2. The arithmetic mean of a distribution is 5. The second and the third moments about the mean are 20 and 140 respectively. Find the third moment of the distribution about 10. Please explain the answer and workings.

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.

well the 1 and -1 are always after the 0 and the 0 is inthe middle. s thats what i would guess. but i dont know what a slope of 2 is sorry

step-by-step explanation:

the correct answer is x/3+10≥32

hope this !