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Chapter 15 Class 11 Statistics
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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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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.) = 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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