Let X and Y be random variable with the following distributions respectively :X 1 2 3 4 5 P 0.1 0.2 0.4 0.2 0.1 Y 10 20 35 50 60 P 0.1 0.2 0.4 0.2 0.1
Skewness of distribution of X and Y are same, Coefficient of Variations are not same
Both distributions are symmetric about their means: X has mean 3 with values 1, 5 and 2, 4 at equal distances and equal probabilities, and Y has mean 35 with 10, 60 (distance 25) and 20, 50 (distance 15) carrying equal probabilities.
A symmetric distribution has all odd central moments equal to 0, so the skewness of X and of Y is 0 in both cases, i.e. the skewness is the same.
For X: \(Var(X) = 0.1(4) + 0.2(1) + 0.4(0) + 0.2(1) + 0.1(4) = 1.2\), so \(\sigma_{X} \approx 1.095\) and \(CV_{X} = 1.095/3 \approx 0.365\).
For Y: \(Var(Y) = 0.1(625) + 0.2(225) + 0.4(0) + 0.2(225) + 0.1(625) = 215\), so \(\sigma_{Y} \approx 14.66\) and \(CV_{Y} = 14.66/35 \approx 0.419\).
The coefficients of variation therefore differ, because Y is not a linear rescaling of X (the gaps 10, 15, 15, 10 are uneven), so any statement claiming equal coefficients or unequal skewness fails.
Check: \(E(X) = 0.1 + 0.4 + 1.2 + 0.8 + 0.5 = 3\) and \(E(Y) = 1 + 4 + 14 + 10 + 6 = 35\), confirming the means used, and 0.365 ≠ 0.419.
Hence, Skewness of distribution of X and Y are same, Coefficient of Variations are not same.
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