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s dy’, gp for ¥ = \= ae 1880, SD 5 V166 = 19 ge, cv for Y = =-* 100 ~ 12.88, y 37 «100, = 47.7%, , ance co-efficient of variance, 3, , isky, jon - 11 (Individual §e,;, , for Pp,, , then Project Y. Hence, pies Toject X jg, , , , , wet, by the two h, , follo :, Find who is better score:, , F jonings- pee vee, yer NEKO: 101 22 0 2 ii is men consistent, , pECcoO: 97 12 40 96 : 45 7 13 65, Computation of Standard Deviation ‘ 8 85 p ee, , atsmen named NEKO and DECO, , , , , , , , , , , , , , , , golution, , is NEKO to be x and DECO to be y, , Table, NECO, , “y | de =(K- X) ae rs, 0 -38.5 1482.25 8 -35.1 1232.01, 7 pos 992.25 8 35.1 | 1232.01, 13 -25.5 650.25 12 $1.2 | 967.21, 4 “24.5 600.25 18 | -30.1. | 906.01, 22 -16.5 272.25 16 | 27.1 | 734.41, 36 -2.5 6.25 40 31 | 9.61, 45 +6.5 42.25 56 +12.9 | 166.41, 65 +26.5 702.25 85 41.9 1755.61, 82 +43.5 1892.25 96 +52.9 2798.41, 101 +62.5 3906.25 97 97 +53.9 | 2905.21, 5385 Suet = 10846.50 [Sy = 431 [Sdy?=12706.9, , , , , , , , , , ye eae ee =32.475
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N CaPpy, , error mr, , o 0 32, CV i x1 38.5, DECO, xy 481, eo = 43.1, n 10, [Say /72706.90 1270.69 = 35.647, ee PSU 6.90, Vienna, Viglen LO, SD 347 100 = 82.71%, CV for Y = x 100 x, , , , y, , > A , consistent, DECO is a better run scorer and more c, , average is more and variation is less), Illustration - 12 (Discrete Series Comparison), , The following table gives the age distribu ;, which of the two groups is more variable in age:, , player than NEKO (because 5, , tion of boys and girls in a school, Fing, , , , , , , , , , , , , , , , , , , , , , , , Age: 10 11 12 13 14 15, No. of Boys: 11 14 7 10 8 5, No. of Girls: 13 15 10 9 5 3, Solution:, for Boys, Age (x) if fx dx=x-% dx fax?, 4g 11 110 -2 4 44, a 14 154 -1 1 14, 12 7 Aa ; i, wo 10 130 1 1 AO, if 8 112 2 A a, 15 5 15 3 in i:, N=55 |x fx= 665 5 fax? =145, oe BOD:, ea ap te, 2, ee - p= = = 2.64 =1.62, CV for X = Boe 100 = om 100 =13.4%
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oa ae =ypgg7i00 = 12.78% 4, Boys groups more variable in age then girls groups because their cv is more- ’, jlustration 13 (Continuous Series Comparison), Following data represents lives of 2 models of refrigerators A and B. ;, rife(No. of Years): 0 - 2 2-4 a6 oe8 @ 2 10 10 - 12, Model A: 5 16 13 7 5 4, Model B: 2 i 12 19 9, a) Find the average life of each model., b) Which model has great uniformity of life?, Solution:, Model A, Life ie Ae fx dx=x-X dx? fax”, 0-2 5 1 5 -5 25 125, 2.4 16 3 48 -2 4 64, noe 13 5 65 =i 13, 6.8 7 49 1 7, 8-10 9 45 5 9 45, get 4 1 44 5 25 100, oe | N= 80 5 fx= 254 = fdx* =354
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ae, , SPROIMEN OOpy, , , , ~, , wo, bo, f, , , , , , , , , , , , 2-4 :, - 60, 4-6 12 3 ea 0 0, 6-8 19 7 2 5 4 26, Seer 2 : a 4 16 16, 10\- 12 1 11 11 gs, N=50 3 = 308 —/*, eo Dy =, yo a, N, 284 —, SD for Y \ 50, = /5.68 =2.38, 2.38, CV for Y 76 7~109 «= 33.24%, Average life of Model A is 6.12 and Model B is 7.16. Model B has gree:, , uniformity of life., Illustration - 14 (Continuous Series Comparison), , The following information is related to the marks scored in a test by girls section, and boys section in a school. From this, you are required to find out: _, , i) Which section scores more marks in an average?, ii) Which section is more consistent in scoring?, , Marks: 0-10 10-20 20-30 30-40 40-50, No. of students, Boys section: 9 13 20 10 3, , Girls section: 12 24 12, , ~
