BBA102_BCM102_Business Mathematics and Statistics

Classification and Tabulation of Data 195 with their strength in 1961-62. In 1963-64 the total strength of the college was 1,250. The number of students in F.Y. Com Class was double the number of students in Sr. B.Com. class. The number of students in 1. Com. class was 330.” 8. Represent the information given below by a suitable table. “The number of students in a college in the year 1941 was 510. Of these 480 were boys and the rest girls. In 1951, the number of boys increased by 100% and that of girls increased by 300% as compared to their strengths in 1941. In 1961, the total number of students in the college was 1200, the number of boys being double than the number of girls.” 9. Present the following information in a concise tabular form and indicate with type of lamp shows the greatest wastage during manufacture: ‘Lamps are rejected at several manufacturing stages for different fault. 12,000 glass tubes are supplied to make 40-watt, 60-watt and 100-watt lamps in the ratio 1:2:3. At the stage I, 10 per cent of the 40-watt, 4 per cent of the 60-watt and 5 per cent of the 100-watt bulbs are broken. At the stage II, about 1 percent of the remainder of the lamps have broken filaments, At the stage III 100 100-watt lamps have badly soldered caps, and half as many have crooked caps; twice as many 40-watt and 60-watt lamps have these faults. In the stage IV, about 3 percent are rejected for bad type-marking and 1 in every 100 are broken in the packing which follows. 9.13 Unit End Questions (MCQ and Descriptive) (A) Descriptive Type 1. Define ‘classification’ and ‘tabulation’ and explain their importance in statistics. 2. What are the different types of tables that you know? Explain them with suitable illustrations. 3. State the precautions that you would take in tabulating statistical data. CU IDOL SELF LEARNING MATERIAL (SLM)

196 Business Mathematics and Statistics 4. Following are the marks secured by 50 candidates at an examination (out of 100 marks). 60 50 49 40 31 23 8 22 0 30 41 40 56 69 66 52 43 41 34 24 19 18 26 35 42 51 53 42 49 53 44 36 25 17 10 29 39 46 44 58 46 38 31 14 21 20 32 33 30 47 Classify the above data in intervals 0-10, 10-20 and so on. 5. Draw up a blank table, with suitable headings, in which the number of workers occupied in five factories, distinguishing males from females and skilled from unskilled, can be entered. 6. Present the following information in a suitable tabular form: TISCOs production for the whole financial year — that is 1963-64 ending March 1964 — of saleable steel, steel ingots and pig iron stood at 1,506,400 metric tonnes, 1,891,700 metric tonnes and 1,811,200 metric tonnes in that order. The corresponding figures for the preceding year were 1,317,700 metric tonnes, 1,799,300 metric tonnes and 1,766,000 metric tonnes, respectively. Source: Quarterly Bulletin of the Eastern Economist Vol. 15, No. 3. 7. Explain the purpose and importance of classification and tabulation of statistical data. What are specific purpose and general purpose-tables? 8. What do you understand by classification of data? Discuss its importance in statistical analysis. 9. From the following observations prepare a frequency distribution table in ascending order starting with 5-10 (exclusive method). CU IDOL SELF LEARNING MATERIAL (SLM)

Classification and Tabulation of Data 197 Marks in English 12 36 40 30 28 20 19 19 27 15 26 20 19 7 26 37 5 20 11 17 37 10 10 16 45 33 21 30 20 5 10. From the following observations prepare a frequency distribution table in ascending order starting with 100-110 (exclusive method). 125 108 112 126 110 132 120 130 136 138 125 111 147 137 145 150 142 135 136 130 149 155 119 125 140 148 137 132 165 154 11. The following are the marks of 100 examinees in statistics out of a maximum of 100. Group them into classes with an interval of 10. 57 44 8 75 0 18 45 14 0 4 64 66 72 51 69 34 56 22 34 8 58 83 20 70 57 28 22 38 5 45 51 88 17 93 64 36 34 37 58 32 64 30 80 73 24 46 48 a 16 65 96 56 20 64 50 63 47 4 32 10 78 48 55 52 66 8 53 50 0 35 28 54 38 33 20 54 52 48 84 50 94 90 38 84 30 58 20 0 99 42 79 33 38 60 61 36 10 34 2 80 CU IDOL SELF LEARNING MATERIAL (SLM)

198 Business Mathematics and Statistics 12. The following are the weekly wages (in rupees) of 40 labourers in a certain factory: 22 21 23 21 19 25 18 20 22 24 28 29 30 35 36 38 34 30 28 26 35 31 34 25 27 31 33 26 27 32 25 26 28 32 24 28 27 31 29 29 Classify the data by taking a class-interval of ` 3. 13. If the class-marks of the first three classes of a frequency distribution are 32.5, 37.5 and 42.5, what are the class boundaries of these classes? If the observations are taken as correct to the nearest integer, what are the class-limits? 14. A record of marks of 180 candidates shows marks ranging in value from 12 to 79. In how many classes would you classify this data? What are the boundaries of the first three classes? What are the class-limits of the first three classes? What are the class- marks of the last three classes? B. Multiple Choice/Objective Type Questions 1. Classifying data under several heads (attributes) is different classes is ________. (a) Simple classification (b) Manifold classification (c) Multiple classification (d) Complex classification (e) None of these 2. The basic requirement of good classifiction is ________. (a) Incomparability (b) Instability (c) Non-flexibility (d) Non-ambiguity (e) None of these 3. ________________ is meant to present the entire original data on a subject. (a) Special purpose table (b) General purpose table (c) Statistical table (d) None of these 4. A one-way table gives answers to questions about ________________ characteristics of the data. (a) Two (b) Three (c) One (d) Four Answers (1) (b); (2) (c); (3) (b); (4) (c) 9.14 References References of this unit have been given at the end of the book. ˆˆˆ CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 199 UNIT 10 MEASURES OF CENTRAL TENDENCY Structure 10.0 Learning Objectives 10.1 Introduction 10.2 Characteristics of an ideal average 10.3 Types of Statistical Averages 10.4 Arithmetic Mean 10.5 Weighted Arithmetic Mean 10.6 Median 10.7 Advantages and Disadvantages of Median 10.8 Mode 10.9 Advantages and Disadvantages of Mode 10.10 Geometric Mean 10.11 Advantages and Disadvantages of Geometric Mean 10.12 Quartiles, Deciles and Percentiles 10.13 Additional Problems (Solved) 10.14 Summary 10.15 Key Words/Abbreviations 10.16 Learning Activity 10.17 Unit End Questions (MCQ and Descriptive) 10.18 References CU IDOL SELF LEARNING MATERIAL (SLM)

200 Business Mathematics and Statistics 10.0 Learning Objectives After studying this unit, you will be able to: z Explain the definition of arithmetic mean, median, mode and the advantages and disadvantages in respect of each of these. z Describe the meaning of dispersion and the type of measures of dispersion that include: Range, Mean Deviation, Quartile Deviation and Standard Deviation. z Grasp the procedure of the calculation of range, mean deviation, Q.D., S.D. and the coefficient of variation in respect of data both discrete and continuous. 10.1 Introduction When statistical data is properly classified and condensed into a frequency distribution, then it is easy for us to study the different characteristics of that data. Further, graphs and diagrams may also be drawn so as to convey a better impression to the mind about the data. But graphs and diagrams are only visual aids which appeal more to the eye than to the mind. Therefore to know more about the data, we feel the necessity of some descriptive measures. Such measures are known as measures of location. Definition Measures which describe, characterise or represent a given complex data, are known as ‘statistical averages’. In other words, every single expression that sums up the characteristics of an entire group of figures or complex data. Descriptive measures such as these generally throw light on the central characteristics of the data aid are therefore present in the central parts. Hence, they are also known as ‘measures of central tendency.’ Main Objects of a Statistical Average The two main objects in calculating a statistical average. Firstly, to indicate precisely the sum and substance of an entire mass of numerical data. Secondly, to serve as a means as well as a measure of comparison with other similar groups of numerical data. CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 201 10.2 Characteristics of an Ideal Average z It should be easily intelligible and workable in the sense that its calculation should be simple. z Its definition should be clearcut, rigid and most appropriate. z It should take into account all the items of the data. z It should not be influenced due to fluctuations of sampling. z It should be useful for the purposes of further algebraic analysis. 10.3 Types of Statistical Averages The different types of Statistical Averages – z Arithmetic Mean, z Median, z Mode, z Geometric Mean, z Harmonic Mean, z Quadratic Mean, z Moving Average, z Progressive Average. These are grouped into three separate groups as: mathematical averages, averages of position, and business averages, arithmetic mean, geometric mean, harmonic mean and quadratic mean are known as mathematical averages since they are calculated by using mainly certain mathematical procedures. It is necessary to consider every item in the data while calculating these averages. Mode and median are known as averages of position since the position of certain items in the data are taken into consideration while finding these. For instance, the middle item in a series of ordered items is a positional item since it occupies the central position. Similarly, an item that is repeated a number of times has a certain position when items are arranged with their corresponding frequencies. CU IDOL SELF LEARNING MATERIAL (SLM)

202 Business Mathematics and Statistics Moving averages and progressive averages are known as business averages since they are of immense use in ascertaining the progress of business. For instance, three-year moving averages of profits or progressive averages of sales may be calculated to determine the trend and progress in business. 10.4 Arithmetic Mean The arithmetic mean (AM.) is the quantity obtained by summing up the values of items in a variable and dividing the sum by the number of items. Individual Items If xl + x2 + x3 … + xn (items) are n items with frequencies. AM = x1  x2  x3 }xn n ¦x AM = n Where x = xl + x2 + x3 … + xn (items) S = ‘Sum of’ or Summation N = Number of Items This method of finding the A.M. is called the direct method. It is very simple and does not involve any mathematical skill, but it is useful only when the items are we and the size of the figure is small. Individual items are many and their size is large it would be difficult to calculate the A.M. with this method. To remove this difficulty a shortcut method is used. We can assume one of the items as an arbitrary (assumed) mean and in how much the items deviate room this assumed mean. We then divide the total number of deviations by the number of items and get a certain quantity. By adding this quantity to the assumed average we obtain the actual A.M. Formula: Arithmetic Mean = {Assumed mean} + {Sum of the deviations of items} From the assumed mean [Total number of items] CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 203 Symbolically, AM. = a + ¦D n Problem 1: Using the shortcut method, in the A.M. of 215, 218, 222, 235 and 270. Let 222 be the assumed mean. Item Deviation (d) 215 – 7 218 – 4 A = 222 222 0 6fd– 50 235 + 13 n = 5 270 + 48 Total + 50 AM. = 222 + 50/5 = 222 + 10 = 232. Discrete Series (Direct Method) If x1 + x2 + x3 … xn (items) are n items with frequencies f1 f2, f3, … fn respectively, then AM = x1f1  x2f2  x3f3 } xnfn f1 f2 f3n }fn AM = ¦in 1xifi n ¦in 1fi ¦x AM = ¦f ¦ fx AM = n where n = Sfn 6 is the summation notation and is pronounced as ‘sigma’. CU IDOL SELF LEARNING MATERIAL (SLM)

204 Business Mathematics and Statistics Direct Method: This method of ending the A.M. of the frequency distribution is called the direct method. Here, the A.M. is obtained by taking the sum of the products of the items and the corresponding frequencies and dividing this sum by the total number of frequencies. Problem 2: Calculate the AM. of the following frequency distribution: Weekly Wages (`) No of Labourers 20 3 25 5 30 8 35 14 40 10 45 6 50 4 Weekly Wages (`) No. of Labourers x X 20 3 60 25 5 125 30 8 240 35 14 490 40 10 400 45 6 270 50 4 200 Total 50 1785 ¦ fx AM = n where n = Sfn Average weekly wages = ` 35.70 Shortcut Method: To minimize calculations in the case of a frequency distribution, it is better to use the shortcut method. In this method the deviations of the quantities room an assumed mean are first calculated; these deviations are then multiplied by their respective frequencies. Then the quantity obtained by dividing the sum of these products (deviation X frequency) by the total number of frequencies is added to the assumed mean. The resulting figure is the actual A.M. CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 205 Symbolically, ¦ fd AM = n Where, D = x – a = deviation, a = assumed mean n = 6f = number of items = total of frequencies The previous example can be worked out using the shortcut method: Weekly No. of Deviation room D Wages (`) X Labourers Assumed Mean D .45 20 3 –15. –50 –40 25 5 –10 0 30 8 –5 +50 +60 [35] 14 0 +60 +35 40 10 +5 45 6 + 10 50 4 + 15 Total 50 Let 35 be the assumed mean. ¦ fd AM = n 35 = 35 + 50 = 35 + 0.70 = 35.70. Average weekly wages = ` 35.70 The calculation can be simplified further by dividing the deviations (D) room the assumed mean by a common actor. The deviations that are divided by the common actor become step deviations (denoted by d). The sum of the products of these step deviations and their respective frequencies (Sfd) are multiplied by this common actor (I). The resulting numerical expression (6fdxi) divided by the total frequencies (n) gives a certain quantity. This quantity, when added to the assumed mean, gives the actual A.M. CU IDOL SELF LEARNING MATERIAL (SLM)

206 Business Mathematics and Statistics ¦ fd Symbolically, Mean = *i n where a = assumed mean. 6fd = sum of the products of the step deviations and their respective frequencies, i = common factor, n = total number of frequencies. In the case of the previous example the calculation is simplified as shown below: Weekly Wages No. of Deviation room d /d `X Labourers Assumed Mean D –3 –9 20 3 –15 –2 –10 25 5 –10 –1 30 8 –8 35 14 –5 0 0 40 10 0 +1 45 6 +2 + 10 50 4 +5 +3 + 12 + 10 + 12 Total 50 + 15 +7 ¦ fd Mean = *i n 7 = 35 + 50 × 5 = 35 + 0.70 = 35.70 Average weekly wages = ` 35.70 Continuous Series In the case of a continuous series, where items are grouped in certain class intervals, the common factor is the length of the class interval, or instance, the length of the class interval 10-15 is 5. The procedure or calculating the A.M. here is: firstly, finding the mid-values of the respective class intervals, so that these mid-values represent the respective classes. Next, one of these mid- values is taken as an assumed mean and deviations (step) of other mid-values room the assumed mean are found, room this step onwards the working procedure and the formula of calculating the A.M. is exactly the same as in the case of the discrete series. CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 207 Formula: /d –18 ¦ fd –22 AM = *i –14 50n 0 Problem 3: Calculate the mean room the following data: + 15 + 18 Marks No. of Students + 15 0-10 6 –6 10-20 11 20-30 14 30-40 20 40-50 15 50-60 9 60-70 5 80 Class Mid values / D D 6 0-10 5 11 –30 –3 10-20 15 –20 –2 20-30 25 14 –10 –1 30-40 35 20 40-50 45 15 0 0 50-60 55 + 10 +1 60-70 65 9 +20 +2 +30 +3 Total 80 Here n = 80. Mean = a+ ¦fd ui n §6 · = 35 + ©¨ 80 ¹¸ × 10 = 35 – 0.75 = 34.25 CU IDOL SELF LEARNING MATERIAL (SLM)

208 Business Mathematics and Statistics Advantages of A.M. Disadvantages of A.M. 1. The common man can easily understand its 1. Omission of even a single item of the data gives meaning and use. an incorrect value of the A.M., unlike median and mode where extreme items can simply be 2. Since its definition is precise and clear, discarded. calculation is easy and its value always determinate. 2. It may not be identical with anyone of the items of the data. That is, it may not be one of the 3. It satisfies most of the conditions laid down or figures that comprise the data. an ideal average. 3. The act that it gives a large weight to extreme 4. Its chief merit consists in the act that it takes into items is its handicap since it then ails to be a consideration every item in the data. good representative. That is, the value of the A.M. of the data consisting of very large and 5. No special arrangement of the data is necessary very small items may lead to. while calculating it, unlike median and mode where we have to arrange and group the data in a 4. Unlike median and mode, it cannot be observed certain manner. by a mere inspection of the data but has to be calculated. 6. It is of much significance in manipulations either of an algebraic or arithmetic nature. 5. It cannot be used in qualitative (or descriptive) studies, unlike the median. 7. It forms a good basis of comparison while comparing different group’s of numerical data. 8. Given the A.M. and the number of items, the total of the items can be calculated at once (by multiplication), and given the total of the items and their number, the A.M. can be found easily. 10.5 Weighted Arithmetic Mean When the different items in a data are assigned weights according to their significance (relatively) the average then calculated is called :a weighted arithmetic average. Symbolically, the weighted average of n items x1, x2, x3 …xn whose weights are w1, w2, w3 … wn respectively is: x1w1  x2w2  x3w3 } xn wn = w1  w2  w3 }wn For example, a person buys three different books or ` 12, ` 18 and ` 30 respectively then the average price of a book that he purchases is: 12 18  30 3 = ` 20 On the other hand, he purchases five copies of the book at ` 12 each, three copies at ` 18 each and two copies at ` 30 each 10 books in all the average price of a book would be: CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 209 12 * 518 *330 * 2 10 This is called the weighted average since it has been calculated by giving the weights 5, 3 and 2 to the respective prices of the books. 10.6 Median The value of the middle item of a series that is ordered either in the ascending or descending order of magnitude is called the median. In other words, to determine the median it is necessary to arrange the items first either in ascending order or in descending order, for example, the median of the following 11 numbers 12, 15, 16, 21, 14, 22, 31, 32, 30, 28, 19, is 21 since 21 is the middle number when the numbers are written in ascending order as under: 12, 14, 15, 16, 19, 21, 22, 28, 30, 31, 32. ª n 1º In symbols, i n stands or number of items in a data then the median is the value of the«¬ 2 »¼th item, when the numbers are written either in ascending or descending order 0 magnitude. I n is an odd number, then the median can be located at once. But i n is an even number, then the median is equal to the value of the A.M. of the two central items. For example, the median of the numbers 17, 18, 21, 26, 28, 31, 33, 36 is: (26 + 28)/2. The median can be obtained in the following way also: Here n = – 8, therefore ª n 1º Median = the value of «¬ 2 »¼th item Median = the value of (4.5)th item Median = the value of 4th item +1/2 (5th item – 4th item) Median = the value of 26 + 1/2(28 – 26) Median = the value of 26 + 1 Median = 27. CU IDOL SELF LEARNING MATERIAL (SLM)

210 Business Mathematics and Statistics Discrete Series The following is the procedure or finding the median in the case of a discrete series: (i) Items are arranged either in ascending or descending order of magnitude and then their respective frequencies are written against them. (ii) These frequencies are cumulated. ª n 1º The value of the «¬ 2 »¼th item, where n = total number of frequencies, is determined by mere inspection of the cumulative frequencies. Problem 4: In the median room the following data: Age 18 19 20 21 22 23 24 25 26 27 28 29 No. of Persons 4 5 7 9 11 16 22 18 15 12 10 6 The following Problem explains the above procedure: Cumulative Frequency Age Frequency 4 9 18 4 16 19 5 25 20 7 36 21 9 52 22 11 74 23 16 92 24 22 107 25 18 119 26 15 129 27 12 135 28 10 29 6 135 ª n 1º Median = the value of ¬« 2 »¼th item ª1352 º Median = the value of «¬ 2 »¼th item Median = the value of 68th item Median = 24. CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 211 Continuous Series The technique of interpolation is used to in the median in a continuous frequency distribution. Interpolation is the process or technique of finding out the appropriate value or a missing or unknown value. Depending on the given data, the technique of interpolation also enables us to in missing or unknown values in the data. Interpolation may not necessarily give the exact value, only an approximate or as far as possible accurate value. The method of interpolation or finding out the median is given by the following formula: Median = L + § N  · ¨© 2 CF¸¹ Where L = the lower limit of the class in which the median item is situated. i = length of the class interval. F = frequency of the median class. N = the total of the frequencies. CF = the cumulative frequency of the class immediately preceding the median class. Procedure: The given frequencies of a continuous series are to be cumulated and these cumulated frequencies are to be written against the respective classes. The class in which the value of the n/2 item aIls is to be determined by inspecting the cumulative frequencies. The class determined in such a manner is called the median class since the median exists in that class. The next step is to in the values of F and CF and then use the given formula. The Problem worked below makes it clearer. Problem 5: Calculate the median room of the following data: Class Frequency Cumulative frequency 0-10 6 6 10-20 11 17 20-30 14 31 30-40 20 51 40-50 15 66 50-60 9 75 60-70 5 80 80 CU IDOL SELF LEARNING MATERIAL (SLM)

212 Business Mathematics and Statistics Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of students 6 11 14 20 15 9 5 Here, N = ?, f = 80, N/2 = 40. By observing the cumulative frequencies it is clear that the median class is 30-40 ? L = 30, i = 10, =20, CF = 31 Median = L+ i§ N  · f ©¨ 2 CF¸¹ Median = 30 + ^ [40 – 31] Median = 30 ± Median = 30 + 4.5 Median = 34.5 10.7 Advantages and Disadvantages of the Median Advantages of the Median Disadvantages of the Median 1. Its meaning is clear and intelligible. 1. An irregular series is characterized by extreme 2. Most of the conditions of an ideal average are variations between the items; in the case of satisfied by it. such a series the median ails to be representative. 3. It is very easy to in it in the case of a discrete series. 2. Arrangement of the data either in the ascending order of magnitude or descending 4. The median is of immense use while estimating order of magnitude is the chief requisite. This qualities such as honesty, intelligence, virtue, process is tedious when the data is vast, etc. morality etc. and proves be a good representative. 3. It absolutely ails to be of any use in cases where it is necessary to give large weights to 5. It’s possible to in the median by knowing only extreme items. the values of the central items and the number of items. That is, the values of the extreme items 4. Sometimes the median may exist between two are not necessary or finding the median. values, thus involving the work of i to estimation. 6. It can also be used algebraically to some extent. 5. Unlike the arithmetic mean, it is not possible to in the total value of all items if we know the value of the median and the number of items. 6. Its field of application through algebraic processes is restricted. CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 213 10.8 Mode The value of the item in a variable that is repeated the greatest number of times is called the mode. In other words, it is the value of the item that occurs most frequently in the data. It is said to be the most prominent item as well as a typical measurement, for instance, when we speak of modal income or modal marks we mean average income or average marks respectively. It is easy to determine the mode in the case of a discrete series that is regular — that is, where frequencies are not distributed irregularly. But i the frequency table has a number of irregularities, the location of the mode becomes mid cult. To overcome this difficultly it is necessary to adjust the frequencies through the grouping process. It’s then possible to locate the mode at once. In the case of a continuous series that is regular the modal class can be determined at once and the interpolation formula can be used. But I the continuous series is irregular it is necessary to adjust the frequencies by grouping, which involves widening the groups into which the frequencies all until the modal group is traced. After tracing the modal class the following interpolation formula is used: Mode = L + f1 f0 * i 2f1 f0 f2 Where L = lower limit of the modal class. F1 = frequency of the modal class. F0 = frequency of the class immediately lower to the modal class. F2 = frequency of the class immediately higher to the modal class. I = length of the class interval. Problems: (Discrete series) in the mode or the following data: X 21 6 22 9 23 12 24 16 25 22 26 22 CU IDOL SELF LEARNING MATERIAL (SLM)

214 Business Mathematics and Statistics 27 24 28 18 29 17 30 15 31 12 32 7 Solution: x123456 21 6 27 15 22 9 21 23 12 28 24 16 37 38 50 25 22 44 60 26 22 68 64 46 27 24 42 59 28 18 50 35 29 17 44 32 30 15 27 31 12 34 19 32 7 CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 215 Column Table of Analysis 28 1 1 x with maximum frequency 2 3 27 4 25 26 5 6 26 27 No. of times 24 25 26 25 26 27 26 27 1354 The modal value of x is 26 since it is repeated the largest number of times according to the table of analysis. By merely observing the given data one may infer that 27 is the mode since it is repeated 24 times. But the analysis proves that this is not correct. Hence 26 is the mode. Problem 7: (Continuous Series) Calculate the mode room the following data: Marks No. of Students 0-10 6 10-20 11 20-30 14 30-40 20 40-50 15 50-60 9 60-70 5 80 Solution: Since the given frequency distribution is regular, we in that the class 30-40 that has the maximum frequency is the modal class. Therefore, i = 10, F0 = 14, F2 = 15, F1 = 20, L = 30. Substituting these values in the formula or the mode, we get: Mode = § F1  F0 · L ¨ ¸* i © 2F1 F0  F2 ¹ = 30 + § 2014 · * 10 ©¨ 401415 ¹¸ CU IDOL SELF LEARNING MATERIAL (SLM)

216 Business Mathematics and Statistics §6· = 30 + ¨©11¹¸ * 10 = 30 + 5.45 = 35.45 Problem 8: The following data relates to the age distribution of 50 persons: Age 20-30 30-40 40-50 50-60 60-70 70-80 No. of Persons 3 7 14 16 8 2 Draw a histogram or the data and in the modal value. Check the value by direct calculation. Solution: The following is the procedure to in the mode graphically: 1. Draw the histogram or the data by representing age along the x-axis and the number of persons along the y-axis. 2. Draw two straight lines diagonally in the inside of the bar relating to the modal class. Join each upper corner of the modal bar to the upper corner of the adjacent bar. 3. From the point of intersection of the two diagonal lines, draw a perpendicular as shown in Fig. (1) to the x axis. The root of the perpendicular shows the modal value, in this case 52. 5 5 5 5 5 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Age Mode 52 Direct calculation to verify the value of the mode is like this: Mode = L + § F1 F0 F2 · ¨ 2F1 F0  ¹¸*1 © CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 217 § 1614 · = 50 + ©¨32148¸¹ * 10 §2· = 560 + ¨©10 ¸¹ * 10 = 50 + 2 ? Mode =52; Hence verified. Problem 9: Calculate the mode room the following data: Class Frequency 0-5 10 5-10 13 10-15 16 15-20 17 20-25 18 25-30 16 30-35 11 35-40 14 40-45 9 124 Since the given frequency is not regular, we have to group the frequencies and prepare the analysis table or ending the modal class. Class Frequency 123456 0-5 10 23 5-10 13 29 39 10-15 16 33 15-20 17 35 46 20-25 18 34 51 25-30 16 51 30-35 11 25 27 35-40 14 34 41 40-45 9 CU IDOL SELF LEARNING MATERIAL (SLM)

218 Business Mathematics and Statistics Table of Analysis Column Cases with Maximum Frequencies 1 15-20 20-25 25-30 2 15-20 20-25 20-25 3 15-20 15-20 25-30 4 20-25 5 2 6 5-10 10-15 1 10-15 Total No. of times 245 It follows from the above tables of analysis that the modal group is 20-25. Therefore, L = 20, i = 18, F0 = 17, F2 = 16, F1 = 5. 10.9 Advantages and Disadvantages of Mode Advantages of Mode Disadvantages of Mode 1. It is quickly intelligible and its calculation is 1. It does not satisfy some of the conditions laid not difficult. down or an ideal average. 2. It is the most predominant item in a discrete 2. It is quite possible that in certain types of data series. it may not be properly defined and hence it may be indeterminate and indefinite. 3. It can be located by a mere inspection of the frequencies in the case of a discrete series. 3. As in the case of the A.M., the value of the mode is not the value obtained by considering 4. It is a continuous series is a regular one — that the value of every item of the data. is, with the maximum frequency in the centre — the mode can be calculated easily without 4. In the case of an irregular series it is not very knowing the frequencies at the two extremities easy to determine the mode. of the series. 5. It is quite likely that in certain instances there 5. It is a very useful average in studying business may be two or more values or the mode. relating to sales, profits etc. 6. It is not amenable or greater Mathematical manipulation. 7. It may fail to be a representative average in the case of certain types of data. 10.10 Geometric Mean X1, X2, … Xn are n quantities, the geometric mean (G.M.) of these quantities is the nth root of the product of these quantities. G.M. = n X1, X2 , X3 }Xn CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 219 For example = 3 2 * 4 *8 = 4 Calculating the G.M. when there are only three or four quantities is easy, but it becomes laborious when there are many quantities. The difficulty can be overcome by using logarithms. Here’s an example Problem 10: Find the geometric mean of the following six quantities: 28, 31, 45 56 63 72. Let G = 9 28,31, 45,56,63, 72 = (28, 31, 45, 56, 63, 72)1/6 1 Log G = 4 (28, 31, 45, 56, 63, 72) 1 = 4 [log28 + log31, log45 + log63 + log72] 1 = 4 [1.4472 + 1.4914 + 1.6532 + 1.7482 + 1.7993 + 1.8573] 1 = 4 (9.9966) Log G = 1.6661 G = Antilog (1.6661) i.e., G = 46.35. 10.11 Advantages and Disadvantages of Geometric Mean Advantages of G.M. Disadvantages of G.M. 1. It has a clear cut and precise definition. 1. Though its definition is clear cut in the 2. Its value is determinate, provided the value of mathematical sense, it is not quickly intelligible to the common man. each of the items is greater than 0. 3. It is just like A.M. it is also based on all the 2. If anyone of the items is 0, it is impossible to calculate it. Further, if anyone of the items is items of the data. negative it becomes imaginary. 4. Its chief merit lies in the act that large items are 3. It is not useful in those cases where less given less weight and small items are given more weightage is to be given to small items and weight. more weightage is to be given to big items. 5. It is of immense use in dealing with percentages, 4. Sometimes it ails to be representative in the rates and ratios and is especially useful in index sense that it may considerably differ (in no. construction and in certain economic magnitude) room most of the items of the data. Problems. 6. It is amendable to algebraic manipulations. CU IDOL SELF LEARNING MATERIAL (SLM)

220 Business Mathematics and Statistics 10.12 Quartiles, Deciles and Percentiles The median being the central value in a given data, it divides the data into two equal parts. While one part consists of all the values less than the median, the other has all the values that are greater. But sometimes a given data may be divided into our, ten or a hundred parts. The values of those items that divide a given data into our, ten or a hundred parts are called quartiles, deciles and percentiles respectively. Hence, a given data has three quartiles denoted by Q1, Q2, Q3, nine deciles (D1, D2, D3, … Dn) and 99 percentiles (P1, P2, P3, …, P99). Problem 12: Calculate the quartiles, seventh decile and the 56th percentile of the following data: 14, 21, 28, 33, 18, 37, 23, 49, 45, 43, 57, 64, 53, 68, 39. These items when written in ascending order of magnitude are: 14, 18, 21, 23, 28, 33, 37, 39, 43, 45, 49, 53, 57, 64, 68. § n 1· Q1 = Value of the ©¨ 4 ¹¸th item §151· Q1 = Value of the ¨© 4 ¹¸th Item Q1 = Value of the 4th item Q1 = 23. § n 1· Q2 = Value of the 2 ©¨ 4 ¹¸ th item §151· Q2 = Value of the 3¨© 4 ¸¹th item Q2 = Value of the 8th item Q2 = 39. § n 1· Q3 = Value of the 3¨© 4 ¸¹thitem §151· Q3 = Value of the 3©¨ 4 ¹¸th item CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 221 Q3 = 53. § n 1· D7 = Value of the 7 ©¨ 10 ¹¸th item D7 = Value of the 7(11.2)th item D7 = Value of the 11th item + 0.2 + (12th item – 11th item) 2 D7 = 49 + 10 (53–49) D7 = 49 + 0.8 D = 49.8 § n1· P56 = Value of the 56¨© 10 ¹¸th item P56 = Value of the (896)th item P56 = Value of the 8th item + 0.96 + (9th item - 8th item) 6 P56= 39 + 100 (43 – 39) P56= 39 + 0.96 × 4 P56 = 42.84. Problem 13: Calculate the quartiles, seventh decile and the 56th percentile or the following data: Size Frequency 21 7 14 3 19 9 24 6 17 15 26 4 15 5 CU IDOL SELF LEARNING MATERIAL (SLM)

222 Business Mathematics and Statistics Size Frequency cf 14 3 3 15 5 8 17 15 23 19 9 32 21 7 39 24 6 45 26 4 49 49 § n1· Q1 = Value of the ¨© 4 ¸¹th item § 491· Q1 = Value of the ¨© 4 ¸¹th item Q1 = Value of the 4th item Q1 = Value of the 4th item Q1 = 17. § n1· Q2 = Value of the 2 ©¨ 4 ¸¹th item Q2 = Value of the 25th item Q2 = 19 § n1· Q3 = Value of the 3¨© 4 ¸¹th item § 491· Q3 = Value of the 3©¨ 4 ¸¹th item Q3 = Value of the (37.5)th item Q3 = 21. § n1· D7 = Value of the 7¨© 10 ¸¹th item D7 = Value of the 35th item D7 = 21 Q3 = Value of the (37.5) th item CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 223 § n1· P56= Value of the 56©¨ 100 ¹¸th item P56 = Value of the 28th item P56 = 19. Problem 14: Calculate the quartiles, seventh decile and the 56th percentile or the following data: Class Frequency 0-10 12 10-20 16 20-30 21 30-40 29 40-50 40 50-60 32 60-70 26 70-80 17 80-90 15 208 Class f cf 0-10 12 12 10-20 16 28 20-30 21 49 30-40 29 78. 40-50 40 118 50-60 32 150 60-70 26 176 70-80 17 193 80-90 15 208 208 Q1 = L+ fi¨©§ N  · 4 CF¹¸ 10 30 Q1 = 30 + 29 (52 – CF) = 30 + 29 CU IDOL SELF LEARNING MATERIAL (SLM)

224 Business Mathematics and Statistics Q1 = 30 + 1.03 Q1 = 31.03 Q3 = L+ i§ 3N  · 6010 (156150) f ©¨ 4 CF¸¹ 26 Q3 = L + (3J – CF) Q3 = 60 + (156 – 150) 60 Q3 = 60 + 26 Q3 = 60 + 2.3 Q3 = 62.3 Q2 is the same as median D7 = L + iª §n ·º f ¬« t©¨10 ¹¸ CF »¼ 10 D7 = 50 + 32 (143.5 – 118) 255 D7 = 50 + 32 D7 = 57.97 P56 = L+ i ª §n ·º f «¬56©¨10 ¸¹CF¼» 10 P56 = 40 + 40 (116.48 – 78) 38.48 P56 = 40 + 4 P56 = 40 + 9.62, P546 = 49.62 CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 225 10.13 Additional Problems (Solved) Problem 15: The total of the ages of three boys a year ago was 42. Find their present average age. Let the total of the ages of the three boys be x. Then, x – 3 = 42 x = 42 + 3 = 45 = 45 45 Therefore their present average age is 3 = 15 years. Problem 16: The average weight of 40 persons is 40 110 lb and that of another 10 persons is 115 lb. What is the average weight of all the 50 persons? The average weight = 110 * 40115*10 4010 4400 1150 = 50 5500 = 50 = 111 lb Problem 17: The average marks secured by a group of 50 students were 44. Later on, it was discovered that a score of 36 marks was misread as 56. Find the correct average marks secured by the students. Correct Average Marks = 44 * 505636 50 2200  20 = 50 2180 = 50 = 43.6 marks CU IDOL SELF LEARNING MATERIAL (SLM)

226 Business Mathematics and Statistics Problem 18: The arithmetic mean of 45 numbers was calculated as 52. On verification it was found that two numbers were misread as 42 and 36 instead of 24 and 63 respectively. Find the correct arithmetic mean. Correct AM = 52  45 42  36  24  63 45 52  45  9 = 45 9 = 52 45 = 52 + 0.20 = 52.20 Problem 19: The average age of a class of boys and girls is 15.2 years. The average age of the boys is 16 and that of the girls is 14. Find the percentage of boys and girls in the class. Let the number of boys be n1 Let the number of girls be n2 16n114n2 15.2 = n1n2 z 152 (n1 + n2) = 160nl + 140n2 z 152nl + 152n2 = 160nl + 140n2 z 152n2 –140n2 = 160nl + 140n2 z 12n2 = 8n1 n2 8 n2 12 n2 8 6 n2 12 4 The percentage of boys in the class is 60 and that of girls is 40. CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 227 Problem 20: Determine the value of the mode by using the formula, Mean – Mode = 3. (Mean-Median) for the distribution of marks given below: No. of Marks No. of Students Less than 10 5 Less than 20 15 Less than 30 98 Less than 40 242 Less than 50 367 Less than 60 405 Less than 70 425 Less than 80 438 Less than 90. 439 Solution: Class M·V. f D d fd cf 0-10 5 5 –40 –4 –20 5 10-20 15 10 12 20-30 25 .83 –30 –3 –30 98 30-40 35 144 242 40-50 [45] 125 –20 –2 –166 367 50-60 55 38 405 60-70 65 20 –10 –1 –144 425 70-80 75 13 438 80-90 85 P 9. 0 439 1 +10 +1 38 +20 +2 40 +30 +3 39 +40 +4 4 439 –239 ¦ fd Median = *i n 239 Median = 45 – 439 *10 Median = 45 – 5.44 = 39.56 CU IDOL SELF LEARNING MATERIAL (SLM)

228 Business Mathematics and Statistics n 439 2 2 = 219.5 Median class is 30-40 i Median = L+ § N  · f¨© 2 CF¸¹ 10 Median = 30 + 144 (219. 5 – 98) 1215 Median = 30 + 144 Median = 30 + 8.44 Median = 38.44 Using the formula: Mode = Mean – 3(Mean – Median), = 39.56 + 3(39.56 – 38.44) = 39.56 – 3(1.12) = 39.56 – 3.36 = 36.20 Problem 21: The arithmetic mean or the data. Given below is 33 in the missing frequency. Class Frequency 0-10 2 10-20 4 20-30 3 30-40 ? 40-50 3 50-60 5 Solution: Let the missing frequencies be k Then the total of frequencies = 17 + k CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 229 Class M.V. fD d Id 0-10 5 2 –20 –2 .4 10-20 15 4 –10 –1 –4 20-30 25 30 30-40 35 k + 10 0 0 40-50 45 3 +20 +1 +k 50-60 55 5 +30 +2 +6 +3 +15 Total 13 + k Here, a = 25, If = 17 + k, 6fd = 13 + k ¦ fd Mean = a + n ,, where i = 10 (13  k ) 33 = 25 + (17k) * 10 (13010k) 33 – 25 = (17 k) 8(17 + k) = 130 + 10k 136 + 8k = 130 + 10k 136 – 130 = 10k – 8k 6 = 2k k= 3 Therefore the missing frequency is 3. 10.14 Summary In one sense, the term dispersion implies that within a given data items may differ in their numerical value; the existence of such a difference indicates dispersion or lack of uniformity in data. Dispersion is said to be considerable or slight depending on whether these differences are large or small. Range is one of the simplest measures of dispersion. It’s the difference between the largest and the smallest items of a given data. Though easier to calculate, it is not an accurate measure of dispersion, only a rough one. CU IDOL SELF LEARNING MATERIAL (SLM)

230 Business Mathematics and Statistics To calculate mean deviation of dispersion it is necessary to find the deviations of the items of a given data from one of the averages in that data, such as mean, median or mode. Disregarding the algebraic signs, the sum total of these deviations divided by the number of items gives us the mean deviation. The measure quartile deviation also called the semi-inter quartile range is based on the third and the first quartiles. It is easy to calculate, because the deviations of individual items of the data need not be determined as in the case of mean deviation. Standard deviation measure, measures dispersion absolutely. The method of finding it is similar to that of the mean deviation method, with one difference: here individual deviations from the arithmetic mean are squared. Here’s how you calculate standard deviation (i) Calculate the deviation of each item from the arithmetic mean and square it. (ii) Divide the sum of these squares by the number of items in the given data. (iii) The square root of this quantity gives you the standard division. 10.15 Key Words/Abbreviations AM = Arithmatic Mean GM = Geometric Mean HM = Harmonic Mean 10.16 Learning Activity Problem 1: Find the mean, median and mode for the following data. Marks Number of Students 10-20 12 10-30 30 10-40 53 10-50 83 10-60 95 10-70 105 10-80 110 CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 231 Solution: fd The given data is arranged in the following manner: –36 –36 Marks Number of Students –23 10-20 12 P 20-30 18 + 12 30-40 23 +20 40-50 30 + 15 50-60 12 –48 60-70 10 70-80 5 110 Class MV. f cf 1) d- 10-20 15 12 12 –30 –3 20-30 25 18 30 –20 –2 30-40 35 23 53 –10 –1 40-50 [45] 30 83 0 Pi 50-60 55 12 95 + 10 +1 60-70 65 10 105 +20 +2 70-80 75 5 110 +30 +3 110 ¦ fd Mean = a + n * 1 (48) Mean = 45 + 110 * 10 48 Mean = 45 – 11 Mean = 42 – 4.36 Mean = 40.64 Median = L+ 1§ N  · f ©¨ 2 CF¸¹ CU IDOL SELF LEARNING MATERIAL (SLM)

232 Business Mathematics and Statistics n 110 2 = 2 = 55 The median class is 40-50 L = 40, i = 10, F = 30, cf = 53 10 Median = 40 + 30 (55 - 53) 2 Median = 40 + 3 Median = 40 + 0.67 Median = 40.67 Mode = L+ F1  F0 *i 2F1 F0 F2 The class containing mode is 40-50 L= 40, i = 10, Fl = 30, F0 = 23, F2 = 12 30  23 Mode = 40 + 602312 *10 7 Mode = 40 + 25 * 10 Mode = 40 + 14 5 Mode = 40 + 2.80 Mode = 42.80 CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 233 Problem 2: Find the mean, median and mode. d –12 Marks in Economics Number of Students –27 –22 0-100 120 –15 10-100 117 20-100 108 0 30-100 97 +26 40-100 82 +34 50-100 64 +42 60-100 38 +20 70-100 21 + 10 80-100 90-100 7 56 2 Solution: Arranging the given data we get: Class M Frequency c/ D D 0-10 53 3 –40 –4 10-20 15 9 12 –30 –3 20-30 25 ii 23 –20 –2 30-40 35 15 38 –10 –1 40-50 [45] 18 56 50-60 55 26 82 0 0 60-70 65 17 99 + 10 +1 70-80 75 14 113 +20 +2 80-90 85 5 118 +30 +3 90-100 95 2 120 +40 +4 +50 +5 120 ¦ fd Mean = a + n 56 Mean = 45 + 120 * 10 Mean = 4 + 4.67 = 49.67 CU IDOL SELF LEARNING MATERIAL (SLM)

234 Business Mathematics and Statistics §n· Median = Value of ¨© 2 ¹¸th item §120 · Median = Value of ©¨ 2 ¹¸th item Median = Value of the 60th item This value lies in the class 50-60 Using the interpolation method, Median = L+ 1§ N · f ©¨ 2  CF¸¹ N Substituting, L = 50, i = 10, F = 26, CF = 56 and 2 = 60 We get 10 Median = 50 + 26 (60 – 56) 10 Median = 50 + 26 Median. = 51.54 F1 F0 Mode = L + 2F1  F0  F2 * i The class containing the mode is 50-60 L = 50, i = 10, F1 = 26, F0 = 18, F2 = 17 2618 Mode = 5 0 + 521817 * 10 80 Mode = 50 + 70 Mode = 50 + 4.71 Mode = 54.71 CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 235 Problem 3: In the median and mode or the following data: Age No. of Persons 18-20 3 21-23 7 24-26 12 27-29 14 30-32 1 33-35 10 36-38 7 39-41 5 42-44 2 78 The class intervals should be written as shown below: Age f cf 17.5-20.5 3 3 20.5-23.5 7 10 23.5-26.5 12 22 26.5-29.5 14 36 29.5-32.5 18 54 32.5-35.5 10 64 35.5-38.5 7 71 38.5-41.5 5 76 41.5-44.5 2 78 78 §n· Median = Value of ©¨ 2 ¹¸th item § 78 · Median = Value of ©¨ 2 ¸¹th item Median = Value of the 39th item This value lies in the class 29.5-32.5 CU IDOL SELF LEARNING MATERIAL (SLM)

236 Business Mathematics and Statistics §n· Therefore, L = 29.5, i = 3, F = 18, CF = 36, ¨© 2 ¹¸ = 39 Median = L+ i§ N · f ©¨ 2  CF¹¸ 3 Median = 29.5 + 18 (39-56) Median = 50 + 0.5 Median = 30 In the given data the highest frequency is 18, therefore the modal class is 29.5-32.5. Taking L = 29.5, Fl = 18, F0 = 14, F2 = 10 and i = 3, we get, Fl  F0 Mode = L + 2F1  F0  F2 * 1 1814 Mode = 29.5 + 361410 * 3 4 Mode = 50 + 12 * 3 Mode = 30.5 Problem 4: An incomplete distribution is given as follows: Variable Frequency 10-20 12 20-30 30 30-40 40-50 ? 50-60 65 60-70 70-80 ? 25 Total 18 229 CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 237 The median value is given as 46. … (i) (a) Using the median formula, ill up the missing frequencies. (b) Calculate the arithmetic mean of the completed table. Let the missing frequencies for the class 30-40 and 50-60 be F0 and F2. Total of the given frequencies = 229 Therefore, the sum of the missing frequencies = F0 + F2 = 229 – 150 = 79 Therefore F0 – F2 = 79 The value of median is given as 46 Median = L+ i§ N · j©¨ 2  CF¹¸ The median lies in class 40-50 Therefore, L = 40, i = 10, F = 65 § n · 229 ¨© 2 ¹¸ 2 = 114.5 = 115 – 42 – F0 Cf = 12 + 30 + F0 = 42 + F0 10 46 = 40 + 65 (115 – 42 – F0) 10 46 = 40 + 65 (73 – F0) 46 = 40 + 73010F0 6565 65 * 6 = 730 – 10F0 390 = 730 – 10F0 39 = 73 – F0 F0 = 73 – 39 = 34 From equation (i) F0 + F2 = 79 Therefore F2 = 79 – F0 = 79 – 34 = 45 Therefore the missing frequencies are 34 and 45 respectively. CU IDOL SELF LEARNING MATERIAL (SLM)

238 Business Mathematics and Statistics Problem 5: Compute the mean and the median of the following distribution of IQ of 309 six- year-old children. 1.Q. Frequency 160-169 2 150-159 3 140-149 7 130-139 19 120-129 37 110-119 79 100-119 69 65 90-99 17 80-89 11 70-79 309 Total The given distribution can be arranged as: Class MI D Dd c/ 69.5 - 79.5 74.5 11 –40 –4 –44 11 79.5 - 89.5 84.5 17 –30 –3 – 5–1– 28 89.5 - 99.5 94.5 65 –20 –2 –130 93 99.5 - 109.5 104.5 69 –10 –1 –69 162 109.5 - 119.5 114.5 79 241 119.5 - 129.5 124.51 37 0 00 278 129.5 - 139.5 134.5 19 ±10 +1 +37 297 139.5 - 149.5 144.5 +20 +2 +38 304 149.5 - 159.5 154.5 7 +30 +3 +21 307 159.5 - 169.5 164.5 3 +40 +4 +12 309 2 +50 +5 + 10 309 –176 ¦ fd Mean = a + n * i 176 Mean = 114.5 – 309 * 10 Mean = 114.5 – 5.7 = 108.8 CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 239 §n· Median = Value of ©¨ 2 ¹¸th item § 309 · Median = Value of ©¨ 2 ¹¸th item Median = Value of the (154.5)th item This value lies in the class 99.5-109.5 §n· Therefore, L = 99.5, i = 10, F = 69, CF = 93, ©¨ 2 ¹¸ = 154.5 Median = L+ fi©§¨ N · 2  CF¸¹ 10 Median = 99.5 + 69 (154.5 – 93) 615 Median = 99.5 + 69 Median = 99.5 + 8.91 = 108.41. 10.17 Unit End Questions (MCQ and Descriptive) A. Descriptive Type: Short Answer Type Questions 1. Define the term ‘average’ and discuss the utility of averages in statistics. 2. Define the term ‘geometric mean’ and state its advantages and disadvantages. 3. How is the harmonic mean computed? What are its merits and demerits? 4. How is the weighted average computed? Discuss the uses of such an average. 5. What are the objects of computing a statistical average? Distinguish between averages of position and mathematical averages. 6. What are the desirable characteristics of an average? State the .averages that have most of these characteristics and illustrate their uses. 7. Discuss the merits and demerits of any three of the following averages: (i) Arithmetic mean (ii) Geometric mean (iii) Harmonic mean (iv) Median (v) Mode. 8. What is a statistical average? Define the diferent kinds of statistical averages, and explain the methods of computing them. CU IDOL SELF LEARNING MATERIAL (SLM)

240 Business Mathematics and Statistics 9. What is meant by the term ‘central tendency’? Define the measures of central tendency and discuss which of the measures is superior in representing a given quantitative data. Elucidate your answer. 10. The arithmetic mean of a series of 1.2 items is 43. 12 is added to each item of the series, does the mean change? I so, by how much? [Ans. Yes, the mean increases by 2, t045] 11. The average age of a class of 50 boys was 12 years. The average age of 20 of them was 11 and that of another 10 was 10. Find the average age of the remaining boys. [Ans: 14 years] 12. The average weight of a group of eight boys is 115Ib. The weights of seven of them are 116, 114, 120, 108, 112, 110 and 118 lb. What is the weight of the eighth boy? [Ans: 1221b] B. Multiple Choice/Objective Type Questions 1. The AM of 15 item is 50. If 5 is subtracted from each item the mean changes by _________________. (a) 10 (b) 5 (c) 15 (d) None of these 2 The AM of 10 item is 12. If two item 21 and 27 are added then the New Mean is _______________. (a) 14 (b) 13 (c) 15 (d) None of these 3. The value of the item in a variable that is repeated the greatest number of times is called the _______________. (a) Mode (b) Median (c) Mean (d) None of these 4. _______________ is indispensable in problems involving time, rates and ratios. (a) Arithmetic Mean (b) Goemetric Mean (c) Harmonic mean (d) None of these Answers (1) (b); (2) (a); (3) (a); (4) (c) 10.18 References References of this unit have been given at the end of the book. ˆˆˆ CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 241 UNIT 11 DISPERSION Structure 11.0 Learning Objectives 11.1 Introduction 11.2 Meaning and Types of Dispersion 11.3 Summary 11.4 Key Words/Abbreviations 11.5 Learning Activity 11.6 Unit End Questions (MCQ and Descriptive) 11.7 References 11.0 Learning Objectives After studying this unit, you will be able to: z Explain the meaning, definition, main objects, and characteristics of an ideal average and the types of statistical average. z Grasp the method of calculation of A.M., median, ‘and mode in respect of the data, both discrete and continuous. z Ellaborate the different types of graphs that include: Histogram, Frequency Polygon and curve and also Ogive curve. 11.1 Introduction Statistical averages such as mean, median and mode indicate some of the main characteristics of a given numerical data. However, these averages may ail to give a correct impression when CU IDOL SELF LEARNING MATERIAL (SLM)

242 Business Mathematics and Statistics comparisons are to be made between different types of data, for instance, consider the following Weekly sales of two salesmen: Salesman A: 130 145 145 180 Salesman B: 145 150 150 155 It is clear that the average sales of both salesmen is 150, even though there is much spread Or scatter in the sales of A than that of B. Salesman B sells more or less an uniform amount during the four weeks. In order to study the nature of spread, scatter or lack of uniformity in a given numerical data, it is necessary to use certain measures, or dispersion. Such measures are also called averages of the second order. 11.2 Meaning and Types of Dispersion In one sense, the term dispersion implies that within a given data items may difer in their numerical value; the existence of such a diference indicates dispersion or lack of uniformity in the data. Dispersion is said to be considerable or slight depending on whether these diferences are large or small. In another sense this term is used absolutely as well as relatively: Here dispersion stands or the deviation of items of a given data from an average such as a mean, median or mode. In short, a measure of such deviation is a measure of dispersion. The following are the measures of dispersion: range, mean deviation, quartile deviation and standard deviation. 11.2.1 Range This is one of the simplest measures of dispersion. It’s the diference between the largest and the smallest items of a given data. Though easier to calculate, it is not an accurate measure of dispersion, only a rough one. 11.2.2 Mean Deviation To calculate this measure of dispersion it is necessary to get the deviations of the items of a given data to form one of the averages is that data, such as mean, median or mode. Disregarding the CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 243 algebraic signs, the sum total of these deviations divided by the number of items gives us the mean deviation. The main advantage of this measure over range is that it takes to consideration every item of data and at the same time is not much influenced by very large items. Though simple and quickly intelligible, it is not as good as standard deviation. Problem 1: Calcuiate the mean deviation room the mean and median from the following data: 12, 20, 39, 46, 54, 61, 78,90. Mean = 12 + 20 + 39 + 46 + 54 + 61 + 78 + 90 8 400 Mean = 8 Mean = 50 Size Deviation (d) 12 38 20 30 39 11 46 54 4 61 4 78 11 90 28 40 166 ¦d 166 Mean Deviation room Mean = 8 8 = 20.75 The value of the median is also 50 Therefore mean deviation of median is also 20.75. Problem 2: Calculate the mean deviation room the mean, median and mode or the following data: CU IDOL SELF LEARNING MATERIAL (SLM)

244 Business Mathematics and Statistics Age No. of Boys 10 2 11 4 12 7 13 4 14 3 20 X f xf dA f,dA d fdm 10 2 20 2.1 4.2 24 14 11 4 44 1.1 4.4 00 14 12 7 84 .1 -, 7 26 13 4 52 ‘.9 3.6 18 14 3 42 1.9 5.7 20 242 18.6 242 Mean = 20 = 12.1 ¦ fdA 18.6 Mean deviation from mean = n 20 = 0.93 Median = 12 ¦fdm 18 Mean deviation from median = n 20 = 0.9 Since mode or the data is also 12, mean deviation room mode is also 0.9. Problem 3: Calculate (a) median coeficient of dispersion, and (b) mean coeficient of dispersion room the following data: Size of Items 4 6 8 10 12 14 16 Frequency 2453214 CU IDOL SELF LEARNING MATERIAL (SLM)