SLM_(1)

frequencies in the case of a discrete series. 3. As in the case of the A.M., the value of the 4. It is a continuous series is a regular one — that mode is not the value obtained by considering 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 a very useful average in studying business 5. It is quite likely that in certain instances there relating to sales, profits etc. may be two or more values or the mode. 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 X , X ,X X 12 3 n 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, rates and ratios and is especially useful in index 4. Sometimes it ails to be representative in the no. construction and in certain economic sense that it may considerably differ (in Problems. magnitude) room most of the items of the data. 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· Q = ¨ ¸ Value of the th item 1© 4¹ §151· Q = Value of the ¨ ¸th Item 1© 4¹ Q1 = Value of the 4th item

Q1 = 23. § n 1· Q = ¨ ¸ Value of the 2 th item 2 ©4 ¹ §151· Q = Value of the 3¨ ¸th item 2 ©4 ¹ Q2 = Value of the 8th item Q2 = 39. § n 1· Q = Value of the 3¨ ¸thitem 3 © 4¹ §151· Q = ¨ ¸ Value of the 3 th item 3 ©4 ¹ CU IDOL SELF LEARNING MATERIAL (SLM)

Measures of Central Tendency 221 Q3 = 53. § n 1· ¨¸ D7 = Value of the 7 © 10 ¹th 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 § n 1· ¨¸ P56 = Value of the 56 © 10 ¹th 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 § n 1· Q= ¨ ¸ Value of the th item 1 © 4¹ § 491· Q= ¨ ¸ Value of the th item 1 © 4¹ Q1 = Value of the 4th item Q1 = Value of the 4th item Q1 = 17. (7) n 1· Q = Value of the 2 ¨ ¸th item 2 © 4¹

Q2 = Value of the 25th item Q2 = 19 § n 1· Q = ¨ ¸ Value of the 3 th item 3 © 4¹ §491· Q = ¨ ¸ Value of the 3 th item 3 © 4¹ Q3 = Value of the (37.5)th item Q3 = 21. § n 1· D = Value of the 7¨ ¸th item 7 © 10 ¹ 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 § n 1· ¨¸ P56= Value of the 56 © 100 ¹th 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 i§N · CF¸ Q= L+ ¨ 1 f©4 ¹ 10 30 Q1 = 30 + 29 (52–CF)=30+29 CU IDOL SELF LEARNING MATERIAL (SLM)

224 Business Mathematics and Statistics Q1 = 30 + 1.03 · 10 Q1 = 31.03 i§ 3N Q = L+ ¨ CF¸ 60 (156150) 3 f © 4 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 iª§n · º D = L + « t¨ ¸CF» 7 f ¬© 10 ¹ ¼ 10 D7 = 50 + 32 (143.5 – 118) 255 D7 = 50 + 32 D7 = 57.97 §n· º iª P 56 = L + f « 56¨ ¸CF»

¬ ©10¹ ¼ 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 44001150 = 50 5500

(vi) 50 (vii) 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 220020 = 50 2180 (x) 50 (xi) 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 = 524542362463 45 52459 = 45 9 ª 52 45 ª 52 + 0.20 ª 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 = n1 n2

 152 (n1 + n2) = 160nl + 140n2   152nl + 152n2 = 160nl + 140n2   152n2 –140n2 = 160nl + 140n2   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. fD d fd cf 0-10 5 5 –40 –4 –20 5 10-20 15 10 –30 –3 –30 12 20-30 25 .83 –20 –2 –166 98 30-40 35 144 –10 –1 –144 242 40-50 9. 0 367 50-60 [45] 125 P +1 38 405 55 38 +10

60-70 65 20 +20 +2 40 425 70-80 39 438 80-90 75 13 +30 +3 4 439 85 1 +40 +4 439 –239 ¦fd Median = *i 239 Median = 45 – 439 *10 Median = 45 – 5.44 = 39.56 CU IDOL SELF LEARNING MATERIAL (SLM)

228 Business Mathematics and Statistics ª 439 2 2 = 219.5 Median class is 30-40 i · Median = L + § N f¨ CF¸ ©2 ¹ Median = 30 + 10 (vi) (219. 5 – 98) Median = 30 + 1215 144 Median = 30 + 8.44 Median = 38.44 Using the formula: Mode = Mean – 3(Mean – Median), (xiv) 39.56 + 3(39.56 – 38.44) (xv) 39.56 – 3(1.12)

(xvi) 39.56 – 3.36 (xvii) 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 0 0 30-40 35 k + 10 +1 +k 40-50 45 3 +20 +2 +6 50-60 55 5 +30 +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 + (17 k) * 10 (130 10k) 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 ? Calculate the deviation of each item from the arithmetic mean and square it. ? Divide the sum of these squares by the number of items in the given data. ? 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: The given data is arranged in the following manner: Marks Number of Students 10-20 12 20-30 18 30-40 23 40-50 30 50-60 12 60-70 10 70-80 5 110 Class MV. f cf 1) d- fd 10-20 15 12 12 –30 –3 –36 20-30 25 18 30 –20 –2 –36 30-40 35 23 53 –10 –1 –23 40-50 [45] Pi P 50-60 55 30 83 0 +1 + 12 60-70 65 +2 +20 70-80 75 12 95 + 10 +3 + 15 10 105 +20 5 110 +30

110 –48 ¦fd Mean = a + n * 1 (48) Mean = 45 + 110 * 10 48 Mean = 45 – 11 · Mean = 42 – 4.36 CF¸ Mean = 40.64 ¹ 1§ N Median = L + ¨ f©2 CU IDOL SELF LEARNING MATERIAL (SLM)

232 Business Mathematics and Statistics « 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 + F1F0 2F F F *i 1 02 The class containing mode is 40-50 L= 40, i = 10, Fl = 30, F0 = 23, F2 = 12 3023 Mode = 40 + 60 23 12 *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. Marks in Economics Number of Students 0-100 120 10-100 117 20-100 108 30-100 40-100 97 50-100 82 60-100 64 70-100 38 80-100 21 7 90-100 2 Solution: Arranging the given data we get: Class M Frequency c/ D Dd 0-10 53 3 –40 –4 –12 10-20 15 9 12 –30 –3 –27 20-30 25 ii 23 –20 –2 –22 30-40 35 15 38 –10 –1 –15 40-50 [45] 18 56 00 0

50-60 55 26 82 + 10 +1 +26 60-70 65 17 99 +20 +2 +34 70-80 75 14 113 +30 +3 +42 80-90 85 5 118 +40 +4 +20 90-100 95 2 120 +50 +5 + 10 120 56 ¦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 ¹th item §120· ¨¸ Median = Value of © 2 ¹th item Median = Value of the 60th item This value lies in the class 50-60 Using the interpolation method, 1§ N · Median = L + ¨ CF¸ f© 2 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 + 52 18 17 * 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 ¸¹th 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, ¨ ¸ = 39 ©2 ¹ i§N · Median = L + ¨ CF¸ f©2 ¹ 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 *1 Mode = L + 2F F F 2 10 1814 Mode = 29.5 + 36 14 10 * 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) (vi) Using the median formula, ill up the missing frequencies. (vii) 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 i§N · Median = L + ¨ CF¸ j© 2 ¹ 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) + 73010F0 46 = 40 + 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 Total 309 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 00 241 0

119.5 - 129.5 124.51 37 ±10 +1 +37 278 129.5 - 139.5 134.5 19 +20 +2 +38 297 139.5 - 149.5 144.5 7 +30 +3 +21 304 149.5 - 159.5 154.5 3 +40 +4 +12 307 159.5 - 169.5 164.5 2 +50 +5 + 10 309 309 –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 (vi) n · ¨¸ Median = Value of © 2 ¹th item (vii) 309· ¨¸ Median = Value of © 2 ¹th 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 i§N · Median = L + ¨ CF¸ f© 2 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 = Define the term ‘average’ and discuss the utility of averages in statistics. = Define the term ‘geometric mean’ and state its advantages and disadvantages. = How is the harmonic mean computed? What are its merits and demerits? = How is the weighted average computed? Discuss the uses of such an average. = What are the objects of computing a statistical average? Distinguish between averages of position and mathematical averages. = What are the desirable characteristics of an average? State the .averages that have most of these characteristics and illustrate their uses. = 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. = 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 = 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. = 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] = 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] z 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 9. 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 z 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 © 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) 241 Dispersion

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:

 Explain the meaning, definition, main objects, and characteristics of an ideal average and the types of statistical average.  Grasp the method of calculation of A.M., median, ‘and mode in respect of the data, both discrete and continuous.   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)