Introduction:
In probability theory and statistics, the Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, given the average rate of occurrence of the events.

Given Information:
We are given that the random variable X follows a Poisson distribution with a mean of 5. We need to find the expectation E[(X^2)^2].

Formula for Expectation:
The expectation of a function g(X) of a random variable X is given by E[g(X)] = Σ[g(x) * P(X = x)], where Σ denotes the sum over all possible values of X.

Calculating E[(X^2)^2]:
To find E[(X^2)^2], we need to find the expression for (X^2)^2 and then calculate the expectation.

Step 1: Finding (X^2)^2:
(X^2)^2 = ((X^2) * (X^2)) = (X^4)

Step 2: Calculating E[(X^2)^2]:
To calculate E[(X^2)^2], we substitute (X^4) into the formula for expectation:

E[(X^2)^2] = Σ[(X^4) * P(X = x)]

Step 3: Using the Poisson Probability Mass Function:
The Poisson distribution probability mass function is given by P(X = x) = (e^(-λ) * λ^x) / x!, where λ is the mean of the distribution.

In our case, λ = 5. Substituting this into the formula, we have:

E[(X^2)^2] = Σ[(X^4) * (e^(-5) * 5^x) / x!]

Step 4: Simplifying the Expression:
We can simplify the expression further by expanding (X^4) and rearranging the terms:

E[(X^2)^2] = Σ[(X^4) * (e^(-5) * 5^x) / x!]
= Σ[(X^2) * (X^2) * (e^(-5) * 5^x) / x!]
= Σ[(X^2) * (X^2) * (e^(-5)) * (5^x) / x!]
= (e^(-5)) * Σ[(X^2) * (X^2) * (5^x) / x!]

Step 5: Applying the Linearity of Expectation:
The expectation operator is linear. This means that E[aX + bY] = aE[X] + bE[Y].

Using this property, we can split the sum and calculate the expectation of each term separately:

E[(X^2)^2] = (e^(-5)) * [Σ[(X^2) * (5^x) / x!]]

Step 6: Using the Moment Generating Function:
The moment generating function of a Poisson distribution with mean λ is given by M(t) = e^(λ(e^t - 1)).