofPersons Solution: Height is a continuous variable, and hence 58\" must be treated as 57.5\"-58.5\", 59\" as 58.5\"-59.5\", and so on. Height Frequency x IX &2 Cumulative (in inches) Frequency f !Tom61 58 10 -3 -30 90 10 59 18 -2 -36 72 28 59.5\"-60-60.5\" 30 -1 -30 30 58 60.5\"-61-61.5\" 42 -96 62 35 28 0 0 0 100 62.5\"-63-63.5\" 16 63.5\"-64-64.5\" 8 1 35 35 135 187 2 56 112 163 65 3 48 144 179 4 32 128 187 171 611 +75 Mean= 61 +7-5 = 61.4 Mode= 60.5 +3-5 = 61.04 187 ' 65 a= ~-(l2.)2 =~3.27-0.16 = ~ = 1.76 187 187

Skewness = 61.4 - 61.04 = 0.36 inches. Wlriance, ~rrrnts, Skewness 0.36 and Kurtosis Coefficient of Skewness = -1.7-6 = 0.205 NOTES Alternatively, we can determine the median, Self-Instructional Material 113 Median = The size of 187 th item = 93.5th item 2 = 60.5 + 1X 35\"5 = 61.35 = 0.15 42 Skewness= 3(61.4- 61.35) = 3(0.05) 0.15 Coefficient of Skewness = 1.76 = 0.09 The two coefficients are different because of the difficulties associated with determination of mode. 5.6.2 Bowley's (Quartile) Measure of Skewness In the above two methods of measuring skewness, the whole series is taken into consideration. But, absolute as well as relative skewness may be secured even for a part of the series. The usual device is to measure the distance between the lower and the upper quartiles. In a symmetrical series, the quartiles would be equidistant from the value of the median, i.e., Median - Q1 = Q3 - Median In other words, the vaiue of the median is the mean of Q1 and Q3. In a skewed distribution, quartiles would not be equidistant from median unless the entire asymmetry is located at the extremes of the series. Bowley has suggested the following formula for measuring skewness, based on above facts. Absolute SK = ( Q3 - Me)- (Me- Q,) = Q3 + Q1 - 2 Me (5.6) If the quartiles are equidistant from the median, i.e., ( Q3 - Md) = (Md- Q1), then SK = 0. If the distance from the median to Q1 exceeds that from Q3 to the median, this will give a negative skewness. If the reverse is the case; it will give a positive skewness. If the series expressed in different units are to be compared, it is essential to convert the absolute amount into the relative. Using the interquartile range as a denominator we have for the coefficient of skewness as follows: <?3 + Q -2Md (5.7) Relative SK = <?3 _ Q (Q -Md)-(Md- Q) or, (<?3- Md)+ (Md- Q) If in the series the median and lower quartiles coincide, then the SK becomes (+1). If the median and upper quartiles coincide, then the SK becomes (-1 ). This measure of skewness is rigidly defined and easily computable. Further, such a measure of skewness has the advantage that it has value limits between (+1) and (-1 ), with the result that it is sufficiently sensitive for many requirements. The only criticism levelled against such a measure is that it does not take into consideration all the item of these series, i.e., extreme items are neglected. Example 5.10: Calculate the coefficient of skewness of the data of table given in example 9 based on quartiles. Solution: With reference to table given in example 9, we have, !Q1 =The size of ~th(= 1 7 = 46.75th) item

lilriance, Ahments, Skewness =59.5 + 318.075 and Kurtosis = 59.5 + 0.63 = 60.13 NOTES ~Q3 =The size of 3:Vth item ( = 3 x 87 =140.25th) item 114 Self-Instructional Material =62.5+ 25.285 = 62.5 + 0.19 = 62.69 (using formula 5.6) Skewness = 62.69 + 60.13- 2 (61.35) = 0.12 (using formula 5.7) 0.12 Coefficient of Skewness = 62 _69 _ 60_13 0.12 = 2.56 = 0.047 5.6.3 Kelly's (Percentile) Measure of Skewness To remove the defect of Bowley's measure that it does not take into account all the values, it can be enlarged by taking two deciles (or percentiles), equidistant from the median value. Kelly has suggested the following measure of skewness: SK = P. _ ~o + lfo 50 2 ~ =~- ~+Ll 2 Though such a measure has got little practical use, yet theoretically this measure seems very sound. Example 5.11: Calculate the Karl Pearson's coefficient of skewness from the following data: Marks NJ. ofStudents Marks NJ. ofStudents above 50 above 0 150 70 , 140 , 60 30 , 10 100 , 70 14 , 80 , 80 0 20 80 30 \" 40 Solution: Marks Frequency Mdpoint X=(X-A)/10 I(X) I(X2) Cumulative Frequency (cf) 0-10 10 5 -3 -30 90 10--20 40 15 -2 -80 160 10 20--30 20 25 -1 -20 20 50 70 30--40 0 35 -130 40--50 10 45 00 0 70 50--60 40 55 1 10 10 80 60--70 16 65 2 80 160 120 70--80 14 75 3 48 144 136 150 4 56 224 150 194 808 +64

Since it is a bimodal distribution Karl Pearson coefficient is appropriate and we need to 111riance, MJments, Ske\"ff/11ess and Kurtosis calculate X, Me and a. NOTES - 64 X = 35 + 150 X 10 = 35 + 4.27 = 39.27 2150 Median = Size of th item }0 X 5 =40+ - - =45 10 (\"f. X'))Standard Deviation (cr) . i x '!:._f(NX'2) - fiN( 2 = 1OX 808 _ ( 64 ) l 150 150 = 10 x .J5.387- 0.182 = 10 X 2.28 = 22.8 3(X- Median) 3(39.27 -45) Skewness = a = 22.8 3(-5.73) -17.19 = 22.8 = ~ =-0.75 Example 5.12: From the following data compute quartile deviation and the coefficient of skewness. Size 5-7 8-10 11-13 14-16 17-19 Frequency 14 24 38 20 4 Solution: Size Frequency Cumulative Frequency 4.5-7.5 14 14 7.5-10.5 24 38 10.5-13.5 38 76 13.5-16.5 20 96 100 16.5-19.5 4 Q! = 7.5 + 23 X411 = 8.87 Q3 = 10.5 + 3x37 = 10.5 + 31181 = 10.5 + 2.92 = 13.42 ~ . = 10.5 + ~3 X 12 = 10.5 + 36 = 10.5 + 0.947 = 11.447 Medtan 38 Quart1.1e D .. = Q-Q = 13.42- 8.87 = -4.5-5 = 2.275 evtatwn 2 22 Q + Q -2Me Skewness = --\"Q'----Q-\"--- Self-Instructional J1.1aterial 115

~riance, MJments, Skewness 13.42 + 8.87- 22.89 and Kurtosis 13.42-8.87 NOTES = -0.6 =- 0.13 4.55 Example 5.13: In a certain distribution the following results were obtained: .X =45.00; Median= 48.00 Coefficient of Skewness =- 0.4 You are required to estimate the value of standard deviation. Solution: Skewness = 3 (Mean - Median) (j - 0.4 = 3 (45 - 48) -0.4a = -9 a =9- =22.5 0.4 Example 5.14: Karl Pearson's coefficient of skewness of a distribution is +0.32. Its standard deviation is 6.5 and mean is 29.6. Find the mode and median of the distribution. Solution: Coefficient of Skewness = -Me-an--M-od-e 0.32 = 29.6- Mode 6.5 or 6.5 x 0.32 = 29.6- Mode Mode = 29.6- 2.08 = 27.52 Coefficient of Skewness = 3 (Mean - Median) (j 0.32 = 3(29.6- Median) 6.5 6.5 x 0.32 = 88.8 - 3 Median Median = 88·8 - 2'08 = 28.91 3 Example 5.15: You are given the position in a factory before and after the settlement of an industrial dispute. Comment on the gains or losses from the point of the workers and that of the management. Before After NJ. ofUOrkers 2440 2359 JIJean Wiges 45.5 47.5 JIJedian Wiges 49.0 45.0 Standard frviation 12.0 10.0 Solution: Employment. Since the number of workers employed after the settlement is less than the number of employed before, it has gone against the interest of the workers. Wiges. The total wages paid after the settlement were 2350 x 47.5 = Rs 1,11,625; before the settlement the amount disbursed was 2400 x 45.5 = Rs 1,09,200. 116 Self-Instructional Material

This means that the workers as a group are better off now than before the settlement, Wlriance, MJments, Skewness and unless the productivity of workers has gone up, this may be against the interest of and Kurtosis management. NOTES Uniformityin the wage structure. The extent ofrelative uniformity in the wage structure before and after the settlement can be determined by a comparison of the coefficient of variation. Coefficient ofVariation, Before= J3_ x 100 = 26.4 45.5 Coefficient of Variation, After= __!_Q_ x 100 = 21.05 47.5 This clearly means that there is comparatively lesser disparity in due wages received by the workers. Such a position is good for both the workers and the management. Pattern ofthe wage structure. A comparison of the mean with the median leads to the obvious conclusion that before the settlement more than 50 per cent of the workers were getting a wage higher than this mean, i.e., (Rs. 45.5). After the settlement the number of workers whose wages were more than Rs. 45.5 became less than 50 per cent. This means that the settlement has not been beneficial to all the workers. It is only 50 per cent workers who have been benefited as a result of an increase in the total wages bill. 5.7 KURTOSIS So far we have characterized a frequency distribution by its central tendency, variability, and the extent of asymmetry. There is one more common type of attribute of frequency distribution viz., its peakedness. Look at the Fig. 5.5 in which are drawn three symmetrical curves A, B and C A A - - LEPTOJCURTIC B - - MESOKURTIC C-- PLATYKUirrlC c 'B' A' 0 A' 'B' C' Fig. 5.5 The three curves differ widely in regard to convexity, an attribute to which Karl Pearson referred as 'Kurtosis.' The measure of kurtosis exhibits the extent to which the curve is more peaked or more flat-topped than the normal curve. A curve to be called a normal curve must have its convexity as shown in curve B [in addition to two other requisites, viz., (1) Unimodal, (Ji) Symmetrical]. When the curve of a distribution is relatively flatter than the normal curve, it is said to have kurtosis. When the curve or polygon is relatively more peaked, it is said to lack kurtosis. Karl Pearson gave the name 'Mesokurtic' to a normal curve or a skewed curve that has the same degree of convexity as the normal curve. In Fig. 5.5 curve B is a mesokurtic curve. If some of the cases about one standard deviation from the mean move in towards the centre and other move out towards the tails, thus making the curve unusually peaked, we say that the result would be a 'Leptokurtic' curve, as curve A If on the other hand, some of the cases around the mode move out a little towards each half of the curve, thus making the curve unusually flat-topped, we say that the result would be a 'Platykurtic' curve, as curve C The kurtosis is measured by, j3 = a = J.14 or J.14 24 2 4 J.12 () Self-Instructional Material 117

Y.Jriance, A-f:Jments, Skewness In a normal curve, P2 will be equal to three. If P2 is greater than three, the curve is and Kurtosis more peaked, if less than three, the curve is flatter at the top than normal. NOTES The above formula may be rewritten as: Check Your Progress K=a4 -3= f.J4 -3 3. The mean of 5 observations is 4.4 and the variance is CT4 8.24. If three of the five observations are 1, 2 and 6, If K is positive, it means that the number of cases near the mean is greater than in fmd the other two. normal distribution. If K is negative, the curve is more flat-topped than the corresponding 4. Defme Lorenz Curve. normal curve. 5. What is skewness? 6. Explain kurtosis. The measures of skewness and kurtosis are also expressed in terms of the Greek letter, 118 Self-Instructional Material gamma (}1. r = .[ii; = J.J3 = a:, I CT3 Piy.=fJ-3=f.J-4...,.3=f.J4--3= fJ2 - 2 3f.J2 2 f.J2 22 CT 4 y1 and y2 are the measures of skewness and kurtosis respectively. If y1 is more than zero, then the conclusion will be the presence of positive skewness; if y1 is less than zero, it will mean negative skewness, and in case y1 is zero, then there will be absence of skewness. Similarly, ifa curve is Leptokurtic, y2 will be positive; ifPlatykurtic, y2 will be negative; and in case Mesokurtic, y2 will be exactly zero. Example 5.16: From the following frequency distribution calculate the first four moments, P1 and P2. Oass 1(}-14 15-19 2(}-24 25-29 3(}-34 35-39 40--44 45-49 5(}-54 f 4 8 19 35 20 7 5 Solution: Let the assumed mean A= 32 Gass Mdpoint f (X) f(X) f(X2) f(X3) ~) Cumulative Frequency 10-14 12 1 -4 -4 16 -64 256 1 15-19 324 5 20-24 17 4 -3 -12 36 -108 128 13 25-29 22 8 -2 -16 32 -64 19 32 30-34 35-39 27 19 -1 --1-9 19 --1-9 0 67 40-44 -51 -255 20 87 45-49 112 94 50-54 32 35 0 -0 0 -0 405 99 256 100 37 20 1 20 20 20 42 7 2 14 28 56 47 5 3 15 45 135 52 1 4 4 16 64 100 +53 212 +275 1520 \"Lf(X) \"LIX2 \"f,/X3 \"LIX4 =+2 =+212 =+20 = +1520 1: !Xx (CJ) =2-X-5= 10 0 1 -= v.I = 1:[ 100 100 . 1;/)(2 X ( CJ}2 212 X 52 212 X 25 V2-= 1:[ =53 100 100 1: /)(3 X ( CJ)3 20 X 53 20 X 125 =---= =25 ~3= 1:[ 100 100 1: tx4 X ( CJ)4 1520 X 54 1520 X 625 = =9500 v4,= 1:[ 100 100

f.J, = v, - v, =0 lilriance, .f\\,f)ments, Skewness and Knrtosis f.J]_ = v2- ~=53-(0.li=53-0.1 =52.99 NOTES vif-lJ = v3 - 3 v2 v1+ 2 =25 -3 X 53 X (0.1) +2 (0.1)3 =25-15.9+0.002=9.102 p4 = v4 - 4 v3 v1+ 6 v2 ~ - 3 v{ =9500 -4 X 25 X (0.1)+ 6 X 53 X (0.1i-3(0.1)4 = 9500-10 + 3.18-0.0003 = 9493.1797 or 9493.18 Corrected p 2 = 112 - J} (where h is the class interval) 12 52 = 52.99- - = 52.99- 22.083 = 50.91 12 Corrected p3 = 113 = 9.I J} f.J2 7h4 Corrected p4 = p4 - - 2 - + 240 = 9493.18- 25 X 52.99 X 7 X ( 5)4 2 240 = 9493.18- 662.375 + 18.23 = 8849.03 2 = (9.1) = 0.000627 (50.91)3 P2 = f.L4 88849.03 2 = (50.91)2 = 3.414 f.Jz r1 = Jji; = ~o.ooo627 = 0.25 r2 =f32 -3 =3.414-3=0.414 Example 5.17: The first four central moments of a distribution are 0, 2.5, 0.7 and 18.75. Test the skewness and kurtosis of the distribution. Solution: Skewness is tested by p 3 which should be equal to zero in a symmetrical distribution. Since in the given problem it is 0.7, we can conclude that the distribution is not syrrnnetrical. But to measure the extent and direction of the skewness we make use of constant, HJIa 3 = 3 = = ~0.031 = 0.18 f.Jz Since ~ = 0.18 we conclude that the distribution is not symmetrical but has a positive skewness = 0.18. Kurtosis is tested by /32. In normal case /32 should be equal to three. If it is greater than three, the curve is more peaked, if less than three the curve is more flat-topped. Pi -p - f.L4 f.L4 - 18.75 - 18.75 - 3 2 - o-4 = (2.5)2 - 6.25 - Since /31 = 3 we conclude that the curve is mesokurtic. Self-Instructional Mlterial 119

larfance, MJments, Skewness 5.8 SOLVED PROBLEMS and Kurtosis Problem 1: Find the standard deviation and the coefficient of variation of the following NOTES data: 120 Self-Instructional Material Oass Interval Frequency Oass Interval Frequency 3.00-3.25 6 4.00-4.25 47 3.25-3.50 19 4.25-4.50 29 3.50-3.75 35 4.50-4.75 15 3.75-4.00 44 4.75-5.00 5 Solution: Oass Frequency Mdvalue X IX JX2 Interval (X) (f) =X-3.875 0.25 3.00-3.25 6 3.125 -3 -18 54 3.25-3.50 19 3.375 -2 -38 76 3.50-3.75 35 3.625 -1 -35 35 3.75-4.00 44 3.875 0 00 4.00-4.25 47 4.125 +1 +47 47 4.25-4.50 29 4.375 +2 +58 116 4.50-4.75 15 4.625 +3 +45 135 4.75-5.00 5 4.875 +4 .,f-20 80 N=200 'f. IX r. JX'2 =+79 =543 X- = A +'f-.Nf-)CX J. = 3.8 7 79 0.25 5+-X 200 = 3.974 Standard Deviation, u= J~ ':2 -(~:r xi 543 - ( 79 ) 2 X0.25 = 0.40 200 200 Coefficient ofVariation (C.V.) = ~ xlOO X = 0.40 X100 = 10.06% 3.974 Problem l: Two workers on the same job show the following results over a long period of time. AB Mean time of completing the job (in minutes) 30 25 S.D. (in minutes) 64 (a) Which worker appears to be more consistent in the time he requires to complete the job? (b) Which worker appears to be faster in completing the job?

Solution. (a) The coefficient of variation = ~ x 100 lilriance, JIJoments, Skewness X and Kurtosis A= 6 100 = 20% NOTES -X Self-Instructional Material 121 30 B= .i.xlOO = 16% 25 The coefficient of variation is more for worker A, hence worker B appears to be more consistent in the time required to complete the job. (b) Worker B appears to be faster in completing the job as he takes on an average 25 minutes as against 30 minutes taken by worker A Problem 3: The A.M. of 5 observations is 4.4. and the variance is 8.24. If 3 of the 5 observations are 1, 2 and 6, find the other two. Solution: Let the two observations be x and y. _ Sum of all the observations X= Number of items Sum of all the observations = 5 x 4.4 = 22 (1) 1 + 2 + 6 + x+ y= 22 x+ y= 13 d2 = rxz -(X)Z /1. 8.24= x2+r+12+22+62 -(4.4)2 (2) 5 r+or Y-=97 (x -.02 = (x+ .02 - 4xy (r= (x+ _02 - 2 [(x+ .02 - + y)] = 169-2 (132 - 97) = 25 x-y=5 Subtracting equation (2) from (1 ), we get 2y= 8 or y= 4 Putting this in equation (2), x= 5 + 4 = 9 Hence, x = 9 andy= 4 :. Other two observations are 9 and 4. Problem 4: Calculate quartile coefficient of skewness from the following: \"Height NJ. ofPersons Cumulative Frequency Under 100 1 15 100-109 14 81 110-119 66 203 120-129 122 348 130-139 145 469 140-149 121 534 150-159 65 565 160-169 31 577 170-179 12 582 180-189 5 584 190-199 2 586 200 and above 2

lilriance, Nbments, Skewness Solution: and Knrtosis Coefficient of Skewness = ~ + Q - 2 ~ NOTES ~-Q 122 Self-Instructional Material Height N:J. ofPersons Cumulative Frequency Under 100 14 15 100-109 66 81 110-119 122 203 119.5- 120-129- 129.5 145 348 129.5- 130-139- 139.5 121 469 139.5- 140-149- 149.5 65 534 150-159 31 565 160-169 12 577 170-179 5 582 180-189 2 584 190-199 2 586 200 and above Calculation of Median N = 293. This item lies in 129.5- 139.5 group. 2 N- cf Q =L+ 2 f x i where Lis the lower limit of class. 2 = 129 5 + 293-203 X 10 . 145 = 129.5 + 900 = 129.5 + 6.2 = 135.7 145 Calculation of Q1 ~ = 146.5. This item lies in 119.5- 129.5 group. N Q = L+ 4-cf Xi= 119.5 + 146.5-81 X 10 If 122 = 119.5 + 655 = 119.5 + 5.37 = 124.87 122 Calculation of Q2 3N = 439.5. This item lies in 139.5 - 149.5 group. 4 3N Q = L+ 4 - cf Xi= 139.5 + 439.5-348 x}O 3f 121 = 147.1 Calculation of Skewness 147.1 + 124.87- 2(135.7) - 271.97-271.4 - 0.57 - 0 147.1-124.87 - 22.23 - 22.33 - ·03 Problem 5: It is known that X and the median of a distribution are 3.0 and 4.0. Is the distribution skewed?

Solution: In the synunetrical distribution the values ofmean and median are alike. Skewness lariance, Aflments, Skewness can be measured in absolute terms by taking the difference between mean and median. As and Kurtosis the mean and median of this distribution are of different sizes, therefore, the distribution is skewed. Moreover, distribution is negatively skewed as Mean < Median. NOTES Problem 6: Consider the following distributions: x A B Median 100 90 90 80 S.D. 10 10 (a) Distribution A has the same degree of variation as distribution B. Do you agree? Give reasons. (b) Both the distributions have same degree of skewness: True/False. Give reasons. Solution: (a) Coefficient ofVariation = ___!!____ x 100 B Mean A 10 C.v. 10 x 1oo= 11.11 C.V. = - X 100 = 10 =9-0 100 Coefficient of variation of distribution B is more than that of A and hence, A does not have the same degree of variation as distribution B. (b) Coefficient of Skewness = 3 .(..:M....__ean_- M_ed_ian..):... B a A 3(1 00 - 90) = 30 = 3 3(90 - 80) = 30 = 3 10 10 10 10 As the coefficient of skewness is same, both distribution have same degree of skewness. Problem 7: The sum of a set of 100 numbers is 4000 and the sum of its square is 162500 and Median is 41. Find the coefficient of skewness. Solution: We are given I:X= 4000 and I: X = 162500 N= 100 Mean= NLX = W400o0 =40 VN-S.D. = /I:X2 2 = 162500 -1600 = ~2500 = .J25 = 5 (A.M.) 100 100 Coefficient of Skewness = -3(::-X_-_M_e_d_ia_n..:....) = 30(40-41) = -3 = -0 6 5 5. Alternatively (solving with the help of moments). Let the moments about origin, i.e., zero are represented by v. Mean or v 1'= NLX = 4000 = 40 100 v' = LX2 = 162500 = 1625 2N 100 Self-Instructional Material 123

Mlriance, J\\lhments, Skewness Variance or Pz = v 2'- (v{)2 = 1625- (40)2 = 25 and Kurtosis Therefore S.D. = J25 = 5 NOTES 3(X- Median) 3(40- 41) 124 Self-Instructional Material Coefficient of Skewness= a = 5 = -0.6 Problem 8: Examine whether the following results ofa piece of computation for obtaining the second (central) moment are consistent or not? N= 120, LX=- 125 and LJt = 128 Solution: Second (central) moment means second moment about arithmetic mean, i.e., square of S.D. If deviations are taken from assumed mean, formula for its calculation is, #2= if= L~- (L~r = 128 - (-112205 ) 2 = 1.067- 1.085 =- 0.018 120 But Pz• i.e., square of standard deviation can never be negative. Hence, there is inconsistency in the figures. Problem 9: The first three moments about the value 2 are 1, 16 and -40. Compute the moments about the mean and the origin. Solution: #I = v1 - v1 = 1 - 1 = 0 ~t2 = v2 -(vi= 16-(li = 16- I= 15 #3 = lJ - 3 V2 VI + 2( VI )3 = -40-3 X 16 X 1 + 2(1)3 = -40-48 + 2 = -86 Now moments about the origin are, LIX v{ = N =X= A+ v1 = 2 + 1 = 3 v.2' = LIX2 = u_ + v'I 2 = 15 +9 = 24 -N- r-L, v.3' = LfX' + 3vI'v2.'- 2vI'3 -N- = JL. rj = - 86 + 3 X 3 X 24 - 2 X 27 =76 Thus, the first three moments about the mean stand at the figures of 0, 15 and - 86 respectively; and about the origin are 3, 24 and 76 respectively. It may be noted that as the first moment about the origin is always equal to the value of the mean, so the mean value of the above distribution is 3. 5.9 SUMMARY In this unit, you have learned about variance, moments, skewness and kurtosis measures of deviation and distribution. Variance or coefficient of variation has the same properties as standard deviation and is the square of standard deviation, represented as cr'l. The Lorenz curve is a graphic method of measuring deviations from the average. It was devised by Dr. Lorenz for measuring the inequalities in the distribution of wealth and can be applied with equal advantage for comparing the distribution of profits amongst different groups of business and such other things. Moment is a mechanics term and is calculated where the 'moment of a force' describes the tendency or capacity of a force to turn a pivoted lever. You have also learned the concept of skewness, which refers to lack of symmetry in

iistribution and the description of kurtosis, which refers to a measure of peakedness of a Ulriance, MJments, Skewness distribution. and Kurtosis 5.10 ANSWERS TO 'CHECK YOUR PROGRESS' NOTES 1. 36.6 approx., 22.6 approx. Self-Instructional Material 125 2. Coefficient of variation A= 6.0% and coefficient of variation B = 6.4%. So brick layer A will continue to be more consistent than brick layer B 3. 4, 9 4. The Lorenz curve is a graphic method of measuring deviations from the average. It was devised by Max 0. Lorenz for measuring the inequalities in the distribution of wealth. It is a cumulative percenmge curve. In it the percentage ofitems are combined with the percentages of such other things as wealth, profits or tum-over, etc. 5. Skewness refers to the lack of symmetry in a distribution. In the symmetrical distribution, the mean, median and mode coincide. 6. Kurtosis is a measure of peakedness of a distribution. It shows the degree ofconvexity of a frequency curve. 5.11 QUESTIONS AND EXERCISES Short-Answer Questions 1. Describe variance. 2. What is a Lorenz Curve? 3. Explain the different types of skewness. 4. How will you measure skewness? 5. Define the term kurtosis. 6. How will you measure the degree of convexity of a frequency curve using kurtosis? 7. When can coefficient of variations be greater than 100 per cent? What can you say about the items of the given data in such a case? 8. The first four moments of a distribution about the value 4 are -1.7, 17, -30, 108. Calculate the moments about the mean. 9. For a distribution standard deviation is 2. What should be the value of f.14 so that it is (a) Mesokurtic (b) Leptokurtic (c) Platykurtic. 10. The first four central moments of a distribution are 0, 2.5, 0.7 and 18.75. Test the skewness and kurtosis of the distribution. 11. The first four moments of a distribution are 1, 4, 10 and 46, respectively. Compute the first four central moments and the Beta constants. Comment upon the nature of the distribution. 12. The first three moments of distribution about the value 3 of a variable are 2, 10 and 30, respectively. Obtain the first three moments about zero. Show also that the variance of the distribution is 6. 13. In a frequency distribution the coefficient of skewness based upon the quartiles is 0. If the sum of the upper and the lower quartiles is 100 and the median is 38, find the value of upper quartile.

'Ulriance, MJments, Skewness Long-Answer Questions and Kurtosis 1. From the following information.about the accidents on a road in 200 days, calculate NOTES the mean number of accidents and the variance of accidents. 126 Self-Instructional Material No. of Accidents Per Day 0 1 2 3 4 5 No. ofDays 46 76 38 25 10 5 2. The following is a record of the number of bricks each day for 20 days by two brick layers A and B. A 725, 700, 750, 650, 675, 725, 675, 725, 625, 675, 700, 725, 675, 800, 650, 675,625,700,650. B 575, 625, 600, 575, 675, 625, 575, 550, 650, 625. 550, 700, 625, 600, 625, 650, 575, 675, 625, 600. Calculate the coefficient of variation in each case and discuss the relative consistency of the two brick layers. If the figures for A were in every case 10 more and those of B in every case 20 more than the figures given above, how would the answer be affected? 3. Why is Lorenz Curve used? Write the steps necessary for drawing a Lorenz Curve. 4. What is skewness? How would you find it in a nonsymmetrical distribution? 5. Distinguish between: (a) Dispersion and Skewness (b) Positive and Negative Skewness (c) Quartile Deviation Method of Measuring Skewness and Pearson's Measure of Skewness 6. Explain the three terms, Dispersion, Skewness and Kurtosis with example. 7. Give a suitable formula for measuring the peakedness (kurtosis) of frequency distribution. 8. Find Bowley coefficient of skewness from the following data. According to the Census of 1941; the following are the population figures in 000' of the 1st 36 cities in India. 2488, 591, 437,20, 213, 143, 1490, 407, 284, 176, 169, 181, 777,387,302,213,204,153,133,391,263,176,178,142,522,360,260,193,131, 92, 672, 258, 239, 160, 147, 151. 9. Compute the coefficient of dispersion and skewness of the following data. Central Size 1 2 34 56 7 8 9 10 Frequency 2 9 11 14 20 24 20 16 52 10. Find mean, median, standard deviation and a coefficient ofskewness from the following data of ages of students of a school. Age 5-7 8-10 11-13 14-16 17-19 No. of Students 7 12 19 10 2 11. Find the coefficient of skewness for the following distribution: Variable 0- 5- 10- 15- 20- 25- 30- 35-40 Frequency 2 5 7 13 21 16 83 12. Find Karl Pearson's coefficient of skewness from the following data: Marks Above 0 10 20 30 40 50 60 70 80 No. of Students 150 140 100 80 80 70 30 14 0 13. Calculate the first four moments for the following data: Marks 10-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 6 28 96 75 56 30 8 1 14. In a continuous frequency distribution grouped in intervals of 2, the first four moments about the mean, i.e., 10 expressed in class interval units are, 0, 2, 8, -2 and 24.5 respectively. Obtain the first four moments about mean and calculate P1 and p2. Also calculate the coefficient of skewness based on moments and comment on kurtosis.

15. Second, third and fourth central moments of a variable characteristics are 19.67, 29.26 lflriance, MJnrnts, Skewness and 866.0 respectively. Calculate the beta coefficients correct to three decimal places. and Kurtosis 16. Find out the mean, mean deviation and kurtosis of the following data: NOTES Class Interval 0--10 10--20 20--30 30--40 Frequencies 1 3 4 2 17. Calculate coefficient of skewness by all methods known for the following data on ages of 250 persons taken in a sample study. Age Below 10 20 30 40 50 60 70 80 No. of Persons 15 35 60 84 96 127 188 250 18. The following figures relate to marks obtained by 20 students of a class in test. Find the first four mClments about asswned mean 10, first four moments about arithmetic mean and the values of PI and P2• 1, 9, 12,2, 10, 15, 3, 16, 19, 5, 18, 17, 6,20, 0, 8,5, ~ 11, 12. 19. 1-12• 113 and 114 of a variable are 19.67, 29, 26 and 866.0. Calculate PI and p2. 20. For a distribution of items, X= 54, u= 3, PI= 0 and p2 = 3. During calculations two items 64 and 50 were taken as 62 and 52. What are the correct constants. 21. (a) Karl Pearson coefficient ofskewness ofdistribution is + 0.32. Its standard deviation is 6.5 and mean is 29.6. Find the mode and median of the distribution. (b) If the mode of the above distribution is 24.8, what will be the standard deviation? (c) For a distribution, Bowley's coefficient ofskewness is -0.36, QI is 8.6 and median is 12.3. What is quartile coefficient of dispersion? (d) Compute quartile deviation and coefficient of skewness for the following given values: Me= 18.8, QI = 14.6, Q3 = 25.2 5.12 FURTHER READING Kothari, C.R. 1984. Quantitative \"Jechniques, 3rd Edition. New Delhi: Vikas Publishing House Pvt. Ltd. Chandan, J.S. 1998. Statistics for Business and Economics. New Delhi: Vikas Publishing House Pvt. Ltd. Chandan, J.S., Jagjit Singh and K.K. Khanna. 1995. Business Statistics, 2nd Edition. New Delhi: Vikas Publishing House Pvt. Ltd. Self-!JJStructional Material 127



UNIT 6 PROBABILITY THEORY, Probability Theory, PERMUTATIONS AND Permutations and Combinations COMBINATIONS NOTES Structure 6.0 Introduction 6.1 Unit Objectives 6.2 Elementary Probability Theory 6.2.1 Probability 6.2.2 Definition ofProbability Axiomatic, Classical and Frequency 6.2.3 Random Experiment 6.2.4 Sample Space 6.2.5 Events 6.2.5 Independent and Dependent Events 6.2.6 Addition Rule 6.2.7 Multiplication Rule 6.3 Conditional Probability 6.4 Bayes' Theorem 6.5 Permutations and Combinations 6.5.1 Permutation ofx out ofn Distinct Items 6.5.2 Combination ofx Items out ofn Distinct Items 6.6 Solved Problems 6.7 Summary 6.8 Answers to 'Check Your Progress' 6.10 Questions and Exercises 6.11 Further Reading 6.0 INTRODUCTION In this unit, you will study different theories ofprobability and understand why probability is considered the most important tool in statistical calculations. The subject ofprobability in itself is a cumbersome one, hence only the basic concepts will be discussed in this unit. The word probability or chance i5 very commonly used in day-to-day conversation, and terms such as possible or probable or likely, all have the similar meanings. Probabilitycan be defmed as a measure ofthe likelihood that a particular event will occur. It is a numerical measure with a value between 0 and 1 of such likelihood where the probability ofzero indicates that the given event cannot occur and the probability of one assures certainty of such an occurrence. The probability theory helps a decision-maker to analyse a situation and decide accordingly. The following are few examples ofsuch situations: • What is the chance that sales will increase ifthe price ofa product is decreased? • What is the likelihoodthat a new machine will increase productivity? • How likelyis it that a given project will be completed on time? • What are the possibilities that a competitor will introduce a cheaper substitute in the market? Self-Instructional Material 129

Business Statistics-H In this unit, you will also study why all these uncertainties require knowledge of probability so that calculated risks can be taken. Since the outcomes ofmost decisions NOTES cannot be accurately predicted because of the impact of many uncontrollable and W1predictable variables, it is necessary that all the known risks be scientifically evaluated. 130 Self-InstructionalJl.fltcrial Probability theory, sometimes referred to as the science ofuncertainty, is very helpful in making such evaluations. It helps the decision-maker with only limited information to analyse the risks and select the strategy ofminimum risk. This unit also discusses the laws ofaddition and multiplication. The law ofaddition states that when two events are mutually exclusive, the probability that either of the events will occur is the sum oftheir separate probabilities. The law ofmultiplication is applicable when two events occur at the same time. Finally, this unit explains probability distributions and Bayes' theorem. Bayes' theorem is a theorem ofprobability theory originally stated by the Reverend Thomas Bayes. It is based on the philosophy of science and tries to clarify the relationship between theory and evidence. 6.1 UNIT OBJECTIVES After going through this unit, you will be able to: • Analyse the various definitions ofprobability theory • Define independent and dependent events • Explainconditional probability • Apply Baye's Theorem for drawing conclusions • Explain permutations and combinations 6.2 ELEMENTARY PROBABILITY THEORY 6.2.1 Probability Probability can be defmed as a measure of the likelihood that a particular event will occur. It is a numerical measure with a value between 0 and 1 of such a likelihood where the probability of zero indicates that the given event cannot occur and the probability of one assures certainty of such an occurrence. For example, if a radio weather report indicates a near-zero probability ofrain, it can be interpreted as no chance of rain and if a 90 per cent probability of rain is reported, then our understanding, it is most likely to rain. A 50 per cent, probability or chance of rain indicates that rain is just as likely to occur as not. This likelihood can be shown as follows: 0 0.5 Event does Event as likely Event definitely not occur to occur as not occurs ----------------------~~ Increasing likelihood of occurrence Probability theory provides us with a mechanism for measuring and analysing uncertainties associated with future events. Probability can be subjective or objective.

Subjective probability is purely individualistic so that an individual can assign a probability Probability Theory, to the outcome of a particular event based upon whatever information regarding this Permutations and Combinations event is available to him along with his personal feelings, experience, judgement and expectations. Two different individuals may assign two different probabilities for the NOTES outcome of the same event. Self-Instructional Mlterial 131 The objective probability of an event, on the other hand, can be defined as the relative frequency of its occurrence in the long run. In other words, the probability ofan outcome in which we are interested, known as favourable outcome or successful outcome can be calculated as the number of favourable outcomes divided by the total number of outcomes. For example, if (s) defmes the number of successful outcomes and (n) is the total number of outcomes, then the probability of a successful outcome is given as (sin). Experiment: An experiment is any activity that generates data. For example, tossing a fair coin is considered a statistical experiment. An experiment is identified by two properties. (I) Each experiment has several possible outcomes and all these outcomes are known in advance. (ii) None of these outcomes can be predicted with certainty. In this example, we are not certain whether the outcome will be a head or a tail. Some of the experiments and their possible outcomes can be given as: Experiment Possible Outcomes 1. Tossing of a fair coin Head, tail 2. Rolling a die 1, 2, 3, 4, 5, 6 3. Selecting an item from a production lot Good, bad 4. Introducing a new product Success, failure 6.2.2 Definition of Probability: Axiomatic, Classical and Frequency Probability theory is also called the theory ofchance and can be mathematically derived using standard formulas. A probability is expressed as a real number, p e [0, 1] and the probability number is expressed as a percentage (0 per cent to 100 per cent) and not as a decimal. For example, a probabilityof0.55 is expressed as 55 percent. When we say that the probability is 100 per cent, it means that the event is certain, while 0 per cent probability means that the event is impossible. We can also express probability of an outcome in the ratio format. For example, we have two probabilities, i.e., 'chance of winning' (1/4) and 'chance ofnot winning' (3/4), then using the mathematical formula of odds, we can say, 'chance ofwinning': 'chance ofnot winning'= 114:3/4 = 1 :3 or 113 We are using probability in vague terms when we predict something for future. For example, we might say it will probably rain tomorrow or it will probably be a holiday the day after. This is subjective probability to the person predicting, but implies that the person believes the probability is greater than 50 per cent. Different types ofprobabilitytheories are: (I) Classical TheoryofProbability (ii) Axiomatic Probability Theory (iii) Empirical Probability Theory

Business Statistics-II Classical Theory of Probability NOTES The classical theory of probability is the theory based on the number of favourable outcomes and the number of total outcomes. The probability is expressed as a ratio of 132 Self-InstructionalMaterial these two numbers. The term 'favourable' is not the subjective value given to the outcomes, but is rather the classical terminology used to indicate that an outcome belongs to a given event ofinterest. Classical definition ofprobability: Ifthe number ofoutcomes belonging to an event Eis ~. and the total number ofoutcomes is N then the probability ofevent Eis defined NE as PE= N. For example, a standard pack ofcards (without jokers) has 52 cards. Ifwe randomly draw a card from the pack, we can imagine about each card as a possible outcome. Therefore, there are 52 total outcomes. Calculating all the outcome events and their probabilities, we have the following possibilities: • Out ofthe 52 cards, there are 13 clubs. Therefore, ifthe event ofinterest is drawing a club, there are 13 favourable outcomes, and the probability ofthis event becomes: 13 1 -=- 52 4 • There are 4 kings (one ofeach suit). The probability ofdrawing a king is: 5: = 1~ . • What is the probability ofdrawing a king or a club? This example is slightly more complicated. We cannot simply add together the number ofoutcomes for each event separately (4 + 13 = 17) as this inadvertently counts one ofthe outcomes twice (the kingofclubs). Thecorrectansweris: 16 from 13 + 4 - 1 · 52 52 52 52 We have this from the probability equation, Aclub) +r{king)- r{king ofclubs). • Classical probability has limitations, because this definition ofprobability implicitly defines all outcomes to be equiprobable and this can be only used for conditions such as drawing cards, rolling dice, or pulling balls from urns. We cannot calculate the probability where the outcomes are unequal probabilities. It is not that the classical theory ofprobability is not useful because ofthese mentioned limitations. We can use this as an important guiding factor to calculate the probability of uncertain situations as mentioned here and to calculate the axiomatic approach to probability. Frequency of occurrence This approach to probability is widelyused in a wide range ofscientific disciplines. It is based on the idea that the underlying probability ofan event can be measured byrepeated trials. Probability as a measure of frequency: Let nA be the number of times event A occurs after n trials. We define the probability ofevent Aas, PA=LimnA nn->oo

It is not possible to conduct an infinite number oftrials. However, it usually suffices to Probability Theory, conduct a large number oftrials, where the standard oflarge depends on the probability Permutations and Combinations being measured and how accurate a measurement we need. NOTES Defmition ofprobability: The sequence 2n in the limit that will converge to the same n Self-Instructional Material 133 result every time, or that it will converge at all. To understand this, let us consider an experiment consisting of flipping a coin an infinite number oftimes. We want that the probability ofheads must come up. The result may appear as the following sequence: HTHHTTHHHHT1TTHHHHHHHH1T1T1T1THHHHHHHHHHHHH HHH1111111111111111... This shows that each run ofkheads and ktails are being followed by another run ofthe n oscillates between, 3I and 32 same probability. For this example, the sequence : which does not converge. These sequences may be unlikely, and can be right. The definition given here does not express convergence in the required way, but it shows some kind of convergence in probability. The problem of formulating exactly can be considered using axiomatic probability theory. Axiomatic Probability Theory The axiomatic probability theory is the most general approach to probability, and is used for more difficult problems in probability. We start with a set ofaxioms, which serve to define a probability space. These axioms are not immediately intuitive and are developed using the classical probability theory. Empirical Experimental Probability Theory Experimentalprobability is another name for empericalprobability. The empirical approach to determining probabilities relies on data from actual experiments to determine approximate probabilities instead ofthe assumption ofequal likeliness. Probabilities in these experiments are defmed as the ratio ofthe frequency ofthe possibility ofan event, 1(E), to the number oftrials in the experiment, n, written symbolically as ~E) = J(E)/n For example, while flipping a coin, the empirical probability ofheads is the number of heads divided by the total number offlips. The relationship between these empirical probabilities and theoretical probabilities is suggested by the Law ofLarge Numbers. The law states that as the number oftrials of an experiment increases, the empirical probability approaches theoretical probability. Hence, ifwe roll a die a number oftimes, each number would come up approximately one-sixth ofthe time. The study ofempirical probabilities is known as statistics. 6.2.3 Random Experiment An experiment is said to be random if we cannot predict the outcome before the experiment is carried out. Arandom experiment is one which can be repeated, practically or theoretically, any number of times. We can toss a coin or roll a die any number of times and study the outcomes. The result ofany outcome may or may not influence that ofsucceeding outcomes. Thus, any throw ofa coin or dice is independent ofall earlier throws. But if a card is drawn from a deck of cards and not replaced, the experiment made for a second draw will be influenced by the result ofthe fust. The consideration ofthe random character ofa phenomenon is inevitable because of the complicated characteristics of the laws of nature, ignorance about the relevant

Business Statistics-n laws and the theoretical or practical difficulty allowing for the effect ofa large number offactors under consideration or disturbing factors or other shortcomings ofour tools of NOTES work. 134 Self-InstructionalMlterial Arandom variable is a probabilityvariable. A specified degree ofprobability is attached to every value of a random variable. What is meant by this statement is that when we conduct an experiment, say by tossing a coin, we cannot definitely say what will come up. Random forces will decide the outcome, although each outcome may have an equal chance ofprobability ofshowing up. A random variable is a quantity which in different observations can assume different values. 6.2.4 Sample Space Tossing a coin: A sample space is the collection of all possible events or outcomes of an experiment. For example, there are two possible outcomes of a toss of a fair coin, which are a head and a tail. Then, the sample space for this experiment denoted by S would be: S=[H,T] so that the probability of the sample space equals 1 or, P[S] = P[H, T] =1 This is so because when the coin is tossed, either a head or a tail must occur. Rolling a dice: Rolling a dice is also an important experiment ofsample space. A common dice is a small cube whose faces show numbers 1, 2, 3, 4, 5, 6 one way or the other. These dice are either marked with the real digits or number ofdots as shown below: ~-- ;-· - •I :···;.-: ~·--··: ;•-;•: ~:-:: _!. ' ~--- '~-·: ~~- • ~!.___·: When we roll a die any of the six faces can come as a result of the roll, since there are a total ofsix faces. Hence, the sample space isS= [1, 2, 3, 4, 5, 6], and P[S] = 1, since one ofthe six faces must occur. Tossing two coins: The experiment of sample space is on tossing two coins. Both the coins can be tossed simultaneously, or one after the other. The difference between the two cases is that in the second case one can easily differentiate between the coins which one is the first and which is second. When the two indistinguishable coins are tossed simultaneously, then there are three possible outcomes, [H, H], [H, T], and [T, T]. If the coins are distinct or different, and if they are thrown one after the other, then there are four possible outcomes: [H, H], [H, T], [T, H], [T, T]. This can be represented in a concise form as, [HH, HT, TH, TT]. Hence, the outcome ofan experiment depends upon its nature. Sample Space for Indistinguishable Coins, S = [H, H], [H, T], [T, T] Sample Space for Distinct Coins, S = [HH, HT, TH, TT] or [H, H], [H, T], [T, H], [T, T] Rolling two dice: Similarly, the sample space can be calculated as follows for rolling oftwodice. s = [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

Note: Smallest number on dice is 1, ifboth the dice show 1, then the sum is 2. Probability Theory, Permutations and Combinations Similarly, largest number on dice is 6, and ifboth the dice show 6, then the sum is 12. NOTES 6.2.5 Events Self-Instructional Material 135 An event is an outcome or a set of outcomes of an activity or a result of a trial. For example, getting two heads in the trial of tossing three fair coins simultaneously would be an event. Elementary event An elementary event, also known as a simple even~ is a single possible outcome of an experiment. For example, if we toss a fair coin, then the event of a head coming up is an elementary event. If the symbol for an elementary event is (E), then the probability of the event (E) is written as P[E]. Joint event A joint event, also known as compound even~ has two or more elementary events in it. For example, drawing a black ace from a pack of cards would be a joint event, since it contains two elementary events of black and ace. Complement of an event The complement of any event A is the collection of outcomes that are not contained in A. This complement of A is denoted as A' (A prime). This means that the outcomes contained in A and the outcomes contained in A' must equal the total sample space. Therefore P[A] + P[A'] = P[S] = 1 or P[A] = 1- P[A'] For example, if a passenger airliner has 300 seats and it is nearly full, but not totally full then event A would be the number of occupied seats and A' would be the number of unoccupied seats. Suppose there are 287 seats occupied by passengers and only 13 seats are empty. Typically, the stewardess will count the number of empty seats which are only 13 and report that 287 people are aboard. This is much simpler than counting 287 occupied seats. Accordingly, in such a situation, knowing event A' is much more efficient than knowing event A. Mutually exclusive events Two events are said to be mutually exclusive if both events cannot occur at the same time as the outcome of a single experiment. For example, if we toss a coin, then either event head or event tail would occur, but not both. Hence, these are mutually exclusive events. Venn Diagram A convenient way to visualize the concept of events, their relationships and sample space is done by using Venn diagrams. The sample space is represented by a rectangular region and the events and the relationships among these events are represented by circular regions within the rectangle. For example, two mutually exclusive events A and B are represented in the Venn diagram as follows:

Business Statistics-II 00 =s NOTES Event P[A u B] is represented in the Venn diagram as follows: Event [AB] is represented as follows: [AB] Union of Three Events The process of combining two events to form the union can be extended to three events so that P[A u B u C] would be the union of events A, B, and C. This union can be represented in a Venn diagram. c Fig. 6.1 Union of1hree Events Example 6.1: Suppose a sample of 50 students is taken and a survey is made of this sample regarding their reading habits. The survey results are shown as follows: Event N.Jmber of Students Magazines they Read [A] 20 '\"lii1r [B] 15 10 ~wsweek [C] 8 6 Hlmfare [AB] 4 Time and 1\\Cwsweek [AC] 2 Time and Hlmfare [BC] ~wsweek and Hlmfare [ABC] Time and ~wsweek and Filmfare 136 Self-InstructionalMaterial

Find out the probability that a student picked up at random from this sample of Probability Theory, 50 students does not read any of these 3 magazines. Permumtions and Combinations Solution: The problem can be solved by a Venn diagram as follows: NOTES c Self-Instructional Material 137 Since there are 21 students who do not read any ofthe three magazines, the probability that a student picked up at random among this sample of 50 students who does not read any of these three magazines is 21150. The problem can also be solved by the formula for probability for union of three events, given as follows: P[AuBuC] = P[A] + P[B] + P[C]- P[AB]- P[AC]- P[BC] + P[ABC] = 20/50 + 15/50 + 10/50- 8/50- 6/50-4/50 + 2/50 = 29/50 This is the probability that a student picked up at random among the sample of 50 reads either lime or ~u-sweekor Filmfare or any combination of the two or all three. Hence, the probability that such a student does not read any of these three magazines is 21150, which is [1 - 29/50]. Independent and Dependent Events Two events, A and B, are said to be independent events ifthe occurrence ofone event is not at all influenced by the occurrence of the other. For example, if two fair coins are tossed, then the result of one toss is totally independent of the result of the other toss. The probability that a head will be the outcome of any one toss will always be 112, irrespective ofwhatever the outcome is ofthe other toss. Hence, these two events are independent. Let us assume that one fair coin is tossed ten times and it happens that the first nine tosses resulted in heads. What is the probability that the outcome ofthe tenth toss will also be a head? There is always a psychological tendency to think that a tail would be more likely in the tenth toss since the first nine tosses resulted in heads. However, since the events oftossing a coin ten times are all independent events, the earlier outcomes have no influence whatsoever on the result of the tenth toss. Hence, the probability that the outcome will be a head on the tenth toss is still 112. On the other hand, consider drawing two cards from a pack of52 playing cards. The probability that the second card will be an ace would depend on whether the first card was an ace or not. Hence, these two events are not independent events. 6.2.6 Addition Rule When two events are mutually exclusive, then the probability that either of the events will occur is the sum of their separate probabilities. For example, if you roll a single die then the probability that it will come up with a face 5 or face 6, where event A refers to face 5 and event B refers to face 6, both events being mutually exclusive events, is given by,

Business Statistics-ll P[A or B] = P[A] + P[B] or P[5 or 6] = P[5] + P[6] NOTES 116 +116 138 Self-Instructional Mlterial = 2/6 = 1/3 P [A or B] is written as P[A u B] and is known as P [A union B]. However, if events A and B are not mutually exclusive, then the probability of occurrence of either event A or event B or both is equal to the probability that event A occurs plus the probability that event B occurs minus the probability that events common to both A and B occur. Symbolically, it can be written as, P[A u B] = P[A] + P[B] - P[A and B] P[A and B] can also be written as P[A n B], known asP [A intersection B] or simply P[AB]. Events [A and B] consist of all those events which are contained in both A and B simultaneously. For example, in an experiment of taking cards out of a pack of 52 playing cards, assume that: Event A = An ace is drawn Event B = A spade is drawn Event [AB] = An ace of spade is drawn Hence, P[A u B] =P[A] + P[B]- P[AB] = 4/52 + 13/52 - 1152 = 16/52 = 4/13 This is because there are 4 aces, 13 cards of spades, including 1 ace of spades out of a total of 52 cards in the pack. The logic behind subtracting P[AB] is that the ace of spades is counted twice---Qnce in event A (4 aces) and once again in event B (13 cards of spade including the ace). Another example for P[AuB], where event A and event Bare not mutually exclusive is as follows: Suppose a survey of 100 persons revealed that 50 persons read India Today and 30 persons read 11me magazine and 10 of these 100 persons read both India Today and 11me. Then: Event [A] = 50 Event [B] = 30 Event [AB] = 10 Since event [AB] of 10 is included twice, both in event A as well as in event B, event [AB] must be subtracted once in order to determine the event [AuB] which means that a person reads India Today or Time or both. Hence, P[AuB] = P [A]+ P [B]- P [AB] = 50/100 + 301100 -10/100 = 70/100 = 0.7 6.2.7 Multiplication Rule The multiplication rule is applied when it is necessary to compute the probability if both events A and B will occur at the same time. The multiplication rule is different if the two events are independent as against the two events being not independent.

If events A and B are independent events, then the probability that they both will Probability Theory, occur is the product of their separate probabilities. This is a strict condition so that Permutations and Combinations events A and B are independent if, and only if, NOTES P [AB] = P[A] x P[B] or = P[A]P[B] To take an example, if we toss a coin twice, then the probability that the first toss results in a head and the second toss results in a tail is given by, p [HT] = P[H] X P[T] = 112 X 1/2 = 114 However, if events A and B are not independent, meaning the probability of occurrence of an event is dependent or conditional upon the occurrence or non- occurrence of the other event, then the probability that they will both occur is given by, P[AB] = P[A] x P[B/given the outcome of A] This relationship is written as: P[AB] = P[A] X P[B/A] = P[A] P[B/A] where P[B/A] means the probability of event B on the condition that event A has occurred. As an example, assume that a bowl has 6 black balls and 4 white balls. A ball is drawn at random from the bowl. Then a second ball is drawn without replacement of the first ball back in the bowl. The probability ofthe second ball being black or white would depend upon the result of the first draw as to whether the first ball was black or white. The probability that both these balls are black is given by, P [two black balls]= P [black on lst draw] x P [black on 2nd draw/black on 1st draw] = 6/10 X 5/9 = 30/90 = 113 This is so because, first there are 6 black balls out of a total of 10, but if the frrst ball drawn is black then we are left with 5 black balls out of a total of 9 balls. 6.3 CONDITIONAL PROBABILITY Check Your Progress In many situations, a manager may know the outcome of an event that has already I. Define the terms 'simple occurred and may want to know the chances of a second event occurring based probability' and 'joint upon the knowledge of the outcome of the earlier event. We are interested in finding probability'. out as to how additional information obtained as a result of the knowledge about the outcome of an event affects the probability of the occurrence of the second event. 2. What is a mutually exclusive For example, let us assume that a new brand of toothpaste is being introduced in the event? market. Based on the study of competitive markets, the manufacturer has some idea about the chances of its success. Now, he introduces the product in a few 3. Explain the concept of selected stores in a few selected areas before marketing it nationally. A highly independent events. positive response from the test-market area will improve his confidence about the success of his brand nationally. Accordingly, the manufacturer's assessment of high 4. Explain the classical theory of probability of sales for his brand would be conditional upon the positive response from probability. the test-market. 5. What is addition rule? Let there be two events A and B. Then, the probability of event A given the outcome of event B is given by: Self-lnstructional.Mlterial 139 P[A/B]= P[AB] P[B]

Business Statistics-H where P[AIB] is interpreted as the probability of event A on the condition that event B has occurred and P [AB] is the joint probability of event A and event B, and P[B] NOTES is not equal to zero. 140 Self-Instructional M!terial As an example, let us suppose that we roll a die and we know that the number that came up is larger than 4. We want to fmd out the probability that the outcome is an even number given that it is larger than 4. Let event A = even and event B = larger than 4 P[even and larger than 4] Then P[even I larger than 4] = P[larger than 41 (,~)P[AIB] = P[AB] = =1/2 (%)or P[B] But for independent events, P[AB] = P[A]P[B]. Thus substituting this relationship in the formula for conditional probability, we get: P[AIB] = P[AB] = P[A]P[B] P[A] P[B] P[B] This means that P[A] will remain the same no matter what the outcome of event B is. For example, if we want to fmd out the probability of a head on the second toss of a fair coin, given that the outcome of the first toss was a head, this probability would still be 112, because the two events are independent events and the outcome of the first toss does not affect the outcome of the second toss. 6.4 BAYES' THEOREM Reverend Thomas Bayes ( 1702-61) introduced his theorem on probability which is concerned with a method for estimating the probability ofcauses which are responsible for the outcome ofan observed effect. Being a religious preacher and mathematician, his motivation for the theorem came from his desire to prove the existence ofGod by looking at the evidence of the world that God created. He was interested in drawing conclusions about the causes by observing the consequences. The theorem contributes to the statistical decision theory in revising prior probabilities of outcomes of events based upon the observation and analysis ofadditional infotjmati?n. Bayes' theorem makes use ofconditional probability formula where the condition can be descnbed in terms ofthe additional information which would result in the revised probabilityofthe outcome ofan event. Suppose there are 50 students in our statistics class, out ofwhich 20 are male students and 30 are female students. Out ofthe 30 females, 20 are Indian students and 10 are foreign students. Out ofthe 20 male students, 15 are Indians and 5 are foreigners, so that out ofall the 50 students, 35 are Indians and 15 are foreigners. This data can be presented in a tabular form as follows: Male lndiB.Il ForeiJJilt:r Total Female 15 5 30 Total 10 50 w 15 35