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Posted

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.

  • 3 weeks later...
Posted

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

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