Ex 15.3, 5 - Chapter 15 Class 11 Statistics (Term 1)
Last updated at May 29, 2018 by
Last updated at May 29, 2018 by
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.) = π/π Μ Γ 100 where π = Standard Deviation π Μ = Mean Finding standard deviation & mean of both length(x) and weight(y) For length (π) : β΄ Mean π₯ Μ = (ββπ₯π)/n where n = number of terms = 50 Mean = 212/50=4.24 Variance = 1/π^2 [πββγππγπ₯πγ^2 γβ(ββπππ₯π)^2 ] = 1/(50)^2 [50 Γ902.8 β(212)^2] = 1/2500[45140 β44944] = 196/2500 = 0.0784 Standard deviation ("Ο") = βππππππππ = β0.0784 = 0.28 C.VX = π/π Μ Γ 100 = 1.28/4.24 Γ 100 = 6.603 For weight (π) : β΄ Mean π¦ Μ = (ββπ₯π)/n where n = number of terms = 50 Mean = 261/50 = 5.22 Variance = 1/π^2 [πββγππγπ₯πγ^2 γβ(ββπππ₯π)^2 ] = 1/(50)^2 [50 (1457.6)β(γ261)γ^2] = 1/2500 [72880 β 68121] = 4759/2500 = 1.9036 Standard deviation = βπ£πππππππ = β1.9036 = 1.37 C.VY = π/π Μ Γ 100 = 1.37/5.22 Γ 100 = 26.24 Since, C.V. of weight (y) > C.V. of length (x) β΄ Weight is more varying.
Ex 15.3
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