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- 1. Session – 9 Measures of Dispersions Standard Deviation Standard deviation is the root of sum of the squares of deviations divided by their numbers. It is also called ‘Mean error deviation’. It is also called mean square error deviation (or) Root mean square deviation. It is a second moment of dispersion. Since the sum of squares of deviations from the mean is a minimum, the deviations are taken only from the mean (But not from median and mode). The standard deviation is Root Mean Square (RMS) average of all the deviations from the mean. It is denoted by sigma (σ). Characteristics of standard deviation 1. Standard deviation and coefficient of variation possesses all these properties which a good measure of dispersion should possess. 2. The process of squaring the deviation eliminates negative sign and makes mathematical computations easy. Merits 1. It is based on all observations. 2. It can be smoothly handled algebraically. 3. It is a well defined and definite measure of dispersion. 4. It is of great importance when we are making comparison between variability of two series. Merits 1. It is difficult to calculate and understand. 2. It gives more weightage to extreme values as the deviation is squared. 3. It is not useful in economic studies. Standard deviation If the variant xi takes the values of x1, x2 ………….. xn the standard deviation denoted by σ and it is defined by σ= ( ∑ xi − x ) 2 N The quantity σ2 is called variance. 1
- 2. 2
- 3. Alternate Expressions For raw data σ2 = ∑ x2 n () − x 2 For a grouped data σ2 = ∑ fx 2 n () − x 2 2 ∑ fd 2 ∑ fd For a grouped data with step deviation method σ = − N N Coefficient of variance It is defined as the ratio to be equal to standard deviation divided by mean. σ The percentage form of CV is given by CV = x 100 x 3
- 4. Problems 1. Ten students of a class have obtained the following marks in a particular subject out of 100. Calculate SD and CV for the given data below. (x) d = (x1 = 38.5) Sl. No. (x1 - x )2 marks d = (x1 - x ) 1. 5 - 33.5 1122.25 2. 10 - 28.4 812.25 3. 20 - 18.5 342.25 4. 25 - 13.5 182.25 5. 40 1.5 2.25 6. 42 3.5 12.25 7. 45 6.5 42.25 8. 48 9.5 90.25 9. 70 31.5 992.25 10. 80 41.5 1722.25 Σ(x1 - x )2 = Σx = 385 Σd2 = 5320.50 ∑x x= N 385 = 10 = 38.5 σ= ( ∑ xi − x ) 2 N 5320.5 σ= 10 = 23.066 σ CV = x 100 x 23 CV = x 100 38.5 CV = 59.9% 4
- 5. 2. Compute standard deviation and coefficient of varience for following data of 100 students marks. Mid Class f Class point d fd fd2 x 1 – 10 3 0.5 – 10.5 5.5 -2 -6 12 11 – 20 16 10.5 – 20.5 15.5 -1 -16 16 21 – 30 26 20.5 – 30.5 25.5 0 0 0 31 – 40 31 30.5 – 40.5 35.5 1 31 31 41 – 50 16 40.5 – 50.5 45.5 2 32 64 51 – 60 8 50.5 – 60.5 55.5 3 24 72 N = Σf = Σfd = 65 Σfd2= 195 100 a = 25.5 x − a x − 25.5 d= = =d h 10 15.5 − 25.5 − 10 d= = = −1 10 10 ∑ fd x=a+h+ N 65 = 25.5 + 10 = 25.5 + 6.5 100 x 32 2 ∑ fd 2 ∑ fd σ=h − N N 2 195 65 σ = 10 − = 12.359 100 100 σ CV = x 100 x 12.359 CV = x 100 = 38.62% 32 5
- 6. 3. The AM and SD of a set of nine items are 43 and 5 respectively if an item of value 63 is added, find the mean and SD. ∑ xi x= N Σxi = x x N Σxi = 43 x 9 Σx = 387 for 9 items Σx = 387 + 63 for 10 item Σx = 450 ∑ x 450 Modified mean x = = N 10 x = 45 x = 43 σ=5 for 9 items σ2 = ∑ x2 N −x () 2 ∑ x2 − ( 43) 2 25 = 9 ∑ x2 25 = − 1849 9 ∑ x2 25 + 1849 = 9 ∑ x2 = 1874 9 Σx2 = 1874 Σx2 = 16866 for 9 items If 63 is added Σx2 = 16866 + (63)2 = 20835 for 10 items Modified σ2 = ∑ x2 N −x () 2 20835 − ( 45) 2 σ2 = σ2 = 7.64 is modified SD. 10 6
- 7. 4. The mean of 5 observations is 4.4. and variance is 8.24 and if the 3 items of the five observations are 1, 2 and 6. Find the values of other two observations. ∑x w.k.t. x = N ∑x 4.4 = N Σx = 22 σ2 = ∑ x2 N −x () 2 ∑ x2 − ( 4.4) 2 8.24 = 5 ∑ x2 8.24 = − 19.36 9 ∑ x2 8.24 + 19.36 = 5 Σx2 = 138 Σx2 = 12 + 22 + 62 + x12 + x22 138 = 1 + 4 + 36 + x12 + x22 97 = x12 + x22 x12 + x22 = 97 ---- (1) Σx = 1 + 2 + 6 + x1 + x2 22 = 9 + x1 + x2 x1 + x2 = - 13 ---- (2) put (2) in (1) x2 = 13 – x1 by (1) & (2) x12 + (13 – x1)2 = 97 x12 + 169 + x12 – 26x1 = 97 2 x12 – 26x1 + 72 = 0 x12 – 13x1 + 36 = 0 7
- 8. -b± b 2 − 49 x1 = 2a - (-13) ± 169 − 4 x 36 x1 = 2 13 ± 5 x1 = 2 13 5 x1 = ± 2 2 x1 = 6.5 ± 2.5 x1 = 9 or x1 = 4 x1 = 9 x2 = 4 8
- 9. 5. The mean and S.D. of the frequency distribution of a continuous random variable x are 40.604 and 7.92 respectively. Change of origin and scale is given below. Determine the actual class interval. d -3 -2 -1 0 1 2 3 4 f 3 15 45 57 50 36 25 9 d f fd fd2 MV CI -3 3 -9 27 22.5 20-25 -2 15 -30 60 29.5 25-30 -1 45 -45 45 32.5 30-35 0 57 0 0 37.5 35-40 1 50 50 50 42.5 40-45 2 36 72 144 47.5 50-55 3 25 75 225 52.5 55-60 4 9 36 144 57.5 N = 240 Σfd = 149 Σfd2 = 695 ∑ fd 149 x=a+h 40.604 = a + h N 240 40.604 = a + 0.62h ----- (1) 2 ∑ fd 2 ∑ fd σ=h − N N 2 695 149 7.92 = h − 240 240 = h 2.895 − 0.620 7.92 = h x 1.584 h = 4.998 h=5 Put h = 5 in equation (1) 40.604 = a + 0.62 x 5 a = 37.5 9
- 10. Combined Standard Deviation Suppose we have different samples of various sizes n1, n2, n3 …….. having means x1, x2, x3 and standard deviation σ1, σ2, σ3 ……. then combine standard deviation can be computed by the following formula. σ2 (n1 + n2) = n1 (σ12 + d12) + n2 (σ22 + d22) d1 = x 1 − x d2 = x 2 − x 1. The mean’s of two samples of sizes 50 and 100 respectively are 54.1 and 50.3 and there standard deviations are 8 and 7 respectively obtain the SD for combined group. n1 = 50 n2 = 100 x 1 = 54.1 x 2 = 50.3 σ1 = 8 σ2 = 7 n1 x1 + n 2 x 2 x= (n 1 + n 2 ) (50 x 54.1) + (100 x 50.3) x= 50 + 100 x = 51.56 σ2 (n1 + n2) = n1 (σ12 + d12) + n2 (σ22 + d22) d1 = x 1 − x d2 = x 2 − x d1 = 94.1 – 51.56 d1 = 2.54 d12 = 6.45 d2 = 50.3 – 51.56 d2 = - 1.26 d22 = 1.56 σ2 150 = 50 (82 + 6.45) + 100 (72 + 1.58) 3σ2 = (64 + 6.45) + 2 (49 + 1.58) 3σ2 = 70.45 + 2 x 50.58 σ = 7.56 10
- 11. 2. The mean wage is Rs. 75 per day, SD wage is Rs. 5 per day for a group of 1000 workers and the same is Rs. 60 and Rs. 4.5 for the other group of 1500 workers. Find mean and standard deviation for the entire group. We have by data, x 1 = 75, σ1 = 5, n1 = 1000 x 2 = 60, σ2 = 450, n2 = 1500 Let x and σ be the mean and SD of the entire group. n1 x1 + n 2 x 2 Consider x = n1 + n 2 1000 x 75 + 1500 x 60 i.e., x = =60 1000 + 1500 Also we have, (n1 + n2) σ2 = n1 (σ12 + d12) + n2 (σ22 + d22), where d1 = x 1 - x = 75 – 66 = 9; d2 = x 2 - x = 60 – 66 = -6 ∴ (1000 + 1500) σ2 = 1000 (52 + 92) + 1500 (4.52 + (-6)2) ∴ σ 2 = 76.15 or σ = 8.73 11
- 12. 3. The runs scored by 3 batsman are 50, 48 and 12. Arithmtic mean’s respectively. The SD of there runs are 15, 12 and 2 respectively. Who is t he most consistent of the three batsman? If the one of these three is to be selected who is to be selected? A B C AM ( x ) 50 48 12 SD(σ) 15 12 2 σA CVA = x 100 xA 15 CVA = x 100 50 CVA = 30% σB CVB = x 100 xB 12 CVB = x 100 48 CVB = 25% σC CVC = x 100 xC 2 CVC = x 100 12 CVC = 16.66% Evaluation Criteria 1. Less CV indicates more constant player and hence more consistent player is (Player C) 2. Highest rune scorer = x A = 50 12
- 13. 4. The coefficient of variation of the two series are 75% and 90% with SD 15 and 18 respectively compute there mean. CVA = 75% CVB = 80% σA = 15 σB = 18 13
- 14. σ CV = x 100 x 15 18 75 = x 100 90 = x 100 xA xA x A = 20 x A = 20 5. Goals scored by two teams A & B in a foot ball season are as shown below. By calculating CV in each, find which team may be considered as more consistent. No. of goals No. of matches Team (A) Team (B) x A-team B-team fx fx 0 27 17 0 0 1 9 9 9 9 2 8 6 16 12 3 5 5 15 15 4 4 3 16 12 N = Σf = 53 Σf = 40 Σfx = 56 Σfx2 = 48 Team (A) Team (B) fx2 fx2 0 0 9 9 32 24 45 45 64 48 Σfx2 = 150 Σfx2 = 126 ∑ fx 56 xA= = = 1.056 N 53 ∑ fx 48 xB= = = 1.2 N 40 σ = A ∑ fx 2 2 N − x () 2 = 150 53 − (1.056 ) = 1.715 = σ = 1.30 2 A σ = B 2 ∑ fx 2 N − x() 2 = 126 40 − (1.2 ) = 1.95 = σ = 1.30 2 B 1
- 15. σA 1.30 CVA = x 100 = x 100 = 123.8% xA 1.056 σB 1.30 CVB = x 100 = x 100 = 109% xB 1.2 Since, CVB < CVA, team B is more consistent player 6. The prices of x and y share A & B respectively state which share more stable in its value. Price A (xi = 53) Price - A (xi = 105) (xi = x )2 (xi = x )2 (x) (xi = x ) (4) (xi = x ) 55 2 4 108 3 9 54 1 1 107 2 4 52 -1 1 105 0 0 53 0 0 105 0 0 56 3 9 106 1 1 58 5 25 107 2 4 52 -1 1 104 -1 1 50 -3 9 103 -2 4 51 -2 4 104 -1 1 49 -4 16 101 -4 16 Σx = 530 Σ(xi= x )2 = 70 Σx = 1050 Σx(xi= x )2 = 40 2
- 16. ∑x 530 xA= = = 53 N 10 ∑ x 1050 xB= = = 105 N 10 70 σ= = σ = 2.64 A 10 A 40 σ= =σ=2 B 10 B σA 2.64 CVA = x 100 = x 100 = 4.98% x 53 σB 2 CVB = x 100 = x 100 = 1.903% x 105 Since, CVB is less share B is more stable. 7. A student while computing the coefficient of variation obtained the mean and SD of 100 observations as 40 and 5.1 respectively. It was later discovered that he had wrongly copied an observation as 50 instead of 40. Calculate the correct coefficient of variation. ∑x ∑x >> x = i.e. 40 = n 100 ∴ Σx (incorrect) = 4000 Now correct Σx = 4000 – 50 + 40 = 3990 3990 ∴ correct x = = 39.9 100 Let us consider σ = ∑ x2 n − x 2 () 2 ∑x2 ( 5.1) 2 = − ( 40) 2 100 ∑x2 ∑ x2 i.e. ( 40 ) + ( 5.1) = 2 2 or = 1626.01 100 100 ∴ Σx2 (incorrect) = 100 x 1626.01 = 162601 Now correct Σx2 = 162601 – (50)2 + (40)2 = 161701 3
- 17. ∴ correct σ2 = correct ∑ x2 n ( − correct x ) 2 161701 − ( 39.9) = 25 2 i.e., correct σ2 = 100 σ Now correct efficient of variation = x 100 x 5 x 100 = 12.56% 39.9 Hence correct C.V. = 12.53% 4
- 18. 8. The mean and SD of 21 observations are 30 and 5 respectively. It was subsequently noted that one of the observations 10 was incorrect. Omit it and determine the mean and SD of the rest. ∑x ∑x >> x = i.e. 30 = or ∑ x = 630 n 21 ∴ incorrect Σx = 630 Now omitting the incorrect value 10, New Σx = 630 – 10 = 620 n = 21 – 1 = 20 620 New x = = 31 20 Next consider σ 2 = ∑ x2 n − x () 2 ∑ x2 ( 5) 2 = − ( 30 ) 2 100 ∑ x2 i.e. 900 + 25 = 21 ∴ incorrect ∑ x 2 = 925 x 21 = 19425 Again omitting the incorrect value 10. New Σx = 19425 –(10)2 = 19325, n = 20 Hence new σ 2 = new ∑ x2 20 ( − new x ) 2 19325 − (31) 2 = 5.25 20 ∴ New σ = 5.25 = 2.29 9. The mean of 200 items was 50. Later on it was discovered that two items were misread as 92 and 8 instead of 192 and 88. Find out the correct mean. ∑x ∑x >> x = i.e. 50 = or ∑ x = 10000 n 200 ∴ incorrect Σx = 10000 Correct Σx = 10000 – 92 – 8 + 192 + 88 = 10180 10180 ∴ Correct mean = = 50.9 200 5
- 19. 10. Find the missing frequencies in the following data given that the median is 137.2. Class 100-1 110-1 120-1 130-1 140-1 150-1 106 170 10 20 30 40 50 00 -17 -18 0 0 Frequency 15 44 133 F1 125 F2 35 16 N=600 >> We prepare the table with the column of cumulative frequencies and use the formula for median. Class Frequency cf 100-110 15 15 110-120 44 59 120-130 133 192 130-140 f1 192 + f1 Median class 140-150 125 317 + f1 150-160 f2 317 + f1 + f2 160-170 35 352 + f1 + f2 170-180 16 368 + f1 + f2 N = 600 h N Median = 1 + − c f 2 We can take the median class as 130-140 since median is given to be 137.2 130 + 130 l= = 130 , h = 10 f = f1, c = 192 2 10 ∴ 137.2 = 130 + (300 - 192) f1 1080 i.e., 137-2 – 130 = i.e., 7.2 f1 = 1080 or f1 150 f1 But the last cumulative frequency must be equal to N = 600 i.e. 368 + f1 + f2 = 600 368 + 150 + f2 = 600 ∴ f2 = 82 Thus f1 = 150, f2 = 82 6
- 20. Relationship between various measures of dispersion We have some of following relationships among the various methods of measures of dispersion 1. Mean ± QD covers 50% of observations of the distribution 2. Mean ± MD covers 57.5% of observations 3. Mean ± 1 σ includes 68.27% of observations 4. Mean ± 2 σ includes 95.45% of observations 5. Mean ± 3 σ includes 99.73% of observations 2 6. QD = 6745 σ = σ 3 2 4 7. MD = xσ= σ A 5 5 8. QD = MD 6 9. Combining the results we get 3 QD = 2 SD and 5 MD = 4 SD that is also equal to 6 QD. 10. Range = 6 times SD. SOURCES AND REFERENCES 1. Statistics for Management, Richard I Levin, PHI / 2000. 2. Statistics, RSN Pillai and Bagavathi, S. Chands, Delhi. 3. An Introduction to Statistical Method, C.B. Gupta, & Vijaya Gupta, Vikasa Publications, 23e/2006. 4. Business Statistics, C.M. Chikkodi and Salya Prasad, Himalaya Publications, 2000. 5. Statistics, D.C. Sancheti and Kappor, Sultan Chand and Sons, New Delhi, 2004. 6. Fundamentals of Statistics, D.N. Elhance and Veena and Aggarwal, KITAB Publications, Kolkata, 2003. 7. Business Statistics, Dr. J.S. Chandan, Prof. Jagit Singh and Kanna, Vikas Publications, 2006. 7

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