Let's work through this problem step by step.

a) Are the terms "mean" and "expected value" synonyms?

In probability theory and statistics, yes, the terms "mean" and "expected value" are synonyms. The mean of a probability distribution is the same as its expected value, representing the average outcome you would expect from a large number of trials.

So, the answer to part a) is: Yes, they are synonyms.

b) Find the expected value $E(X)$ for the given probability density function.

The expected value of a continuous random variable $X$ with probability density function (PDF) $f(x)$ is given by:

$E(X) = \int_{a}^{b} x f(x) \, dx$

Here, $f(x) = \frac{1}{24} x$, and the interval is $[4, 8]$.

Let's set up the integral:

$E(X) = \int_{4}^{8} x \cdot \frac{1}{24} x \, dx = \frac{1}{24} \int_{4}^{8} x^2 \, dx$

Now, calculate the integral:

$\int x^2 \, dx = \frac{x^3}{3}$

Evaluating this from 4 to 8:

$E(X) = \frac{1}{24} \left( \frac{8^3}{3} - \frac{4^3}{3} \right) = \frac{1}{24} \left( \frac{512}{3} - \frac{64}{3} \right)$

$E(X) = \frac{1}{24} \cdot \frac{448}{3} = \frac{448}{72} = \frac{112}{18} \approx 6.222$

So, the expected value $E(X)$ is approximately 6.222, rounded to three decimal places.

Would you like further details on any step? Here are five related questions to deepen understanding:

1. What is the significance of the expected value in real-world applications?
2. How do you interpret the PDF $f(x) = \frac{1}{24} x$ in this context?
3. How would you compute the variance for this probability density function?
4. What happens to the expected value if the interval changes from $[4, 8]$ to a wider range?
5. How does the expected value differ for discrete and continuous random variables?

Tip: When dealing with continuous probability distributions, always ensure the function integrates to 1 over the specified interval.