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

Ex 15.3,  5 - Chapter 15 Class 11 Statistics - Part 2
Ex 15.3,  5 - Chapter 15 Class 11 Statistics - Part 3 Ex 15.3,  5 - Chapter 15 Class 11 Statistics - Part 4 Ex 15.3,  5 - Chapter 15 Class 11 Statistics - Part 5

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Question 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.) = 𝜎/𝒙 Μ… Γ— 100 where 𝜎 = Standard Deviation 𝒙 Μ… = Mean Finding standard deviation & mean of both length(x) and weight(y) For length (𝒙) : ∴ Mean π‘₯ Μ… = (βˆ‘β–’π‘₯𝑖)/n where n = number of terms = 50 Mean = 212/50=4.24 Variance = 1/𝑁^2 [π‘βˆ‘β–’γ€–π‘“π‘–γ€–π‘₯𝑖〗^2 γ€—βˆ’(βˆ‘β–’π‘“π‘–π‘₯𝑖)^2 ] = 1/(50)^2 [50 Γ—902.8 βˆ’(212)^2] = 1/2500[45140 βˆ’44944] = 196/2500 = 0.0784 Standard deviation ("Οƒ") = βˆšπ‘‰π‘Žπ‘Ÿπ‘–π‘Žπ‘›π‘π‘’ = √0.0784 = 0.28 C.VX = 𝜎/𝒙 Μ… Γ— 100 = 1.28/4.24 Γ— 100 = 6.603 For weight (π’š) : ∴ Mean 𝑦 Μ… = (βˆ‘β–’π‘₯𝑖)/n where n = number of terms = 50 Mean = 261/50 = 5.22 Variance = 1/𝑁^2 [π‘βˆ‘β–’γ€–π‘“π‘–γ€–π‘₯𝑖〗^2 γ€—βˆ’(βˆ‘β–’π‘“π‘–π‘₯𝑖)^2 ] = 1/(50)^2 [50 (1457.6)βˆ’(γ€–261)γ€—^2] = 1/2500 [72880 – 68121] = 4759/2500 = 1.9036 Standard deviation = βˆšπ‘£π‘Žπ‘Ÿπ‘–π‘Žπ‘›π‘π‘’ = √1.9036 = 1.37 C.VY = 𝜎/𝒙 Μ… Γ— 100 = 1.37/5.22 Γ— 100 = 26.24 Since, C.V. of weight (y) > C.V. of length (x) ∴ Weight is more varying.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.