Solution:

Given:

First moment (about 3) = -2

Second moment (about 3) = 10

Third moment (about 3) = -25

Fourth moment (about 3) = 50

To find:

Fourth moment (about 5)

Formulae:

Let X be a random variable with probability distribution f(x).

The k-th moment of X about a constant c is given by:

Mk = E[(X - c)^k], where E denotes the expected value operator.

M1 = E[X - 3] = E[X] - 3

M2 = E[(X - 3)^2] = E[X^2] - 6E[X] + 9

M3 = E[(X - 3)^3] = E[X^3] - 9E[X^2] + 27E[X] - 27

M4 = E[(X - 3)^4] = E[X^4] - 12E[X^3] + 54E[X^2] - 108E[X] + 81

The k-th moment of X about a constant c can also be expressed in terms of the central moments of X:

Mk = ∑(n C k) (-c)^(n-k) * μn, where μn is the n-th central moment of X.

M1 = E[X - 3] = E[X] - 3

E[X] = M1 + 3 = -2 + 3 = 1

M2 = E[(X - 3)^2] = E[X^2] - 6E[X] + 9

E[X^2] = M2 + 6E[X] - 9 = 10 + 6(1) - 9 = 7

M3 = E[(X - 3)^3] = E[X^3] - 9E[X^2] + 27E[X] - 27

E[X^3] = M3 + 9E[X^2] - 27E[X] + 27 = -25 + 9(7) - 27(1) + 27 = 11

The third central moment of X about 3 is:

μ3 = E[(X - 3)^3] = -25

The fourth central moment of X about 3 is:

μ4 = E[(X - 3)^4] = 50

The fourth moment of X about 5 is:

M'4 = E[(X - 5)^4]

M'4 = E[((X - 3) + (3 - 5))^4]

M'4 = E[(X - 3)^4 + 4(X - 3)^3(3 - 5) + 6(X - 3)^2(3 - 5)^2 + 4(X - 3)(3 - 5)^3 + (3 - 5)^4]

M'4 = E[(X - 3)^4] + 4(3 - 5)E[(X - 3)^3] + 6(3 - 5)^2E[(X - 3)^2] + 4(3 - 5)^3E[X - 3] + (3 -

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