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Let's address each problem one by one.

### Problem 3:
Given:
- \( X \) is a continuous random variable with the probability density function (pdf) \( f_X(x) = \frac{8}{x^3} \) for \( x > 2 \).
- We need to find \( E[W] \), where \( W = \frac{X}{3} \).

**Solution:**
1. Since \( W = \frac{X}{3} \), we can express \( E[W] \) as:
   \[
   E[W] = E\left[\frac{X}{3}\right] = \frac{1}{3}E[X].
   \]

2. To find \( E[X] \), we use:
   \[
   E[X] = \int_{2}^{\infty} x \cdot f_X(x) dx = \int_{2}^{\infty} x \cdot \frac{8}{x^3} dx.
   \]

3. Simplify the integral:
   \[
   E[X] = 8 \int_{2}^{\infty} \frac{1}{x^2} dx = 8 \left[ -\frac{1}{x} \right]_2^{\infty} = 8 \left(0 + \frac{1}{2}\right) = 4.
   \]

4. Therefore, \( E[W] = \frac{1}{3} \cdot 4 = \frac{4}{3}. \)

### Problem 4:
Given:
- We need to find the expected value of the function \( g(x) = x^2 \) where \( X \) is a random variable with pdf \( f_X(x) = ae^{-ax}u(x) \), where \( a \) is a constant, and \( u(x) \) is the unit step function.

**Solution:**
1. \( E[g(X)] = E[X^2] \) can be found by:
   \[
   E[X^2] = \int_{0}^{\infty} x^2 \cdot ae^{-ax} dx.
   \]

2. This is a standard expectation for an exponential distribution. We know:
   \[
   E[X^2] = \frac{2}{a^2}.
   \]

### Problem 5:
Given:
- We need to find the mean and variance of the random variable whose pdf is represented by a triangular distribution (as per the graph).

Let's assume the graph indicates a triangular distribution on the interval \( [0, 4] \) with a peak at \( x = 2 \). For this type of distribution:
- \( f_X(x) = \frac{1}{2}x \) for \( 0 \leq x \leq 2 \),
- \( f_X(x) = \frac{1}{2}(4 - x) \) for \( 2 < x \leq 4 \).

**Solution:**
1. **Mean** \( E[X] \) for a triangular distribution can be calculated as:
   \[
   E[X] = \int_{0}^{2} \frac{x^2}{2} dx + \int_{2}^{4} \frac{x(4-x)}{2} dx = 2.
   \]

2. **Variance** is found by:
   \[
   \text{Var}(X) = E[X^2] - (E[X])^2 = \int_{0}^{2} \frac{x^3}{2} dx + \int_{2}^{4} \frac{x^2(4-x)}{2} dx - 4 = \frac{4}{3}.
   \]

### Problem 6:
Given:
- The pdf \( f_X(x) \) is defined as:
  \[
  f_X(x) = 
  \begin{cases}
    x, & 0 < x < 1, \\
    2 - x, & 1 < x < 2, \\
    0, & \text{elsewhere}.
  \end{cases}
  \]
- We need to find the moment generating function (MGF) \( M_X(t) \).

**Solution:**
1. The MGF is defined as:
   \[
   M_X(t) = E[e^{tX}] = \int_{-\infty}^{\infty} e^{tx}f_X(x) dx = \int_{0}^{1} xe^{tx} dx + \int_{1}^{2} (2 - x)e^{tx} dx.
   \]

2. Evaluate these integrals to find the complete MGF.

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Would you like me to provide more detailed solutions to any of these problems, or any clarifications? Here are some related questions to explore:

1. How do you determine the expected value and variance for other types of continuous distributions?
2. What is the significance of the moment generating function (MGF) in probability theory?
3. Can you generalize the process for finding the mean of piecewise functions like in Problem 5?
4. How do you interpret the unit step function \( u(x) \) when defining pdfs?
5. What are some other examples of real-world problems that use triangular distributions?

**Tip:** Understanding how to manipulate integrals is crucial in solving problems involving continuous random variables, especially when calculating moments or expected values.

The original question submitted by the user as text.

This page explores how to compute expected values, variance, and moment generating functions for random variables using probability density functions. It covers exponential and triangular distributions, with step-by-step solutions for integration and mathematical concepts.

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