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- 1. Unit - 4 MEASURES OF CENTRAL TENDENCY AVERAGES Mass data are colleted, classified, tabulated systematically. The data so presented is analyzed further to bring its size to a single representative figure. Descriptive statistics which describes the presented data in a single number. It is concerned with the analysis of frequency distribution or other from of presentation mathematically by which a few constants or representative numbers are arrive. CONCEPT OF CENTRAL TENDENCY: One of the main characteristics of numerical data is central tendency. It is found that the observations tend to cluster around a point. This point is called the central value of the data. The tendency of the observations to concentrate around a central point is known as central tendency. The statistical measures which tell us the position of central value or central point to describe the central tendency of the entire mass of data is known as measure of central tendency or Average of first order Meaning of Statistical Averages:- An average is a figure which represents the large number of observations in a concise or single numerical data. If is a typical size which describes the central tendency. According to CLARK AND SCHAKADE “Average is an attempt to find one single figure to describe whole group of figures” According to COXTON & COWDEN “An average is a single value within the range of the data that is used to represent all of the values in the series” A single simple expression in which the net result of huge mass of unwieldy numerical data or a frequency distribution is concentrated and which is used to represent the whole data is called a statistical average. Objects of Statistical average: An average is of great significance in all the fields of human knowledge, because it depicts the characteristics of the whole group of data understudy. Following are the objectives of objectives of computing the statistical averages 1. To give or present the complex data in a simple manner and concise form. 2. To facilitate the data for comparative study of two different series. 3. To study the mass data from the sample 4. To establish relationship between the two series 5. To provide basis for decision making 6. To calculate the representative single value from the given data. Requisites of a good and ideal Average. Any statistical average to be good and ideal average must possess some of the characteristics as it is a single value representing a single value representing a group of values. Following are the requisite properties of a good and ideal average. 1. It should be easily understood . 2. It should be simple in calculation. 3. It should be based on all the observations. 4. It should not be unduly affected by the extreme values. 5. It should be rigidly defined. 6. It should be capable of further algebraic treatment. 7. It should have sampling stability. Thus a statistical average should have all the above requisites to be an ideal and good average. Limitations of Averages: Although an average is useful in studying the complex data and is very widely used in almost all the spheres of human activity, it is not without limitation that restrict scope and applicability. Following are the limitations of statistical averages. 49
- 2. 1. The extreme values, if any, will affect the averageble figure disproportionately. 2. The composition of the data cannot be viewed with the help of the average. 3. The average does not represent always the characteristics of individual items. 4. The average gives only a representative figure of the mass, but fails to depict the entire picture of the data. 5. An average may give us a value that does not exist in the data. 6. some times an average might give very absurd results. In spite of the limitations, the statistical averages still are useful measures which play an important role in analyzing the mass data. TYPES OF STATISTICAL AVERAGES: Broadly speaking, there are five types of statistical averages which are commonly used in practice. They are. 1. Arithmetic Mean. 2. Geometric Mean. 3. Harmonic Mean. 4. Median. 5. Mode. Arithmetic Mean Arithmetic Mean is the most widely used measurement which represents the entire data. Generally it is termed as an average to a layman. It is the quantity obtained by dividing the sum of the values of the items in a variable by their number. Arithmetic Mean may be two types 1. Simple Arithmetic Mean. 2. Weighted Arithmetic [Not included in your syllabus] Simple Arithmetic Mean It is the quotient of the sum of the values divide by their number. Symbolically 0= x1 +x2 +x3 -----------+ xn = Σx n n Where Σx = The sum of observations ‘n’ = number of observations 0 = is symbol of arithmetic mean INDIVIDUAL SERIES: Calculation of arithmetic mean in individual series. DIRECT METHOD The process of computing the arithmetic mean in this case involves the following steps:- a. Add all the values of the variable x and obtain Σx b. Divide this total of observations Σx by the number of observations ‘n’ Illustration = 01 The monthly income of 5 persons is as given below Rs.1132, 1140, 1144, 1136, 1148 find out arithmetic mean. Solution Calculation of Arithmetic Mean Monthly of income Serial Number In Rs. 0 = Σx 1 1132 n 2 1136 = 5700 3 1140 5 4 1144 0 = 1140 5 1148 n=5 Σx 5700 50
- 3. Illustration = 02 Calculate mean from the following data:- Roll No. 1 2 3 4 5 6 7 8 9 10 Marks 40 56 55 78 58 60 73 35 43 48 Solution Calculation of Mean Roll Number Marks 1 35 2 40 0 = Σx 3 43 n 4 48 = 540 5 50 10 6 55 0 = 54 7 58 8 60 9 73 10 78 N = 10 540 SHORT CUT METHOD The arithmetic mean can also be calculated by shortcut method. This method reduced the amount of calculation. It involves the following steps. 1. Assume any one value as an assumed mean which is also arbitrary mean [‘A’ = Assumed mean] 2. Take the deviations of each item from the assumed mean (A) and denote this deviations by dx. ie dx =x – A. 3. Obtain the sum of these deviations i.e. Σdx 4. Use the formula 0 = A+ Σdx N Illustration = 03 The marks obtained by ten students in an examination are as given below. Calculate mean by shortcut method. Roll No. 1 2 3 4 5 6 7 8 9 10 Marks 3 4 7 43 48 65 57 31 60 59 7 8 8 Solution A= 48, dx = x - A x – 48 Roll No Marks dx 1 31 -17 0 = A + Σdx 2 37 -11 n 3 43 -5 = 48 +46 4 48 0 5 48 0 10 6 57 09 = 48+4.6 7 59 11 0 = 52.6 8 60 12 9 65 17 Average marks of 10 students 10 78 30 = 52.6 Σdx =46 -33 n = 10 +79 51
- 4. Illustration = 04 Calculate mean wage of workers working in a factory. Sl.No. 1 2 3 4 5 6 7 8 9 10 Wages in Rs. 400 500 550 780 580 600 730 650 430 480 Solution A = 550 dx = x - A Wages x – 550 0 = A + Σdx Sl. No In Rs. dx n 1 400 - 150 = 550 + 200/10 2 430 - 120 = 550 + 20 3 480 - 70 = 570 4 500 - 50 Average wages = Rs.570 5 550 0 6 580 30 7 600 50 8 650 100 9 730 180 10 780 230 -390 + 590 200 n = 10 Σdx DISCRETE SERIES Calculation of mean in discrete Series Direct Method The process of computing arithmetic by direct method in case of discrete series involves the following steps 1. Represent the variable by x and the frequency by ‘f’ 2. Multiply each frequency with the corresponding value of the variable i.e. find fx 3. obtain the total of these products i.e. obtain Σfx 4. Find the total frequency i.e. N = Σf 5. Divide the total obtained by the total frequency. i.e. 0 =3fx N Illustration = 5 Calculate mean from the following data Variable: 50 52 55 56 60 63 64 65 67 70 Frequency 5 6 10 15 20 12 8 11 9 10 Variable Frequency fx x f 50 5 250 52 6 312 0 =Σfx 55 10 550 N 56 15 840 = 6438 60 20 1200 106 63 12 756 = 60.72 64 8 512 65 11 715 67 9 603 70 10 700 =Σfx 52
- 5. SHORT – CUT METHOD:- The process of computing arithmetic mean by short – cut method in case of discrete series involves the following steps. 1. Take an assumed mean (A) out of variable x 2. Take the deviations of the values of x from the assumed mean (A) and denote the deviations by dx – x –A 3. Multiply dx with the respective frequency and obtain the total of these products. That is obtain 0 =Σfdx 4. Use the formula 0 =A +Σfdx N ILLUSTRATION = 06 From the following frequency distribution, find out mean height of the students. Height in inches x: 64 65 66 67 68 69 70 71 72 73 No. of Students f 1 6 10 22 21 17 14 5 3 1 Solution A= 68, dx = x – A Height x – 68 F fdx X dx 64 1 -4 -4 0 = A +Σfdx 65 6 -3 -18 N 66 10 -2 -20 = 68 + 13/100 67 22 -1 -22 = 68 + 0.13 68 21 0 0 = 68.13 69 17 1 17 70 14 2 28 Average height of student = 68.13inches. 71 5 3 15 72 3 4 12 73 1 5 5 N = 100 Σfdx 13 -64 +17 STEP – DEVIATION METHOD Steps involved:- 1. Take an assumed mean (A) out of variable x 2. Take deviations of the values x from the assumed mean 3. Divide each dx by c, where c is the common factorial each x i.e. dx = x –A c 4. Multiply this dx with the respective frequency and take and total Σfdx 5. Use the formula. 0 = A =Σfdχ xc N ILLUSTRATION = 07 Calculate the arithmetic mean for the following data. Wages in Rs: 150 200 250 300 350 400 450 500 550 600 Workers: 10 20 30 20 40 10 20 10 8 2 53
- 6. Solution A= 350 C=50 dx = x – A C Wages Workers x – 350 Fdx in Rs. x f 50 dx 150 10 -4 -40 0 = A + Σfdx xC 200 20 -3 -60 N 250 30 -2 -60 = 350 + -58 x50 300 20 -1 -20 170 350 40 0 0 = 350 – 2900/170 400 10 1 10 = 350 – 17.05 450 20 2 40 = 332.95 500 10 3 30 Mean 550 8 4 32 age 600 2 5 10 170 Σfdx -58 -180 + 122 CONTINUOUS SERIES Calculation of arithmetic mean in case of continuous series Direct Method Steps involved:- 1. Obtain the mid – points or mid x of each class 2. Multiply these mid – points by the respective class frequency and obtain the total 3fm 3. Use the formula 0 = Σfm N ILLUSTRATION = 08 Calculate arithmetic mean for the following frequency distribution. Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 –80 80 - 90 obtained No. of 5 4 6 20 12 10 10 8 5 Student Solution Marks f mid x Fm 0.x m 0 = Σfm 0 – 10 5 5 25 N 10 -20 4 15 60 0 = Σfm 20 – 30 6 25 150 N 30 – 40 20 35 700 = 3700/80 40 – 50 12 45 540 = 46.25 50 – 60 10 55 550 60 – 70 10 65 650 Mean Marks = 46.25 70 – 80 8 75 600 80 – 90 5 85 425 N = 80 3700 Σfm SHORT – CUT METHOD CUM STEP – DEVIATION The process of computing arithmetic mean by short – cut method in case of continuous series involves the following steps. 1. Obtain the mid – point of each class 2. Take the assumed mean out of midx 3. Deduct assumed mean from the mid – point of each class & find out the deviations dx = x –A 4. Multiply the respective frequencies of each class by deviations and obtain the total Σfdm 5. Use the formula 0 = A + Σfdm x C N 54
- 7. ILLUSTRATION = 09 Find arithmetic mean from the following frequency distribution. Profit per Shop 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 –70 70 -80 No. of shops ‘f’ 12 20 18 40 25 10 12 13 Solution A = 45, C = 10, dxn = x – A C x– 45 /10 X f Dxn fdm mid x 0 = A + Σfdm x C 0 – 10 12 5 -4 -48 N 10 - 20 20 15 -3 -60 = 45 + -111 x 10 20 – 30 18 25 -2 -36 150 30 – 40 40 35 -1 -40 = 45 – 1110/150 = 45 – 7.4 40 – 50 25 45 0 0 0 = 37.6 50 – 60 10 55 1 10 60 – 70 12 65 2 24 70 – 80 13 75 3 39 N = 150 Σfdm -111 -184 + 73 ‘C’ = Common factor or class interval. ILLUSTRATION = 10 The following frequency distribution represents the age of the persons interviewed, calculate average age. Age in years x 80 – 89 70 – 79 60 – 69 50 – 59 40 – 49 30 – 39 20 – 29 10 - 19 No. of persons f 2 2 6 20 56 40 42 32 Solution Note : Given values are arranges in ascending order A= 44.5, C= 10, dm = x – A C x = 44.5/10 X f midx Dm fdm 10 - 19 32 14.5 -3 -96 0 = A + Σfdm x C 20 – 29 42 24.5 -2 -84 N 30 – 39 40 34.5 -1 -40 = 44.5 + 1740/200 40 – 49 56 44.5 0 0 = 44.5 – 8.7 50 – 59 20 54.5 1 20 0 = 35.8 60 – 69 6 64.5 2 12 70 – 79 2 74.5 3 6 80 – 89 2 84.5 4 8 N = 200 Σfdm -174 -220 +46 ILLUSTRATION = 11 Calculate arithmetic mean from the following data Variable (x) Frequency (f) Less than 05 7 Less than 10 20 Less than 15 45 Less than 20 75 Less than 25 95 Less than 30 99 Less than 35 100 Solution First of all we have to convert the given cumulative frequency table into an ordinary frequency Table as under. 55
- 8. A = 175, C = 5, dm = x – A/C x – 17.5/5 X f mid x Dm fdm 0 = A + Σfdm x C 0 –5 7 2.5 -3 -21 N 5 –10 13 7.5 -2 -26 = 17.5 + -41 x5 10 - 15 25 12.5 -1 -25 100 15 – 20 30 17.5 0 0 = 17.5 – 205/100 20 – 25 20 22.5 1 20 = 17.5 – 2.05 25 – 30 4 27.5 2 8 = 15.45 30 – 35 1 32.5 3 3 N = 100 Σfdx -41 -72 +31 ILLUSTRATION : 12 From the following data calculate the average marks of 65 students. Marks: Less than 10 20 30 40 50 60 70 80 No. of Students 4 10 15 25 30 35 45 65 Solution Note: we will first convert the dat in simple series from the given cumulative frequencies. A= 35, C = 10, dm= x – A C fdm 0 = A + Σfdm x C Marks Students (x – 35)/10 Midx X f dx N 0 – 10 4 5 -3 -12 = 35 + 96 x 10 10 - 20 10-4=6 15 -2 -12 65 20 – 30 15 –10 =5 25 -1 -5 = 35 +(960/65) 30 – 40 25 –15=10 35 0 0 = 35 + 14.77 40 – 50 30 –25=5 45 1 5 = 49.77 Average Marks 50 – 60 35 –30=5 55 2 10 60 – 70 45 –35=10 65 3 30 70 – 80 65 –45=20 75 4 80 65 Σfdm 96 -29 + 125 ILLUSTRATION : 13 From the following information pertaining to 150 workers. Calculate average wage paid to workers. Wages in Rs. more than 75 85 95 105 115 125 135 145 No of Workers 150 140 115 95 70 60 40 25 Solution The lower limit of the first class in 75, the class interval would be 75 – 85, 85 – 95 and soon A = 110, C =10, dm = x - A C Wages 0 = A + Σfdm xC No. of Workers (x – 110)/10 In Rs. Mid x fdm N f dxn X = 110 + 95 x 10 75 – 85 150 –140 = 10 80 -3 -30 150 85 –95 140 – 115 =20 90 -2 -50 = 110 + (950/150) 95 – 105 115 –95 = 20 100 -1 -20 = 110 + 6.33 105 –115 95 – 70 = 25 110 0 0 = 116.33 Average wage 115 –125 70 – 60 = 10 120 1 10 125 – 135 60 – 40 = 20 130 2 40 135 – 145 40 – 25 = 15 140 3 45 145 – 155 25 – 0 = 25 150 4 100 N = 150 Σfdm 95 -100 + 195 ILLUSTRATION: 14 Calculate average marks of 70 students 56
- 9. Marks more than 20 30 35 40 50 60 65 70 80 No. of Students 70 63 55 40 30 18 10 7 0 Solution:- Convert the more than frequency table into ordinary frequency distribution & then proceed as usual. A = 45, C = 5, dm = x – A C Marks Frequency (x – 45)/5 0 = A +Σfdm xC Mid x fdm X f dm N 20 – 30 70 – 63 = 07 25 -4 -28 = 45 + 37 x 5 30 – 35 63 – 55 = 08 32.5 -2.5 -20 70 35 – 40 55 – 40 = 15 37.5 -1.5 -22.25 = 45 + (185/70) 40 – 50 40 – 30 = 10 45 0 0 = 45 +2.643 50 – 60 30 – 18 = 12 55 2 24 = 47.643 Marks 60 – 65 18 – 10 = 8 62.5 3.5 28 65 – 70 10 – 7 = 3 67.5 4.5 13.5 70 – 80 7–0=7 75 6.0 42.0 N =70 Σfdm 37 - 70.5 + 107.5 Open – End Class In a grouped frequency distribution, if the lower limit of the first class and the upper limit of last class are not known, it is difficult to find the Arithmetic Mean under such circumstances, we have to assume that the width of the open classes are equal to the common width of closed classes. It will be clear from the following example. ILLUSTRATION = 15 Calculate Arithmetic Mean from the following data Wages in Below 100 100 – 109 110 – 119 120 – 129 130 – 139 140 – 149 150 & above Rs. No of 10 15 25 40 28 12 20 Workers Solution: A= 124.5, C= 10, dm = x – A C X f mid x dx fdm 0 = A + Σfdm x C 90 – 99 10 94.5 -3 -30 N 100 – 109 15 104.5 -2 -30 = 124.5 + 27 x 10 110 – 119 25 114.5 -1 -25 15 120 – 129 40 124.5 0 0 = 124.5 + (270/150) 130 – 139 28 134.5 1 28 = 124.5+ 1.8 140 – 149 12 144.5 2 24 = 126.3 Average wage 150 – 159 20 154.5 3 60 N = 150 Σfdm 27 - 85 + 112 MERITS OF ARITHMETIC MEAN DEMERITS OF ARITHMETIC MEAN 1. It is easy to understand and easy to calculate. 1. The mean is unduly affected by the extreme 2. It is based on all the observations. items. 3. It is capable of further algebraic treatment. 2. It is un realistic. 4. It is rigidly defined and determinate . 3. It may lead to a false conclusion. 5. It is least affected by fluctuations of sampling. 4. It cannot be calculated in case of open – end AM is as stable as possible. classes. 5. It may not be represented in the actual data. GEOMETRIC MEAN 57
- 10. The Geometric mean is obtained by finding out the products of the different items of a series and getting the root of the product corresponding to the number of items. The Geometric mean of a set of N observations is the nth root of their product It may be defined as the nth root of the product of N values. This can be put in the following formula G.M = √The product of N values G.M = n√ χ1 x χ2 x χ3 …………xχn When the number of observations exceeds two the computations are simplified through the use of logarithms then. GM = Antilog of Σlogx n USES OF GEOMETRIC MEAN 1. This G.M. is highly useful in averaging ratios percentages and rate of increase between two periods 2. G.M. is important in the construction of index Numbers. 3. In economic and social sciences, where we want to give more weight to smaller items and smaller weight to large items, G.M is appropriate. Calculation of Geometric Mean – individual series Steps 1. Find out the logarithms of each value 2. Add all the values of logx i.e. Σlogx 3. The sum of log (Σlogx) is divided by the number of items - Σlogx n 4. Find out the antilog of the quotient this is the GM of the data. ILLUSTRATION : 01 Calculate the Geometric mean of the following Values : 50 72 54 82 93 Solution: Values Logx G.M = Antilog of Σlogx X n 50 1.6990 = -------“------ 9.1710/5 72 1.8573 = ------“ ------ 1.8342 54 1.7324 = 68.26 82 1.9138 93 1.9685 N=5 9.1710 ILLUSTRATION – 2 Calculate geometric mean from the following data. Values x: 6.5 169.0 11.0 112.5 14.2 75.5 35.5 215.0 58
- 11. Solution Values Log G.M = Antilog of Σlogx X x n 6.5 0.8129 = ---------“------- 13.0461 169.5 2.2279 8 11.0 1.0414 = --------“--------1.6308 112.5 2.0511 = 42.74 14.2 1.1523 75.5 1.8779 35.5 1.5502 215.0 2.3325 N=8 13.0461 ILLUSTRATION = 03 : Calculate Geometric Mean of the following Values: 15 250 15.7 1.57 105.7 10.5 1.06 25.7 0.257 10 Solution Values:x logx G.M = Antilog of Σlogx 15 1.1761 n 250 2.3979 = ----------“-------- 9.8561 15.7 1.1959 10 1.57 0.1959 = -----------“----------0.98561 105.7 2.0240 G.M = 9.674 10.5 1.0212 1.06 0.0253 25.7 1.4099 0.257 1.4099 10.0 1.0000 Σlogx 9.8561 n = 10 10 –1 = 9 ILLUSTRATION :04 Calculate Geometric Mean of the following data Values x 3834 382 63 8 0.4 0.03 0.009 0.0005 Solution: Values x log x G.M = Antilog of Σlogx 3834 3.5893 n 382 2.5821 = --------“------- 1.6062 63 1.7993 8 8 0.9031 = --------“ ------- 0.2008 0.4 1.6021 GM = 1.588 0.03 2.4771 0.009 3.9542 0.0005 4.6990 N=8 1.6062 5 + 6 = 11 –10 = 1 ILLUSTRATION :05 Find the geometric mean from the following data Values 0.8974 0.0570 0.0081 0.5677 0.0002 0.0984 59
- 12. Solution Values Log G.M = Antilog of Σlogx X x N 0.8974 1.9530 -12 + 3 0.0570 2.7559 = -----------“------ -9.6655 0.0081 3.9085 6 0.5617 1.7541 = ------------“--- 2.6109 0.0002 4.3010 GM = 0.04083 0.0984 2.9930 N=5 9.6655 4 – 13 = - 9 ILLUSTRATION :06 Calculate the G.M of the following Values 375 0.05 0.5672 0.0854 0.005 Solution Values x Logx GM = Antilog of Σlogx 375 2.5740 n (-5 + 2) 0.05 2.6990 = ------“---- 3.6573 0.5672 1.7538 5 0.0854 2.9315 = -------“------ -1.5314 0.005 3.6990 = GM = 0.3399 N=5 3.6573 5 – 8 = -3 ILLUSTRATION = 07 Find the average rate of increase in population which in the first decade has increased by 20%, in the next by 30% and in the third by 45% Solution Geometric mean is more appropriate to find out an average increased in population over three decades. Population at the GM = A. Log of 3logx Rate of Decade end each decade Log x n increase [Base =100] = A.L of 6.3545 First 20% 120 2.0792 3 Second 30% 130 2.1139 = A.L of 2.1181 Third 45% 145 2.1614 = 131.3 n=3 6.3545 The average percentage increase in population over three decades is 31.3% (131.3 – 100) Merits of Geometric Mean: 1. It is rigidly defined. 2. It is based on all the observations. 3. It is suitable for further mathematical treatment. 4. It is not much affected by fluctuations of sampling. 5. It gives comparatively more weight to small values. Demerits 1. It is comparatively difficult to calculate and not simple to understand. 2. If any observation is zero, then geometric mean becomes zero. 3. It is not defined for negative values. 4. It cannot be obtained by inspection. 5. It may not be represented in the actual data. HARMONIC MEAN Harmonic mean like geometric mean is a measure of central tendency in solving special types of problem involving variable expressed in time rates. Harmonic Mean is the reciprocal of the arithmetic mean of the reciprocal of the given observations. The reciprocal of a number is that value which is obtained dividing one by the value. It is defined as the reciprocal of the average of the individual items. Symbolically HM = . n . 1/x +1/x2 +1/x3 …………+ 1/xn 60
- 13. USES: It is useful for computing the average rate of increase of profit of a concern or average speed at which a journey has been performed. The rate usually indicates the relation between two different types of measuring units that can be expressed reciprocally. INDIVIDUAL SERIES: Calculation of Harmonic Mean is individual series. Steps 1. Find out the reciprocal of each size i.e. 1/x 2. Add all the reciprocals of all values 31/x 3. Apply the formula HM = n n = no of observations Σ1/x Σ1/x = Total reciprocal value of given variables. ILLUSTRATION – 01 Calculate the Harmonic Mean from the following data relating to incomes of 10 families Family 1 2 3 4 5 6 7 8 9 10 Income in Rs. 85 70 10 75 500 8 42 250 40 36 Solution Income in Rs. Reciprocals x 1/x 85 0.01176 70 0.01426 H.M = n . 10 0.10000 Σ1/x 75 0.01333 = 10/0.34631 500 0.00200 = 28.87 8 0.12500 42 0.02318 250 0.00400 40 0.02500 36 0.02778 Σ1/x = 0.34631 ILLUSTRATION = 02 Calculate Harmonic Mean from the following data. Values 2574 475 75 5 0.8 0.08 0.005 0.0009 Solution Values Reciprocals H.M = n . X 1/x Σ1/x 2574 0.00038 =. 8 . 475 0.00210 1324.96581 75 0.01333 = 0.0060378 5 0.20000 0.8 1.25000 0.08 12.50000 0.005 200.00000 0.0009 1111.00000 n=8 Σ1/x = 1324.96581 ILLUSTRATION = 03 Calculate the Harmonic Mean for the following data Variable x 35 250 18.7 234.6 1.06 98.72 0.987 61
- 14. Solution Variable Reciprocal X 1/x 35 0.02857 H.M = n . 250 0.00400 Σ1/x 18.7 0.05348 =. 7 . 234.6 0.00426 2.05701 1.06 0.94340 = 3.402998 98.72 0.01013 0.987 1.01317 n=7 Σ1/x 2.05701 Merits of Harmonic Mean Demerits of Harmonic Mean 1. It is rigidly defined. 1. It is difficult to calculate and is not 2. It is based on all the observation of the series understandable. 3. It is suitable in case of series having wide 2. All the values must be available for dispersion. computation. 4. It is suitable for further mathematical 3. It is usually a value which does not exist in treatment. series. 5. It gives less weight to large items and more 4. It cannot be used when any one of the item is 0 weight to small items. or negative, because when the item 0 is used as a divider the items will be zero. MEDIAN Median may be defined as the value of that item which divides the series into two equal parts, one – half containing values greater than it and the other half containing values less that it. Therefore the series has to be arranged in ascending or descending order, before finding the median. The Median is a positional average and the term position refers to the place of a value in a series. The definitions of Median given by different authors are as follows. “Median of a series is the value of the item actual or estimated when a series is arranged in order of magnitude which divides the distribution into two parts” – SECRIST “ The Median is that value of the variable which divides the group into two equal parts, one part containing all values greater and the other all values less than median. – By L.R. Conner. The Median, as its name indicates, is the value of the middle item in a series, when items are arranges according to magnitude. Calculation of Median: INDIVIDUAL SERIES Steps 1. Arrange the observations in ascending or descending order of magnitude 2. Locate the middle value 3. use the formula Median = the size of n + 1th item 2 ILLUSTRATION = 01 Calculate Median from the following data Marks Obtained 53 48 69 78 82 93 45 68 64 75 62
- 15. Solution Arrange the values in ascending order Marks Median = the size of n + 1th item Obtained 2 X = -----“------ 10 + 1 = 5.5th item 45 2 48 = 5th item is 68+ difference of 68 & 69 53 50% below = 68 + 0.5 of 69 – 68 64 = 68 + 0.50 of 1 68 68.5 Median = 68.5 69 75 50% above 78 82 93 N = 10 ILLUSTRATION =02 Calculate the Value of Median from the following data relating to wages in Rs. Wages in Rs. 125 150 120 160 175 148 182 134 145 Solution Arrange the values in ascending order Wages Median = the size of n + 1th item In Rs. x 2 120 = -------“------- 9 + 1th item 125 2 134 = -------“------- 10/2 = 5th item 145 M = 148 148 Median 150 160 175 182 N=9 ILLISTRATION = 03 In a batch of 15 students, 5 students failed in a test the marks of 10 students, who got passing marks were as follows. 9 6 7 8 8 9 6 5 5 7 What was the median Marks of all the 15 students 63
- 16. Solution: In case of only 10 student marks were given but failed student marks were given. We have to assume that failed candidates must have secured less than 5 marks i.e. minimum marks for pass in the test. Marks Obtained By Students x Median = the size of n + 1th item 0 2 2 = ------“----- (15+1)/2 = 16/2 = 8th item 3 = 8th item is 6 3 3 Median Marks of all the 15 students is 6 5 5 6 Median = 6 6 7 7 8 8 9 9 n = 15 DISCRETE SERIES Calculation of Median in case of discrete Series involves the following steps. 1. Obtain the total frequency Σf = N 2. Obtain cumulative frequency 3. Locate median by using the formula Median = The size of N + 1th item’s value 2 ILLUSTRATION = 04 Calculate median for the following data. Wages in Rs. 255 257 258 260 265 268 270 275 280 281 No. of Workers 10 15 12 24 30 26 15 12 10 6 Solution Cumulative Frequency No. of Median = The size of n + 1th items Value Wages Workers Cf 2 In Rs x F = ------“------ 160 + 1 = 161 255 10 10 2 2 257 15 25 = 80.5 258 12 37 Median = 265 260 24 61 265 30 91 80.5 268 26 117 270 15 132 275 12 144 280 10 154 281 6 160 N = 160 ILLUSTRATION = 05 Calculate the value of Median from the following data, Marks Obtained 45 50 55 60 65 70 75 80 85 90 No. of students f 15 28 15 12 25 35 25 15 20 10 64
- 17. Solution Marks M = The size of n + 1th items Value f cf x 2 45 15 15 = The size of 200 + 1 50 28 43 2 55 15 58 = 200/2 = 100.5 60 12 70 M = 70 65 25 95 70 35 130 100.5 Median = 70 75 25 155 80 15 170 85 20 190 90 10 200 N= 200 CONTINUOUS SERIES Steps 1. Make the classes in exclusive form, if they are in intensive form. 2. Find total frequency N = Σf 3. Find out median number m = N/2 4. Obtain cumulative frequency 5. Find out with the help of m median class 6. Apply the formula to locate the median Median = l1+ l2 – l1 (m – c) f Where m = N/2 Where l1 & l2 represents lower and upper class limits of median class f = Frequency of the median class c = Cumulative frequency of the class preceding the median class ILLUSTRATION – 06 Calculate Median for the following data. Income in Rs 250 – 300 300 – 350 350 – 400 400 – 450 450 – 500 500 – 550 No. of persons 18 12 24 36 22 8 Solution:- Income in Median = l1+ l2 – l1 (m –c) f cf Rs. x f 250 –300 18 18 = Where m = N/2 = 120/2 = 60 300 – 350 12 30 = 400 + 450 –400(60 –54) 350 – 400 24 54 36 400 – 450 36 90 = 400 + 50 x 6 450 – 500 22 112 36 500 – 550 8 120 = 400 + 300 N = 120 36 = 400 + 8.33 M = 408.33 ILLUSTRATION = 07 Calculate the value of Median from the following data Monthly 501-600 601- 700 701- 800 801 –900 901-1000 1001-1100 1101-1200 1201-1300 Income Number of 5 15 80 120 200 100 60 20 families 65
- 18. Solution: Note: Since the variable are in the inclusive form, it has to be convert into exclusive form L1 = ½ of 1 ‘1’ is the difference between the upper limit of preceding class & the lower limit of next L2 + ½ of 1 class X f cf Median = l1+ l2 – l1 (m –c) 500.5 – 600.5 5 5 f 600.5 – 700.5 15 20 Where m = N/2 = 600/2 = 300 700.5 – 800.5 80 100 Median class = 900.5 – 1000.5 800.5 – 900.5 120 220 = 900.5 + 1000.5 + 900.5 (300 – 200) 900.5 – 1000.5 200 420 300 200 1000.5 – 1100.5 100 520 = 900.5 + 100 x 100 1100.5 – 1200.5 60 580 200 1200.5 – 1300.5 20 600 = 950.5 N = 600 ILLUSTRATION = 08 From the following data the median marks Marks 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 Frequency 7 15 18 25 30 20 16 7 2 Solution Note: Convert the inclusive series into exclusive form. Marks f cf Median = l1+ l2 – l1 (m –c) X f 9.5 – 19.5 7 7 Where m=N/2 = 140/2 =70 19.5 – 29.5 15 22 Median = 49.5 – 59.5 29.5 – 39.5 18 40 = 49.5 + 59.5 + 49.5 (70 – 65) 39.5 – 49.5 25 65 30 49.5 – 59.5 30 95 70 = 49.5 + 10 x 5 59.5 – 69.5 20 115 30 69.5 – 79.5 16 131 = 49.5 + 1.67 79.5 – 89.5 7 138 Median = 51.17 Marks 89.5 – 99.5 2 140 N= 140 ILLUSTRATION = 9 Compute Median from the following data. Mid Value 115 125 135 145 155 165 175 185 195 Frequency 6 25 48 72 116 60 38 22 3 Solution: Note: The given problem is a continuous frequency distribution Mid – values of the class – intervals are given. The Difference between two mid – values is 10 Divide this 10 by 2, to get half class interval i.e. 10/2 = 5 This 5 is to be deducted from each mid value to get lower limit and 5 is added to get the upper limit of class. 66
- 19. Values F cf Median = l1+ l2 – l1 (m –c) x f 110 – 120 6 6 Where m = N/2 = 390/2 = 195 120 – 130 25 31 Median Class = 150 – 160 130 – 140 48 79 = 150 + 160 – 150 (195 –151) 140 – 150 72 151 116 150 – 160 116 267 185 = 150 + 10 x 44 160 – 170 60 327 116 170 – 180 38 365 = 150 +(440/116) 180 – 190 22 387 = 150 + 3.79 190 – 200 3 390 = 153.79 N = 390 ILLUSTRATION = 10 Marks Obtained less than 10 20 30 40 50 60 70 80 No. of Students 4 16 40 76 96 112 120 125 Solution Let us convert the data from less than frequency distribution into normal distribution. Marks F cf Median = l1+ l2 – l1 (m –c) x f 0 – 10 4 4 Where m = N/2 = 125/2 = 62.5 10 - 20 12 16 = 30 + 40 – 30 x (62.5 – 40) 20 – 30 24 40 36 30 – 40 36 76 62.5 = 30 + (10/36) x 22.5 40 – 50 20 96 = 30 + (225 / 36) 50 – 60 16 112 = 30 + 6.25 60 – 70 8 120 M = 36.25 70 – 80 5 125 N = 12 5 ILLUSTRATION = 11 The marks obtained by 65 students in statistics are shown in the table given, calculate Median marks. Marks No. of Students More than 20% 65 More than 30% 63 More than 40% 40 More than 50% 40 More than 60% 18 More than 70% 7 Solution Marks Students cf Median = l1+ l2 – l1 (m –c) X F f 20 – 30 2 2 Where m = N/2 = 65/2 = 32.5 30 – 40 23 25 Median class = 50 – 60 40 – 50 0 25 = 50 + 60 – 50 (32.5 – 25) 50 – 60 22 47 32.5 22 60 – 70 11 58 = 50 + (10/22)x 7.5 70 – 80 7 65 = 50 + 3.4 N= 65 = 53.4 marks 67
- 20. Merits of Median 1. It is easy to understand easy to compute. 2. It is quite rigidly defined. 3. It is eliminates the effect of extreme items. 4. It is amenable to further algebraic process. 5. Median can be calculated even from qualitative phenomena 6. Its value generally lies in the distribution. Demerits 1. Typical representative of the observations cannot be computed if the distribution of items is irregular 2. It ignores the extreme items. 3. It is only a locative average, but not computed average. 4. It is more effected by fluctuations of sampling than in Mean. 5. Median is not amenable to further algebraic manipulation. QUARTILES The quartiles are also positional averages like the median. As the median value divides the entire distribution into two equal parts, the quartile divide the entire distribution into four equal parts. A measure while divides an array into four equal parts is known as quartile. Basically there are three such points – Q1, Q2 & Q3 – termed as three quartiles. The first quartile (Q1) or lower quartile has 25% of the items of the distribution below it and 75% of the items are greater than it. Incidentally Q2 the second quartile, coincide with the median, has 50% of the observations above it and 50% of the observations below it. The upper quartile or third quartile (Q3) has 75% of the items of the distribution below it and 25% of the items are above it. These quartiles are very helpful in understanding the formation of a series. They tell us how various items are spread round the median. Their special utility Rs. in a study of the dispersion of items. The working principle for computing the quartile is basically the same as that of computing the median. Calculation of Quartiles Q1 = The size of (n + 1)th item 4 Q3 = The size of 3(n +1)th item 4 INDIVIDUAL SERIES:- ILLUSTRATION – 01 From the following data given below, calculate the first quartile and third quartiles Values: 6 10 8 14 12 16 22 20 18 Solution Value rearranged in ascending order Values Q1 = The size of n + 1th item X 4 6 = The size of (9 + 1) = 10/4 = 2.5th item 8 4 10 = 2nd item + 0.5 of difference between 8 & 10 12 = 8 + 0.5 of 2(10 – 8) 14 16 Q1 = 9 18 Q3 = The size of 3(n + 1)th item 20 4 22 = -------“------- 3(9+1) = (3x10)/4 = 30/4 = 7.5th item n=9 4 = 7 item + 0.5 of difference between 18 & 20 = 18 + 0.5 of 2 (20 - 18) Q3 = 19 68
- 21. ILLUSTRATION : 02 Find the quartiles from the following data Values 19 27 24 39 57 44 56 50 59 67 62 Solution Values arranged in ascending order Values Q1 = The size of n + 1th item x 4 19 = The size of (11 + 1)/4 = 12/4 = 3rd item 24 Q1 = 27 27 Q1 39 Q2 or Median = the size of n + 1th item 44 2 50 Q2 or M = the size of (11 + 1)/2 = 12/2 = 6th item 56 Q2 or M = 50 57 59 Q3 Q3 = The size of 3(n+1)th item 62 n 67 = The size of 3 (11 + 1) 3 x 12 = 9th item n = 11 4 4 Q3 = 59 ILLUSTRATION = 03 A manager while in specting his departments found the number absentees in 10 Departments as follows calculate Q1 & Q3 21 12 15 17 18 18 20 19 0 8 Solution Given values re – arranged in ascending order Values Q1= The size of n + 1th item x 4 0 = The size of (10 + 1)/4 = 11/4 = 2.75th item 6 = 6 + 0.75 (12 –6) = 6 + 0.75 of 6 + 4.5 12 = 10.5 15 17 Q3 = The size of 3(n +1)th item 18 4 18 = ----“--- 3(10 + 1) = 3 x 11 = 8.25th item 19 4 4 20 = 8th item + ¼ (9th item – 8th item) 21 = 19 + ¼ of 1 (20 – 19) N = 10 Q3 = 19.25 DISCRETE SERIES Calculation of Quartiles from discrete series. Steps 1. Find the less than type cumulative frequencies 2. Calculate quartiles using the formula Q1 = The size of N+1th item’s value 4 Q2 = The size of N + 1th items value 2 Q3 = The size of 3(N + 1)th items value 4 ILLUSTRATION = 04 Locate median and quartile from the following data Size of Shoe 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 No. of pairs 20 36 44 50 80 30 20 16 14 69
- 22. Solution Variable Median = The size of N + 1th item’s value f Cf X 2 4.0 20 20 = The size of 310 + 1 = 311 = 155.5 4.5 36 56 77.75 2 2 5.0 44 100 M = 6.0 5.5 50 150 155.5 Q1 = The size of N+1th item’s values 6.0 80 230 233.25 4 6.5 30 260 = The size of (310 + 1)/4 = 77.75 7.0 20 280 Q1 = 5.0 7.5 16 296 Q3 = The size of 3(N+ 1)th item’s values 8.0 14 310 4 N 310 = The size of 3(310 +1) = 233.5 4 Q3 = 6.5 ILLUSTRATION = 05 Calculate the value of the median and quartiles from the following data Marks 25 30 35 40 45 50 55 60 65 70 75 80 85 No. of Students 12 18 22 18 43 17 20 10 30 20 10 15 5 Solution Marks Students M = The size of N + 1th items values cf X f 2 25 12 12 = The Size of (240 + 1)/ 2 = 241/ 2 = 120.5 30 18 30 M = 50 35 22 52 Q1 = The size of N+1th item’s values 40 18 70 60.25 4 45 43 113 = The Size of (240+1)/4 = 241/4 = 60.25 50 17 130 120.5 Q1 = 40 55 20 150 60 10 160 Q3 = The size of 3(N + 1)th items values 65 30 190 180.7 4 70 20 210 5 = The size of 3(240 + 1) = 3 x 241 = 180 .75 75 10 220 4 4 80 15 235 Q3 = 65 85 5 240 240 CONTINUOUS SERIES Steps 1. Find out the cumulative frequency 2. Find out the first quartile item q1 = N/4 3. Find out the third quartile item q3 = 3N/4 Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 f Q2 or M = l1 + l2 – l1 (m – c) where m = N/2 f Q3 = l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 f ILLUSTRATION = 06 Calculate the value of the Median and Quartiles from the following data. Values x 4–6 6–8 8–10 10–12 12-14 14–16 16-18 18-20 20-22 22-24 24-26 26-28 Frequency f 10 5 20 15 12 25 14 26 32 18 20 10 70
- 23. Solution X F cf Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 = 207/4 = 51.75 4–6 10 10 f 6–8 5 15 Lower quartile class 12 – 14 8 – 10 20 35 = 12 + (14-12) (51.75 – 50) 10 - 12 15 5051.75 12 12 – 14 12 62 = 12 +(2/12) x 1.75 = 12 + (3.5/12) = 12 + 0.29 14 – 16 25 87 Q1 = 12.29 16 – 18 14 101103.5 Q2 or M = l1 + l2 – l1 (m – c) where m = N/2 18 – 20 26 127155.25 f = 207/2 = 103.5 20 – 22 32 159 Median class 18 – 20 22 – 24 18 177 M = 18 + 20 –18 (103.5 – 101) 24 – 26 20 197 26 26 – 28 10 207 M = 18 + (2/26) x 2.5 = 18 + (5/26) = 18 + 0.19 207 M = 18.19 Q3 = l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 = (3x207)/4 = 155.25 f Upper Quartile lies between 20 – 22 = 20 + 22 – 20 (155.25 – 127) 32 = 20 + (2/32) x 28.25 = 20 + (56.50/32) = 20 + 1.765 = 21.765 ILLUSTRATION = 07 Find the median and quartiles from the following data. Size 11 – 15 16 –20 21 – 25 26 –30 31 –35 36 – 40 41 – 45 46 – 50 Frequency 7 10 13 26 35 22 11 5 Solution: Convert inclusive series into exclusive series X f cf Q1 = Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N = 129 = 32.25 10.5 – 15.5 7 7 f 2 4 15.5 – 20.5 10 17 = 25.5 + 30.5 – 25.5 (32.25 – 30) 20.5 – 25.5 13 30 32.25 26 25.5 – 30.5 26 56 64.5 Q1 = 25.93 30.5 – 35.5 35 91 96.75 35.5 – 40.5 22 113 Median = l1 + l2 – l1 (m – c) where m = N/2 40.5 – 45.5 11 124 f 45.5 – 50.5 5 129 = 30.5 +35.5 – 30.5(64.5 – 56) N 129 35 M = 31.71 Q3 =l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 = (3x129) = 96.75 f 4 = 35.5 + (40.5 – 35.5) (96.75 – 91) 22 Q3 = 36.81 ILLUSTRATION = 08 Form the following data, calculate median & quartiles Wages Below 6 6 – 10 11 – 15 16 – 20 21 – 25 26 – 30 31 – 35 36 – 40 Total Workers 03 10 20 30 20 9 5 3 100 71
- 24. Solution: Convert inclusive series into exclusive series X f cf Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 = 100/4 = 25 0.5 – 5.5 03 03 f 5.5 – 10.5 10 13 25 = 10.5 + 15.5 – 10.5 (25 – 13) = 10.5 + (5/20) x 12 10.5 – 15.5 20 33 50 20 15.5 – 20.5 30 63 75 Q1 = 13.5 20.5 – 25.5 20 83 Median = l1 + l2 – l1 (m – c) where m = N/2 = 100/2 = 50 25.5 – 30.5 9 92 f 30.5 – 35.5 5 97 = 15.5 +20.5 – 15.5(50 – 33) 35.5 – 40.5 3 100 30 N 100 =15.5 +(5/30) x 17 M = 18.33 Q3 =l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 = (3x100)/4 = 75 f = 20.5 + 25.5 – 20.5 (75 – 63) 20 = 20.5 + (5/20) x 12 Q3 = 23.5 ILLUSTRATION = 09 Calculate median and quartiles from the following data Expenses Less than 5 10 15 20 25 30 35 40 45 No. of Students 5 8 28 37 67 70 90 96 100 Solution Convert the cumulative frequency distribution into an ordinary frequency distribution. Expenditure Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 = 100/4 = 25 f cf X f 0–5 5 5 = 10 + 15 – 10 (25 – 8) = 10 + (5/20) x 17 5 – 10 3 8 25 20 10 - 15 20 28 Q1 = 14.25 15 – 20 9 37 50 20 – 25 30 67 Median = l1 + l2 – l1 (m – c) where m = N/2 = 100/2 = 50 25 – 30 3 70 75 f 30 – 35 20 90 = 20 +25 – 20(50 – 37) 35 - 40 6 96 30 40 – 45 4 100 = 20 +(5/30) x 13 N 100 M = 22.16 Q3 =l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 = (3x100)/4 = 75 f = 30 + 35 – 30 (75 – 70) 20 = 30 + (5/20) x 5 Q3 = 31.25 ILLUSTRATION = 10 Calculate the Values of Median and quartiles from the following data Age More than 0 10 20 30 40 50 60 70 80 90 100 No. of persons 80 77 72 65 55 43 28 16 10 8 0 72
- 25. Solution Convert the given cumulative frequency distribution into an ordinary table. X f Cf 0 – 10 3 3 10 - 20 5 8 Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 = 80/4 = 20 20 – 30 7 15 f 30 – 40 10 25 = 30 + 40 – 30 (20 – 15) = 30 + (10/10)x 5 40 – 50 12 37 20 10 50 – 60 15 52 Q1 = 35 60 – 70 12 64 70 – 80 6 70 40 60 Median = l1 + l2 – l1 (m – c) where m = N/2 = 80/2 = 40 80 – 90 2 72 f 90 – 100 8 80 = 50 + 60 – 50(40 – 37) 100 – 110 0 80 30 N 80 = 50 + (10/15) x 3 M = 52 Q3 =l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 = (3x80)/4 = 60 f = 60 + 70 – 60 (60-52) 20 = 60 + (10/12) x 8 Q3 = 66.67 MODE Mode is the most common item of a series. Mode is the values which occurs the greatest numbers of frequency in a series. It is derived from the French word ‘Lamode’ meaning the fashion. Mode is the most fashionable or typical value of a distribution, because it is repeated the highest number of times in a series. According to Croxton and Cowden “ The mode of a distribution is the value at the point around which the item tend to be most heavily concentrated”. Mode is defined as the value of the variable which occurs most frequently in a distribution. The chief feature of mode is that it is the size of that item which has the maximum frequency and is also affected by the frequencies of the neighboring items. Calculation of Mode – Individual series Mode can often be found out by mere inspection in case of individual observations. The data have to be arranged in the form of an array so that the value which has the highest frequency can be known. For example : 10 persons have the following income Rs.850, 750, 600, 825, 850, 725, 600, 850, 640, 850 hear Rs.850 repeats four times. Therefore model salary is Rs.850. In certain cases that there may not be a mode or there may be more than one more for example. a. 40, 44, 57, 78, 48 -------------No mode b. 45, 55, 50, 45, 55 -------------Bimodal When we calculate the mode from the given data, if there is only one mode in the series, it is called unimodal, if there are two modes, it is called Bimodal. DISCRETE SERIES:- In a discrete series mode can be located in two ways 1. By Inspection 2. By Grouping Inspection Method:- When there is a regularity and homogeneity in the series, then there is a single mode which can be located at a glance by looking into the frequency column for having maximum frequency. 73
- 26. ILLUSTRATION = 01 Find mode from the following series. Height in cm 150 160 170 180 190 200 210 No. of persons 2 4 8 15 6 5 3 Solution: By inspection of the frequency, it is noted that the maximum frequency is 15 which corresponds to the value 180. Hence, Mode is 180cm. GROUPING METHOD: When there are irregularities in the frequency distribution or two or more frequencies are equal then it is not obvious that which one is the maximum frequency. In such cases, we use the method of grouping to decide which one may be considered as maximum frequency. That is we try to find out single mode by using grouping method. This method involves the following steps. a. Prepare grouping table b. Prepare analysis table c. Find mode a. Preparing a grouping table: Steps: construct a table of 6 columns. Column 1:- Given frequencies Column 2:- The given frequencies are added in two’s Column 3:- The frequencies are added in two’s leaving out the first frequency Column 4:- The given frequencies are added in three’s Column 5:- The given frequencies are added in three’s leaving out the first frequency Column 6:- The given frequencies are added in three’s leaving out the first two frequencies. After making these columns, the maximum frequency is underlined or round off b. Construction of analysis Table:- The values containing the maximum frequencies are noted down for each column and are written in a table called analysis Table. c. Location of Mode – The value of the variable which occurs maximum number of times in the analysis table, is the value of mode. ILLUSTRATION = 02 Calculate mode from the following data Variable 8 9 10 11 12 13 14 15 Frequency 5 6 8 7 9 8 9 6 Solution Here maximum frequency 9 belongs to two values of the items 12 & 14. However due to irregular distribution of frequencies, we use the method of grouping to decide which one may be considered maximum frequency. Sum of Frequencies Analysis Table Values F1 2 3 4 5 6 Tally bars Repeat 8 5 9 6 11 14 19 1 1 10 8 21 1 1 2 11 7 15 16 24 11 11 1 5 12 9 11 11 4 13 8 17 17 24 26 1 1 1 3 14 9 23 15 6 15 74
- 27. Separate analysis Table Columns no Size of items containing maximum frequency 10 11 12 13 14 Since 12 occurs maximum number of times. Hence 1 12 14 the value 12 is the mode. 2 12 13 3 12 13 14 4 11 12 13 5 -- 13 14 6 10 11 12 1 2 5 4 3 ILLUSTRATION = 03 Calculate the mode from the following Size : 10 11 12 13 14 15 16 17 18 Frequency: 10 12 15 19 20 8 4 3 2 Solution: Grouping Table Size F Sum of 2 Fre Sum of 3 Fre Analysis Table 1 2 3 4 5 6 Tally bars Repeat 10 10 11 12 22 27 37 1 1 12 15 46 1 11 3 13 19 34 39 54 11111 5 14 20 47 1 11 1 4 15 8 28 12 32 1 1 16 4 9 15 17 3 7 5 18 2 ANALYSIS TABLE Size of item containing maximum frequency Column 11 12 13 14 15 The mode is 13 as size of item repeats five times. 1 14 But through inspection we say the mode is 14, 2 12 13 because the size 14 occurs 20 times. But this wrong 3 13 14 decision is revealed by analysis Table. 4 13 14 15 5 11 12 13 6 12 13 14 1 3 5 4 1 CONTINUOS SERIES – CALCULATION OF MODE: Steps: - 1. Determine the model class (in exclusive). The class having the maximum frequency is called model class. This is done either by inspection or by grouping method. 2. Determine the value of mode by applying the formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2 Z = Mode L1 = Lower limit of the model class L2 = Upper limit of the model class f1 = Frequency of the model class f0 = Frequency of the pre – model class f2 = Frequency of the class succeeding the model class. Note: - This formula is applicable only in case of equal and exclusive class – intervals in ascending order. If the mode lies in the first class interval, than f0 is taken as zero. If mode lies in the last class interval then f2 is taken as zero. 75
- 28. ILLUSTRATION = 04 The distribution of wages in a factory is as follows calculate the mode. Wages in Rs. 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 No. of workers 6 9 10 16 12 8 7 Solution: - By inspection, the maximum frequency is 16, Hence the model class is 30 – 40 Formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2 = 30 + 16 - 10 (40 –30) 2 x 16 – 10 – 12 = 30 + (6/32 – 22) x 10 = 30 + (60 / 10) = 36 ILLUSTRATION = 05 Determine the mode in the following frequency distribution. Income in Rs 100-200 200-300 300-400 400-500 500-600 600-700 700-800 No. of persons 30 70 80 100 20 30 20 Solution By inspection, the maximum frequency is 100, Hence the modal class is 400 – 500 Formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2 = 400 + 100 - 80 (500 - 400) 2 x 100 – 80 – 20 Mode = 420 ILLUSTRATION = 06 Calculate mode from the following series Variable 0–5 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 Frequency 1 2 10 4 10 9 2 Solution: Since the frequencies of class 10 – 15 and 20 – 25 are 10 each, we have to decide the model class by grouping f Sum of two Sum of three Analysis table Value 1 2 3 4 5 6 Tally bar Repeat 0–5 1 5 – 10 2 3 12 13 10 – 15 10 16 1 1 2 15 – 20 4 14 14 24 11 1 3 20 – 25 10 23 1 1111 1 6 25 – 30 9 19 11 21 1 11 3 30 – 35 2 1 1 Separate Analysis CN 10 – 15 15- 20 20 – 25 25 – 30 30 – 35 1 10 –15 20 –25 2 20 – 25 25 – 30 3 15 –20 20 – 25 4 15 –20 20 – 25 25 – 30 5 20 – 25 25 – 30 30 – 35 6 10 –15 15– 20 20 – 25 2 3 6 3 1 Therefore model class is 20 – 25, with maximum number of times; Formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2 = 20 + 10- 4 (25 - 20) 2 x 10 – 4 – 9 = 20 + (6/ 20 - 13) x 15 = 24.28 Calculate Mode in inclusive series 76
- 29. ILLUSTRATION = 07 Calculate mode for the following data Classes 0–9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 - 79 Frequency 1 2 6 6 12 8 5 3 Solution – Convert inclusive series into exclusive series Class F By Inspection, the maximum frequency is 12, hence the modal class in 39.5 - 0. 5 – 9.5 1 – 49.5 9.5 – 19.5 2 Formula Z = L1 + f1 – f0 (L2 – L1) 19.5 – 29.5 6 2f1 – f0 – f2 29.5 – 39.5 6 = 39.5 + 12 - 6 (49.5 – 39.5) 39.5 – 49.5 12 2 x 12 – 6 – 8 49.5 – 59.5 8 = 39.5 + (6/ 2x 12 – 6 - 8) x 10 59.5 – 69.5 5 = 39.5 + (60/ 10) 69.5 – 79.5 3 Z = 45.5 To Calculate mode in case of un – equal class intervals. ILLUSTRATION = 08 Calculate the value of mode from the following data Classes 0–5 5 – 7 7 – 9 9 – 10 10 – 13 13-15 15-16 16-19 19-20 20-21 21 -25 Frequency 2 3 4 2 10 5 3 4 3 2 3 Solution:- Re – arrange the above series into a series with equal class intervals. Classes 0–5 5 – 10 10 – 15 15 – 20 20 - 25 Frequency 5 9 15 10 5 By inspection, the maximum frequency is 15, hence the modal class is 10 – 1. Formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2 = 10 + 15- 9 (15 – 10) 2 x 15 – 19 – 10 = 10 + (6/ 30 -19) x 8 Z = 12. 73 ILLUSTRATION – 09 Find the mode of the following data. Also calculate the value of Median – Height in inches 56 –62 63 – 65 66 – 68 69 – 71 72 – 84 Frequency 15 54 60 81 24 Solution: - Since the class intervals are not equal and cannot be converted into equal class – intervals. So we use empirical formula Z = 3m – 2Mean to find mode. X f cf mid value dx = x – 67 fdx 0 = A + Σfdx x c 55.5 – 62.5 15 15 59 -8 -120 N 62.5 – 65.5 54 69 64 -3 -162 = 67 + (225/234)x1 65.5 – 68.5 60 129 67 0 0 = 67 + 0.96 68.5 – 71-5 81 210 70 3 243 0 = 67.96 71.5 – 84.5 24 234 78 11 264 N= 234 Σfdx= 225 Median = L1 + L2 – L1 (m – c) Where m = N/2 = 234/2 = 117.0 f Median class 65.5 – 68.5 = 65.5 + 68.5 – 65.5 (117.0 – 69) 60 = 65.5 + (3/60) x 48 = 67.9 Mode = 3 Median – 2 Mean Z = 3M - 20 77
- 30. = 67.9 x 3 – 2x 67.96 Z = 67.78 ILLUSTRATION = 10 Find mode of the following frequency distribution Class 14 –15 16 – 17 18 – 20 21 – 24 25 – 29 30 - 34 35 – 39 Frequency 6 14 15 11 11 10 9 Solution Since the class intervals are not equal, we may use the empirical formula to find mode. Z = 3M - 20 Convert into exclusive series A = 22.5, C = 1, dx(x – A)/c Class f cf Mid value dx =(x – A)/c Fdx X 13.5-15.5 6 6 14.5 -8 -48 0 = A + Σfdx x c 15.5-17.5 14 20 16.5 -6 -84 N 17.5-20.5 15 35 19.5 -3 -45 = 22.5+ (98/76)x1 20.5-24.5 11 46 22.5 0 0 0 = 23.78 24.5-29.5 11 57 27.0 4.5 49.5 29.5-34.5 10 67 32.0 9.5 95.0 34.5-39.5 9 76 37.0 14.5 130.5 N= 76 -177 + 275.9 Σfdx =98 Median = L1 + L2 – L1 (m – c) Where m = N/2 = 76/2 = 38 f Median class 20.5 – 24.5 = 20.5 + 24.5 – 20.5 (38 – 35) 11 = 20.5 + (4/11) x 3 = 21.59 Mode = 3 Median – 2 Mean Z = 3M - 20 = 3 x 21.59 – 2x 23 .78 Z = 17.21 ILLUSTRATION = 11 Calculate mode of the following frequency distribution Mid Value 55 65 75 85 95 105 115 Frequency 8 10 16 14 10 5 2 Solution Prepare class intervals with the help of the mid value. By inspection maximum frequency is 16 Hence Modal class is 70 - 80 X f .’. Z = L1 + f1 – f0 (L2 – L1) 50 – 60 8 2f1 – f0 – f2 60 – 70 10 = 70 + 16 – 10 (80 – 70) 70 – 80 16 2 x16 – 10 –14 80 – 90 14 = 70 + (6/ 32 –24)x 10 90 – 100 10 Z = 77.5 100 – 110 5 110 – 120 2 To calculate mode in case of cumulative frequency distribution 78
- 31. ILLUSTRATION = 12 Calculate mode from the following data. Marks more than 0 10 20 30 40 50 60 70 80 90 100 Students 80 77 72 65 55 43 28 16 10 8 0 Solution: Convert the given cumulative frequency distribution to ordinary table. X No. of Students Since maximum frequency is 15, Hence modal class is 50 – 60 0 – 10 80 – 77 = 3 10 - 20 77 – 72 = 5 Z = L1 + f1 – f0 (L2 – L1) 20 – 30 72 – 65 = 7 2f1 – f0 – f2 30 – 40 65 – 55 = 10 = 50 + 15 – 12 (60 – 50) 40 – 50 55 – 43 = 12 2 x 15 – 12 –12 50 – 60 43 – 28 = 15 = 50 + (3 / 30 –24)x10 60 – 70 28 – 16 = 12 Z = 55 70 – 80 16 – 10 = 6 80 – 90 10 – 8 = 2 90 – 100 8–0=0 ILLUSTRATION = 13 Calculate Median and mode from the following data Marks less than 20 30 40 50 60 70 80 90 100 No. of Students 15 40 72 128 206 246 267 290 300 Solution Convert the given cumulative frequency distribution into an ordinary table x No. of Students cf Median = L1 + L2 – L1 (m – c) Where m = N/2 = 300/2 =150 10 – 20 15 15 f 20 – 30 40 – 15 = 25 40 Median class is 50 – 60 30 – 40 72 – 40 = 32 72 = 50 + 60 – 50 (150 – 128) = 50 + (10 /78)x 22 = 52.82 40 – 50 128 – 72 = 56 128 78 50 – 60 206 –128 = 78 206 By inspection maximum frequency is 78. Hence modal class is 50 - 60 – 70 246 – 206 = 40 246 60 70 – 80 267 – 246 = 21 267 Z = L1 + f1 – f0 (L2 – L1) 80 – 90 290 – 267 =23 290 2f1 – f0 – f2 90 – 100 300 – 290 = 10 300 = 50 + 78 – 56 (60 – 50) N = 300 2 x 78 – 56 – 40 = 50 + (220/60) = 53.67 Merits of Mode Mode as a measure of central tendency has some merits. 1. It is simple to understand and it is easy to calculate. 2. Generally it is not affected by end values. 3. It can be determined graphically & it can be found out by inspection. 4. It is usually an actual value as it occurs most frequently in the series. 5. It is the most representative average. 6. Its value can be determined in an open end class interval without ascertaining the class limit. Demerits of mode 1. It is not suitable for further mathematical treatment. 2. It may not give weight to extreme items. 3. In a bimodal distribution there are two modal classes and it is difficult to determine the value of the mode. and it is difficult to determine the value of the mode. 4. It is difficult to compute, when are both positive and negative items in a series. 5. It is not based on all the observations of a given series. 6. It will not give the aggregate value as in average. RELATIONSHIP BETWEEN 3’M’S OR EMPIRICAL RELATIONSHIP In a symmetrical distribution. Mean Median and Mode will coincide i.e. 0 = M = Z. But in an asymmetrical (skewed) distribution, these values will be different when the distribution is moderately skewed and 79
- 32. has greater concentration in the lower values 0 >Median >Z, it means the distribution is positively skewed. If the distribution concentrated in higher values, the tail is towards the lower values, then it is gentatively skewed. In a moderately asymmetrical distribution the difference between 0 and Z is three times of the differences between 0 and median. Symbolically (Karl Pearson’s) Mode = 3 Median – 2 Mean Z = 3M - 20 When the value of mode is ill – defined for a given series in such case, Mode value can be estimated by using the above formula. ILLUSTRATION – 15 If in a moderately asymmetrical frequency distribution, the values of Median and Mean are 72 and 78 respectively estimate the value of the mode. Solution. The value of the mode is estimated by applying the following formula Mode = 3 Median – 2 Mean = 3 x 72 – 2 x78 = 216 – 156 Z = 60 ILLUSTRATION = 16 The mean and mode in a moderately symmetrical distribution are 24.6 and 26.1 respectively. Find the median. Solution :- Formula Z = 3M - 20 26.1 = 3M – 2 x 24.6 26.1 = 3M – 49.2 -3M = -49.2 – 26.1 M = -75.3/-3 M = 25.1 ILLUSTRATION = 17 If mode is 15.99 and median is 15.73, find the most probable value of the Mean, by using 3M’s formula Z = 3M - 20 15.99 = 3 x 15.73 - 20 15.99 = 47.19 – 20 20 = 47.19 - 15.99, 0 = 31.20/2, 0= 15.60 CALCULATION OF ARITHMETIC MEAN, MEDIAN, MODE AND QUARTILES PROBLEM = 01 From the following data calculate the values of Mean, Median, Mode and Quartiles. Classes 0 –3 3 –6 6 –10 10–11 11-15 15-19 19-20 20-24 24-30 30-33 33-37 37-40 Frequency 04 12 24 32 44 52 52 28 20 08 04 10 Solution :A = 17, C = 1, dx = (x – c)/c 80
- 33. x F Cf midx dx fdm 0 = A + Σfdm x c 0 –3 04 04 1.5 -15.5 -62 N 3 –6 12 16 4.5 -12.5 -150 = 17 + (49/290) x 1 6 – 10 24 40 8.0 -9.5 -228 0 = 17.168 10 - 11 32 72 10.5 -6.5 -208 11 –15 44 116 13.0 -4.0 -172 Median = L1 + L2 – L1(m –c ) 15 –19 52 168 17.0 0.0 0 f 19 – 20 52 220 19.5 2.5 130 Where m = N/2 = 290/2 = 145 20 –24 28 248 22.0 5.0 140 Median class is 15 - 19 24 –30 20 268 27.0 10.0 200 30 –33 08 276 31.5 14.5 116 33 – 37 04 280 35.0 18.0 72 37 – 40 10 290 38.5 21.5 215 N= 290 Σfdm= 49 -824 +873 Note: To Calculate the Value of mode, the given series Median = L1 + L2 – L1(m –c ) must be re – arranged with equal class intervals f Values M = 15 + 19 – 15 (145 – 116) F X 52 By Inspection, the highest = 15 + (4/52)x29 = 15 + 2.23 = 17.23 0 – 10 40 frequency is 180 so modal class is Q1= L1+ L2 – L1 (q1–c) Where q1=N/4 = 290/4=72.5 10 - 20 180 10 –20 f 20 –30 48 30 – 40 22 Lower quartile class = 11 –15 N= 290 = 11 + 15 – 11 (72.5 – 72) Formula 44 Z = L1 + f1 – f0 (L2 – L1) = 11 + (4/44)x0.5 = 11 + (2.0/44) = 11.05 2f1 – f0 – f2 Q3 = L1+ L2 – L1 (q1–c) = 10 + 180 – 40 (20 – 10) f 2 x 180 – 40 –48 Where q3 = 3N/4 = (3x290)/4 =217.5 = 10 +(140/360-85)x10 = 10+ (1400/275) = 15.09 Upper quartile class = 19 – 20 = 19 + 20 – 19 (217.5 –168) 52 = 19 + (1/52)x49.5 = 19+0.95 = 19.95 Problem = 02 From the following data of the weekly wages of workers employed in a certain factory, construct a frequency table with 0 – 10, 10-20 and soon. Find Mean, Median, Mode and quartile Value 3 5 1 3 5 3 1 6 5 4 3 6 6 3 4 8 7 5 8 15 s 5 6 0 7 0 5 2 1 3 0 0 2 4 1 9 1 3 2 1 2 2 5 6 4 2 6 1 4 4 5 4 2 4 9 37 2 5 4 3 2 8 3 9 5 0 0 9 4 0 7 8 7 1 5 3 6 3 1 4 2 4 5 5 3 5 3 3 7 2 5 4 Solution A = 35, C=10, dx = (x – A)/c Fdm 0 = A + Σfdm xc Values Tally bars f cf midx Dx X N 0 – 10 1111 1 06 6 05 -3 -18 = 35 + (-2/50) x 10 =35-(20/50) 10 - 20 1111 111 08 14 15 -2 -16 0= 34.6 20 –30 1111 1 06 20 25 -1 -6 30 –40 1111 1111 09 29 35 0 0 Where m = N/2 = 50/2 = 25 40 –50 1111 1111 09 38 45 1 9 Median class is 30 – 40 50 –60 1111 11 07 45 55 2 14 60 –70 1111 5 50 65 3 15 N=50 Σfdm= -2 Where m = N/2 = 50/2 = 25 Q1= L1+ L2 – L1 (q1–c) Where q1=N/4 = 50/4=12.5 Median class is 30 – 40 f Median = L1 + L2 – L1(m –c ) Lower quartile class = 10 – 20 f 81

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