To find the mean (expected value) of the probability density function (PDF) described by the graph, we follow these steps:

1. Understand the PDF:

• The graph consists of four linear segments:
  • Segment 1: From the negative x-axis to the origin (does not contribute to the integral).
  • Segment 2: From $(0, 0)$ to $(6, 0.2)$, passing through $(3, 0.1)$.
  • Segment 3: From $(6, 0.2)$ to $(10, 0)$, passing through $(8, 0.1)$.
  • Segment 4: From $(10, 0)$ to the positive x-axis (does not contribute to the integral).

2. Define the linear functions:

• For Segment 2, the function $f_1(x)$ is linear and can be written as: $f_1(x) = \frac{0.2 - 0}{6 - 0}x = \frac{1}{30}x \text{ for } 0 \leq x \leq 6$
• For Segment 3, the function $f_2(x)$ is linear and can be written as: $f_2(x) = \frac{0 - 0.2}{10 - 6}(x - 6) + 0.2 = -\frac{1}{20}(x - 6) + 0.2 \text{ for } 6 \leq x \leq 10$

3. Calculate the mean (expected value):

The mean $\mu$ of the PDF is given by: $\mu = \int_{-\infty}^{\infty} x f(x) \, dx$ For the given PDF, we split the integral across the segments where the PDF is nonzero: $\mu = \int_0^6 x \left(\frac{1}{30}x\right) dx + \int_6^{10} x \left(-\frac{1}{20}(x - 6) + 0.2\right) dx$

4. Evaluate the integrals:

• First integral: $\int_0^6 x \left(\frac{1}{30}x\right) dx = \frac{1}{30} \int_0^6 x^2 \, dx = \frac{1}{30} \cdot \frac{x^3}{3} \bigg|_0^6 = \frac{1}{90} \cdot 216 = \frac{216}{90} = 2.4$

• Second integral: $\int_6^{10} x \left(-\frac{1}{20}(x - 6) + 0.2\right) dx$ This integral can be split into two parts: $\int_6^{10} x \left(-\frac{1}{20}x + \frac{6}{20} + 0.2\right) dx = \int_6^{10} \left(-\frac{1}{20}x^2 + \left(\frac{1}{10}\right)x\right) dx$ Evaluate this integral and add it to the previous result.

5. Final Mean Calculation:

Adding both integrals gives the mean value of $x$.

Let me calculate the second integral to find the mean.The mean (expected value) of the given probability density function (PDF) is approximately $\mu = 5.33$.

Would you like more details on this calculation, or do you have any other questions?

Here are some related questions for further exploration:

1. How is the variance of a PDF calculated?
2. What is the significance of the mean in a probability distribution?
3. How would the mean change if the PDF graph were shifted horizontally?
4. What if the graph were nonlinear? How would that affect the mean?
5. How do you calculate the mean for a discrete probability distribution?
6. Can the mean of a PDF be negative?
7. How do you interpret the mean in the context of a real-world application?
8. What role does symmetry in the PDF play in determining the mean?

Tip: When dealing with piecewise functions, always break down the function into manageable parts and integrate over each relevant interval.