To tackle this problem, we will go through each part step by step. Here's a breakdown of the process:

1. Probability Density Function and the Cumulative Distribution Function

Given: $f(x) = \frac{3(8x - x^2)}{256}, \quad 0 < x < 8$

This is a probability density function (PDF) for the continuous random variable $X$ in the interval $(0, 8)$.

To find the cumulative distribution function (CDF) $F(x)$, we need to integrate $f(x)$ from $0$ to $x$: $F(x) = \int_0^x f(t) \, dt$

2. Solving for Each Probability

For each part, we will evaluate the CDF at the required points.

1. (a) $P(X < 2)$

   • Calculate $F(2)$ to find this probability.

2. (b) $P(X < 9)$

   • Since the function is defined only for $0 < X < 8$, $P(X < 9)$ will actually be $P(X < 8) = F(8)$, which should be 1 (the total probability for the range).

3. (c) $P(2 < X < 4)$

   • This is calculated as $F(4) - F(2)$.

4. (d) $P(X > 6)$

   • This is $1 - F(6)$.

5. (e) $P(X < x) = 0.95$

   • Here, we need to solve for $x$ such that $F(x) = 0.95$.

Let's compute each of these

I'll proceed with the necessary integrations and calculations.Here are the results for each part:

1. (a) $P(X < 2) = F(2) = \frac{5}{32} \approx 0.15625$

2. (b) $P(X < 9) = P(X < 8) = 1$ (since the PDF is defined for $0 < X < 8$)

3. (c) $P(2 < X < 4) = F(4) - F(2) = \frac{11}{32} \approx 0.34375$

4. (d) $P(X > 6) = 1 - F(6) = \frac{5}{32} \approx 0.15625$

5. (e) Solving $P(X < x) = 0.95$ yields $x \approx 6.917$ within the defined interval $(0, 8)$.

Would you like more details on any step, or do you have questions about specific parts?

Here are five related questions for further exploration:

1. How would the PDF change if the interval were different?
2. How can we verify that this function is a valid probability density function?
3. What is the interpretation of the value found in part (e)?
4. How would these probabilities change if the function were scaled?
5. Can you compute the mean of $X$ given this PDF?

Tip: When solving for cumulative probabilities, understanding the PDF's bounds can simplify interpretation, especially when values exceed the range.