SLM_(1)

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. 12+20+39+46+54+61+78+90 Mean = 8 400 Mean = 8 Mean = 50 Size Deviation (d) 12 38 20 30 39 11 46 4 54 4 61 11

78 28 90 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 11 4 44 1.1 4.4 14 00 12 7 84 .1 -, 7 14 13 4 52 ‘.9 3.6 26 14 3 42 1.9 5.7 18 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 68 10 12 14 16 Frequency 2 45 3 214 CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 245 Solution: XF xf dA fdA Cf dm fdm 42 248 64 8 5.7 11.4 628 85 11 0 0 10 3 14 3.7 14.8 14 2 6 12 2 16 4 8 14 1 40 1.7 85 17 6 6 16 4 21 8 32 30 0.3 09 68 24 2.3 46 14 4.3 43 64 6.3 2.2 Total 21 4 f.7 ¦xf 204 Mean = n 21 = 9.7 ¦fdA 69.7 Mean deviation from mean = n 21 = 3.3 3.3 Mean coeficient of dispersion = 9.7 = 0.34 § n 1· Median = Value of ¨ ¸th item ©2 ¹ § 211· Median = ¨ ¸ th item Value of 2

©¹ Median = Value of the 11th item = 8 ¦fdm 68 Mean Deviation from Median = 21 n 3.238 Mean coefficient of dispersion = 8 = 0.40475 CU IDOL SELF LEARNING MATERIAL (SLM)

246 Business Mathematics and Statistics Problem 4: Calculate the mean deviation from mean for the data given below. Class f 0-5 3 5-10 7 10-15 12 15-20 18 20-25 13 25-30 10 30-35 7 70 dA Deviation from Mean 18.86 Solution: Class M.V. fD d fd dA / dA 0-5 2.5 3 –15 –3 –9 16.36 49.08 5-10 7.5 7 –10 –2 –14 11.36 79.52 10-15 12.5 12 –5 –1 –12 6.36 76.52 15-20 [17.5] 18 0 0 0 1.36 24.48 20-25 22.5 13 +5 +1 + 13 3.64 47.32 25-30 27.5 10 + +2 +20 8.64 86.40 30-35 32.5 7+ +3 +21 13.64 95.48

Total 70 + 19 458.60 ª fd Mean = a + n * i 19 Mean = 17.5 + * 5 70 Mean = 17.5 + ª fdA n 458.60 Mean = 70 MD for Mean = 6.55 CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 247 Problem 5: Calculate mean deviation from mean, median and mode from the following data: Class 3 7 0-10 15 10-20 12 20-30 8 30-40 5 40-50 50-60 50 Solution: Class M.V. f cf D d fc\\ 0-10 5 3 3 –20 –2 –6 10-20 15 7 10 –10 –1 –7 20-30 25 15 25 0 00 30-40 35 12 37 +10 +1 :t 12 40-50 45 8 45 +20 +2 + 16 50-60 55 5 50 +30 +3 + 15 ¦fd 50 +30

Mean = a n * i 30 Mean = 25 + 50 * 10 Mean = 25 + 6 Mean = 31 ª n· Median = Value of ¨© 2 ¸¹th item ª 50· Median = Value of ¨© 2 ¸¹th item Median = Value of the 25th item Median lies in the class 20-30 CU IDOL SELF LEARNING MATERIAL (SLM)

248 Business Mathematics and Statistics i§N · Median = L + ¨ CF¸ f© 2 ¹ 10 Median = 20 + 15 (25 - 10) Median = 20 + 10 Median = 30 F1 F0 Mode = L + 2F1 F0 F2 *l 157 Mode = 20 + 30 7 12 =10 8 Mode = 20 + 11 * 10 80 Mode = 20 + 11 Mode = 20 + 7.27 Mode = 27.27

Class M.V. / d a^ dm /dm dm a3 0-10 5 3 CD CM* +78 25 75 22.27 66.81 10-20 15 7 + 16 + 112 15 105 12.27 85.89 20-30 2% 15 +6 -’-90 5 75 2.27 34.05 30-40 35 12 +4 5 60 7.73 95.76 40-50 45 8 + 14 +48 15 120 17.73 141.84 50-60 55 5 +24 + 112 25 125 27.73 138.65 + 120 560 560 (vii) fd MD from Mean n = 11.2 MD from mode and median is also 11.2. CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 249 Problem: In the mean deviation room median and mode in the.following data: Age No. of Persons 16 30 17 32 18 35 19 46 20 51 21 60 22 43 23 32 24 19 Solution: By inspection, it follows that the median is 20 and mode is 21. Median Value § n 1· Median = ¨ ¸ th item © 2¹ § 3481· Median = ¨ ¸ th item

© 2¹ cf Median = Value of the (174-.5)th item = 20 30 62 (by inspection of the cumulative frequencies). 97 143 Age F 194 254 16 30 297 17 32 329 18 35 348 19 46 20 51 21 60 22 43 23 32 24 19 CU IDOL SELF LEARNING MATERIAL (SLM)

250 Business Mathematics and Statistics Age f Deviation fdm Deviation room 1 dm room 0 dm 21 dm 5 16 30 4 120 4 150 3 128 17 32 3 96 2 105 18 35 2 70 92 19 46 10 46 20 51 1 01 51 21 60 2 60 0 0 22 43 3 86 1 43 23 32 4 96 2 64 24 19 76 3 57 650 690 Mean Deviation from Median = (xviii) fdm n 650 Mean Deviation from Median = 348 ? Antilog (log 650 – log 348) ? Antilog (2.8129 – 2.5416) ? 1.867 « fdm Mean Deviation from Median =

n 650 Mean Deviation from Median = 348 Antilog (log 690 - log 348) Antilog (2.8388 - 2.5416) Antilog (0.2972) 1.867 11.2.3 Quartile Deviation This measure 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. Further, it provides a rough idea about the dispersion in given CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 251 data. Apparently, the greater the distance between the quartiles, the greater is the dispersion. If two diferent frequency distributions have the same values or quartiles, it follows that the quartile deviation or each distribution is also the same. This in turn implies that the dispersion of both distributions is identical. Hence, because of this disadvantage, quartile deviation is not an accurate measure of dispersion. Problem 7: Calculate the quartile deviation and its coeficient or the following data: 8, 3, 15, 4, 18, 6, 11. Solution: Arranging the data in ascending order: 3, 4, 6, 8, 11, 15, 18. n = 7. § n 1· ¨ ¸ First Quartile: Ql = Value of the th item © 4¹ §71· ¨¸ Q1 = Value of © 4 ¹th item Q1 = Value of the 2nd item Q1= 4 ? n 1· Third Quartile: Q3 = Value of 3¨© 4 ¸¹th item (viii)71·

Q3 = Value of ¨© 4 ¸¹th item Q3 = Value of 6th item Q3= 15 Quartile Deviation = Q3 Q1 2 154 11 = 22 =5.5 Coefficient of Quartile Deviation is Q 154 11 254 29 = 0.58 3 Q1 QQ 31 CU IDOL SELF LEARNING MATERIAL (SLM)

252 Business Mathematics and Statistics Problem 8: Calculate the quartile deviation and its coeficient or the following data: Marks No. of Students 10 4 11 6 12 7 13 8 14 6 15 4 35 Marks 4 cf 10 6 11 7 4 12 8 10 13 6 17 14 4 25 15 31 35 35 Solution:

§ n 1· First Quartile: Q1 = Value of the ¨ ¸th item Third Quartile: © 4¹ §351· Q1 = ¨ ¸ Value of th Item ©4 ¹ Ql = Value of the 9th item Q1= 11 § n 1· Q3 = ¨ ¸ Value of 3 th item ©4¹ § 351· Q3 = ¨ ¸ Value of 3 th Item ©4 ¹ Q3 = Value of 27th item CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 253 Q3= 14 Quartile Deviation = Q3 Q1 2 Q 3 Q1 1411 3 = 3 2 = 1.5 QQ 1411 4 1411 25 = 0.12 31 Problem 9: Calculate the quartile deviation of quartile deviation or the following data, and the coeficient. Class Frequency 0-10 3 10-20 7 20-30 15 30-40 12 40-50 8 50-60 5 50 Class f cf

0-10 3 3 10-20 7 10 20-30 15 25 30-40 12 37 40-50 8 45 50-60 5 50 Solution: §n· ¨¸ First Quartile: Ql = Value of the © 4 ¹th item §50· ¨¸ Q1 = Value of © 4 ¹th item Q1 = Value of the (12.5)th item CU IDOL SELF LEARNING MATERIAL (SLM)

254 Business Mathematics and Statistics Q1 Lies in the class 20-30 i§ N · Q1= L+ ¨ CF¸ f©4 ¹ Where L = Lower limit of the first quartile class (L = 20) i = Length of the class interval (i = 10) F = Frequency of the first quartile class (F = 15) CF = Cumulative Frequency of the class preceding the first quartile class (CF = 10) 10 Q1 = 20 + 15 (12.5 – 10) 10 Q1 = 20 + 15 (2.5) 2 Q1 = 20+1 13 Q1 = 20 + 1.67 Q1 = 21.67 ª§n · º Third Quartile: Q3 =Value of « 3 CF ¨ ¸ »th item ¬©4¹ ¼ Q3 = Value of (37.5)th item Q3 = Lies in the class 40-50 iª§N· º Q3 = L « 3¨ ¸CF»

f¬ © 4 ¹ ¼ Where L = Lower limit of the third quartile class (L = 40) i = Length of the class interval (i = 10) F = Frequency of the third quartile class (F = 8) CF = Cumulative Frequency the third quartile class (CF = 37) iª§N· º Q3 = L « 3¨ ¸CF» f¬ © 4 ¹ ¼ CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 255 10 Q3 = 40 37.537 Q3 = 40 + 0.625 40.63 approximately Quartile Deviation = Q3 Q1 2 40.6321.67 18.96 =2 2 = 9.48 Q3 Q1 Coefficient of Quartile Deviation = Q3 Q1 40.6321.67 (viii) 40.6321.67 18.96 (ix) 62.30 (x)Antilog (log 18.96 – log 62.30) (xi) Antilog (1.2779 – 1.7945) (xii) Antilog (1.4834) (xiii) 0.3044.

11.2.4 Standard Deviation This 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 deviation. Its main advantages are that it is rigidly defined and, secondly, individual deviations are squared in its calculations. However, this measure also has certain drawbacks. Since the individual deviations are squared, its value is affected a great deal by the value of each item in the data. Extreme items, CU IDOL SELF LEARNING MATERIAL (SLM)

256 Business Mathematics and Statistics or example, push up its value considerably. But that does not detract from its value as a good dispersion and standard deviation is commonly used. Variance It is the term used to denote the square of the standard deviation. Symbolically a - stands or variance. Coeficient of variation: lt is possible to compare diferent types of data by finding out the percentage of the standard deviation to the arithmetic mean or each type of data. In short the ratio when multiplied by 100 gives a certain percentage this is known as the coeficient of variation. Problem 10: Calculate the standard deviation from the following data: 2, 3, 7, 8, 10. Solution: Xd d2 2 –4 16 3 –3 9 7 +1 1 8 +2 4 10 +4 16 30 = x 30 x = 30AM = =6

S.D. = ¦d2 55 5 46 9.5 = 3.033 5 Problem 11: Calculate the standard deviation from the following data: Marks No. of Students 2 2 5 3 6 8 7 5 9 2 CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 257 Solution: fd.2 32 x f xf d d2 3 0 2 2 4 –4 16 5 18 5 3 15 –1 1 58 6 8 48 0 0 7 5 35 +1 1 9 2 18 +3 9 20 120 27 z xf 120 Mean = n 20 = 6 SD = ¦d2 58 5 2.9 = 1.703 20 Problem 12: Find the standard deviation from the following data: Class f 0-5 2 5-10 3 10-15 7 15-20 5

20-25 3 20 Solution: Class M.V. f D D fd fd2 0-5 2.5 2 –10 –2 –4 8 5-10 7.5 3 –5 –1 –3 3 10-15 12.5 15-20 17.5 70000 20-25 22.5 5 +5 +1 +5 5 3 + +2 +6 12 20 +4 28 S.D. = ¦d2 §fd ·2 ¨ ¸ *i n © n¹ CU IDOL SELF LEARNING MATERIAL (SLM)

258 Business Mathematics and Statistics S.D. = 28 §4·2 ¨ ¸ *5 20 ©20¹ S.D. = 1.400.04 *5 S.D. = (1.166) * 5 S.D. = 5.830 Problem 13: Calculate the standard deviation and the coeficient of variation from the following data. Marks Obtained No. of Students 0-10 3 10-20 7 20-30 11 30-40 15 40-50 12 50-60 8 60-70 4 Solution: 60

Class Mid Frequency Deviation Step - fd fd2 Values (f) D deviation d 0-10 27 10-20 5 3 –30 –3 –9 28 20-30 15 7 –20 –2 –14 11 30-40 25 11 –10 –1 –11 0 40-50 35 15 0 0 o. 12 50-60 45 12 + 10 +1 + 12 32 60-70 55 8 +20 +2 + 16 36 65 4 +30 +3 + 12 146 Total 60 +6 fd2 § fd ·2 SD = ¨ ¸ *1 n ©n¹ CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 259 146 §46·2 S.D. = ¨ ¸ *10 60 © 60 ¹ S.D. = 2.430.01*10 S.D. = (1.556) * 10 S.D. = 15.56 ¦fd Mean = a + n * 1 6 Mean = 35 + 60 * 10 Mean = 36 S.D. Coeficient of Variation = Mean * 100 15.56 10. *100 36 1556 z 36 z 43.22 Problem 14: Which of the following two series ismore variable? Give reasons.

X deviation d2 y deviation d2 d d 4 86 100 from 144 225 83 from 96 169 576 64 1024 136 –8 729 75 –10 441 129 –15 –13 120 –24 –32 324 117 –27 0 21 78 121 –18 225 134 –10 100 0 729 140 –4 16 96 1681 156 144 107 +11 4814 165 +12 441 111 + 15 170 +21 676 123 +27 173 +26 841 137 +41 +29 1440 3812 960 CU IDOL SELF LEARNING MATERIAL (SLM)

260 Business Mathematics and Statistics X: 136 129 120 117 134 140 156 165 170 173 Y: 86 83 64 75 78 96 107 111 123 137 Solution: For X series: ¦fx 1440 Mean = n 10 = 144 © fd2 SD = n 3812 SD = 10 §1 · SD = ¨ log381.2¸ Antilog ©2 ¹ § 2.5811· SD = ¨ ¸ Antilog © 2¹ SD = Antilog (1.2905) SD = 19.52

SD Coefficient of Variation = Mean *100 19.52 z 144 * 100 1952 17. 144 18. Antilog (log 1952 – log 144) 19. Antilog (3.2904 – 2.1584) 20. Antilog (1.1320) 21. 13.55 CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 261 For Y Series: 960 ¦y Mean = n 10 = 96 6.fd2 SD = n 4814 SD = 10 §1 · SD = ¨ log 481.4¸ Antilog ©2 ¹ § 2.6825 · SD = ¨ ¸ Antilog 2¹ © SD = Antilog (1.3412) SD = 21.94 SD Coeficient of Variation = Mean * 100 21.94 = 96 * 100

2194 = 96 z Antilog (log 2194 – log 96) z Antilog (3.3412 – 1.9823) z Antilog (1.3589) z 22.85 Comparing the coefficient of variation in X series with that of Y series we find that there is a greater variation in Y series than in X series. Therefore Y series is more variable than X series. CU IDOL SELF LEARNING MATERIAL (SLM)

262 Business Mathematics and Statistics Problem 15: The following is the record of goals scored by team A in a football season: Number of Goals Scored by Number of Matches Team A in a Match 1 0 9 1 7 2 5 3 3 4 For team B, the average number of goals scored per match was 2.5 with a standard deviation of 1.25 goals. Find out which team is more consistent. Solution: ¦fx 50 Mean = n 25 = 2 ¦fd2 SD = where d stands for deviation from mean n SD = 30 25 SD = 6 5 = 1095

Number of goals Number of xf Deviation fd f d2 scored matches from mean 2d X f –1 4 0 –2 –9 9 0 1 9 –1 0 0 1 9 14 0 5 5 2 7 15 +1 +6 12 3 5 12 +2 30 4 3 Total 25 50 SD Coeficient of Variation of Team A = Mean * 100 CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 263 1.095 9. * 100 2 109.5 = 96 = 54.75 SD Coefficient of Variation of Team B = Mean * 100 1125 2.5 * 100 \\emdash 112.5 = 2.5 The coeficient of variation for team B is less than that for team A. Therefore team B is more consistent. Problem 16: Given below are the share prices of A and B. State which is stable in value. A: 55 54 52 53 56 58 52 50 51 49 B: 108 107 105 105 106 107 104 103 104 101

Solution: d2 B deviation d2 A deviation d d from 105 from 533 4 108 +3 9 55 +2 1 107 +2 4 54 +1 1 105 0 0 52 –1 0 105 0 0 53 0 9 106 +1 1 56 +3 25 107 +2 4 58 +5 1 104 –1 1 52 –1 9 103 –2 4 50 –3 4 104 –1 1 51 –2 16 101 –4 16 49 –4 530 70 1050 40 CU IDOL SELF LEARNING MATERIAL (SLM)

264 Business Mathematics and Statistics For shares A: = xf 530 Mean = n 10 = 53 H fd2 SD = n 70 SD = 10 SD = 7 = 2.646 (From table of Sq. roots) SD Coefficient of Variation = Mean * 100 2646 5 53 * 100 6 4.99 For shares B: 18. xf 1050 Mean = = 105

n 10 z fd2 SD = n 40 SD = 10 SD= 4 =2 (From table of Sq. roots) SD Coefficient of Variation = Mean * 100 2 = 105 * 100 CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 265 200 ? 105 ? 1.905 approx. Comparing the respective coeficients of variation in the prices of shares A and B, we find 1.905 is less than 4.99. Therefore the prices of share B are more stable than that of A. Problem 17: The scores of two batsmen A and B in ten innings during a certain season are as under. Which of the batsmen is more consistent in scoring? A: 32 28 47 63 71 39 10 60 96 14 B: 19 331 48 53 67 90 10 62 40 80 Solution: d d2 B d d2 A –14 196 19 –31 961 –18 324 32 +1 31 –19 361 28 +17 1 47 ±25 289 48 –2 4 63 625 71 –7 53 +3 9 39 –36 49 10 + 14 1296 67 + 17 289 60 +50 96 –32 199 90 +40 2600 14 2500 1024 10 –40 1600 62 +12 144 40 –10 100 80 +30 900

460 6500 500 5968 For Batsman A: 8. x 460 AM = n 10 = 46 ? fd2 SD = n 6500 SD = 10 CU IDOL SELF LEARNING MATERIAL (SLM)

266 Business Mathematics and Statistics SD = 650 = 25.49 SD Coefficient of Variation = Mean * 100 25.49 = 46 * 100 2549 = 46 = 55.40 For Batsman A: = x 500 AM = n 10 = 50 ? fd2 SD = n 5968 SD = 10 SD = 596.8 = 24.43

SD Coefficient of Variation = Mean * 100 24.43 = 50 * 100 2443 = 50 = 48.86 The coefficient of variation in the scores of batsman B is less than that of batsman A. Hence batsman B is more consistent in scoring. CU IDOL SELF LEARNING MATERIAL (SLM)

Dispersion 267 Problem 18: Given the sets numbers 2, ?, 8, 11, 14 and 2, 8, 14, find – = = The mean of each set. = The variance of each set. = The coefficient of variation of the combined or pooled sets. Solution: Let the two sets of numbers be denoted by A and B respectively. Set A: Variance = Mean = Variance = = = Set B: Mean =

)2 + (88)2 + (118)2 + (148)2 2+5+8+11+14 40 8 5 3690936 5 90 5 zf 5 =18 d = 2 2814 24 n = fd2 =8 ( n 2 3 8 ) V2 (28)2 (88)2 (148)2 3 2 + 36036 72 ( 5 8 = =24 33 Coefficient of Variation of the combined set A and set B: n x 'n x ' Combined Mean X' = 1 1 2 2 n1 n2 CU IDOL SELF LEARNING MATERIAL (SLM)