Step-by-step explanation:
The arithmetic mean of a distribution is 5. The second and the third moments about the mean are 20 and 140 respectively. What is the third moment of the distribution about 10?
Call the random variable x.
Now, define a new variable y = x - 5. Note that x - 10 = y - 5.
So, it is clear that (x-10)^3 = (y-5)^3
Also, note that (y-5)^3 can be expanded as follows:
Expand (y-5)³
Result ; y³-15 y²+75 y - 125
Letting E denote expectation with respect to the random variable x, we see that
E[(y-5)^3 ] = E(y^3) -15 E(y^2) + 75 E(y) - 125
Again, recalling that y = x - 5, have
E(y^3) = 140
E(y^2) = 20
E(y) = E(x) - 5 = 5 - 5 = 0
Thus,
E[(y-5)^3 ] = 140 -15(20) + 75(0) -125 = -285
Finally, note that
E[(y-5)^3] = E[({x-5} -5)^3] = E[(x-10)^3]
So, we get E[(x-10)^3] = -285.
