kidia Posted April 12, 2006 Share Posted April 12, 2006 I have a question here I will appreciate for any idea,The mean of distribution is 5.The second and third moments about the mean are 20 and 140 respectively.Find the moment of the distribution about 10. Link to comment Share on other sites More sharing options...
Tartaglia Posted April 28, 2006 Share Posted April 28, 2006 The second and third moment about the mean are the variance and skewness respectively, but in order to answer the question you need to specify which moment you want about 10. This should help E(X) = 5 E(X^2) - 25 =20 therefore E(X^2) = 45 E(X^3) - 3E(X^2)E(X) + 2 *(E(X))^3 = 140 therefore E(X^3) = 140 +3*45*5 -2*5^3 = 565 Link to comment Share on other sites More sharing options...
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