To solve this question, let's break down the tasks step-by-step.

1. Identify the Probability Density Function (PDF):

The PDF $f(x) = \frac{3(8x - x^2)}{256}$ is defined over the interval $0 < X < 8$.

2. Compute the Cumulative Distribution Function (CDF), $F(x)$:

The CDF, $F(x)$, is obtained by integrating $f(x)$ from the lower bound (0) to $x$: $F(x) = \int_0^x f(t) \, dt = \int_0^x \frac{3(8t - t^2)}{256} \, dt$ This integral will give us $F(x)$, the cumulative distribution function, which we will use to calculate the probabilities.

3. Calculate the Given Probabilities:

Using the CDF $F(x)$, we can evaluate each probability:

• (a) $P(X < 2) = F(2)$
• (b) $P(X < 9)$ (Note: Since $X$ is defined only on $0 < X < 8$, $P(X < 9) = P(X < 8) = 1$)
• (c) $P(2 < X < 4) = F(4) - F(2)$
• (d) $P(X > 6) = 1 - F(6)$

4. Determine $x$ for $P(X < x) = 0.95$:

For part (e), we need to find $x$ such that $F(x) = 0.95$. This involves solving for $x$ in the equation: $F(x) = 0.95$

I'll proceed with these calculations to provide the exact values for each part.Here are the calculated probabilities for each part:

1. (a) $P(X < 2) = \frac{5}{32} \approx 0.15625$
2. (b) $P(X < 9) = 1$
3. (c) $P(2 < X < 4) = \frac{11}{32} \approx 0.34375$
4. (d) $P(X > 6) = \frac{5}{32} \approx 0.15625$
5. (e) $P(X < x) = 0.95$ yields $x \approx 6.92$ (only the real positive solution is meaningful within the interval)

Would you like further details on any of these calculations?

Here are some related questions for further exploration:

1. How would the cumulative distribution function change if the interval was $0 < X < 10$ instead of $0 < X < 8$?
2. What is the significance of the CDF reaching 1 at $x = 8$?
3. How do we interpret the result $P(X < x) = 0.95$ in terms of a quantile?
4. Can the probability $P(X > 6)$ be determined directly from properties of the distribution without using the CDF?
5. What steps would you take to verify that this function $f(x)$ is indeed a valid PDF over the interval?

Tip: For any continuous distribution, remember that the area under the curve of the PDF over its defined interval should equal 1, a property essential for verifying any probability density function.