Ex 15.3, 5 - Sum and sum of square corresponding to length x - Co-efficient of variation

Slide18.PNG
Slide19.PNG Slide20.PNG Slide21.PNG

  1. Chapter 15 Class 11 Statistics
  2. Serial order wise
Ask Download

Transcript

Ex 15.3, 5 The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below: Which is more varying, the length or weight? The value having more Coefficient of Variation will be more variable. Coefficient of Variation (C.V.) = ๐œŽ/๐’™ย ฬ… ร— 100 where ๐œŽ = Standard Deviation ๐’™ย ฬ… = Mean Finding standard deviation & mean of both length(x) and weight(y) For length (๐’™) : โˆด Mean ๐‘ฅย ฬ… = (โˆ‘โ–’๐‘ฅ๐‘–)/n where n = number of terms = 50 Mean = 212/50=4.24 Variance = 1/๐‘^2 [๐‘โˆ‘โ–’ใ€–๐‘“๐‘–ใ€–๐‘ฅ๐‘–ใ€—^2 ใ€—โˆ’(โˆ‘โ–’๐‘“๐‘–๐‘ฅ๐‘–)^2 ] = 1/(50)^2 [50 ร—902.8 โˆ’(212)^2] = 1/2500[45140 โˆ’44944] = 196/2500 = 0.0784 Standard deviation ("ฯƒ") = โˆš๐‘‰๐‘Ž๐‘Ÿ๐‘–๐‘Ž๐‘›๐‘๐‘’ = โˆš0.0784 = 0.28 C.VX = ๐œŽ/๐’™ย ฬ… ร— 100 = 1.28/4.24 ร— 100 = 6.603 For weight (๐’š) : โˆด Mean ๐‘ฆย ฬ… = (โˆ‘โ–’๐‘ฅ๐‘–)/n where n = number of terms = 50 Mean = 261/50 = 5.22 Variance = 1/๐‘^2 [๐‘โˆ‘โ–’ใ€–๐‘“๐‘–ใ€–๐‘ฅ๐‘–ใ€—^2 ใ€—โˆ’(โˆ‘โ–’๐‘“๐‘–๐‘ฅ๐‘–)^2 ] = 1/(50)^2 [50 (1457.6)โˆ’(ใ€–261)ใ€—^2] = 1/2500 [72880 โ€“ 68121] = 4759/2500 = 1.9036 Standard deviation = โˆš๐‘ฃ๐‘Ž๐‘Ÿ๐‘–๐‘Ž๐‘›๐‘๐‘’ = โˆš1.9036 = 1.37 C.VY = ๐œŽ/๐’™ย ฬ… ร— 100 = 1.37/5.22 ร— 100 = 26.24 Since, C.V. of weight (y) > C.V. of length (x) โˆด Weight is more varying.

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail