To solve this problem, we need to address the following:

1. Find the cumulative distribution function (CDF) for the probability density function (PDF) given as: $f(x) = \frac{3(8x - x^2)}{256}, \quad 0 < x < 8.$ The cumulative distribution function $F(x)$ is found by integrating $f(x)$ from 0 to $x$: $F(x) = \int_{0}^{x} f(t) \, dt = \int_{0}^{x} \frac{3(8t - t^2)}{256} \, dt.$

2. Evaluate the probabilities for the specific cases:

   • $P(X < 2)$
   • $P(X < 9)$ (Note: $X$ cannot exceed 8, so this will need consideration)
   • $P(2 < X < 4)$
   • $P(X > 6)$
   • Find $x$ for $P(X < x) = 0.95$.

Let's go through each part step-by-step.

Step 1: Calculate the CDF $F(x)$

To compute the CDF $F(x)$, we integrate the PDF: $F(x) = \int_{0}^{x} \frac{3(8t - t^2)}{256} \, dt.$ This integral will yield the cumulative distribution function, which we can use to evaluate each probability in parts (a) through (e).

Step 2: Solve each Probability

1. (a) $P(X < 2)$: Substitute $x = 2$ in the CDF $F(x)$ once we find it.

2. (b) $P(X < 9)$: Since $X$ is only defined for $0 < X < 8$, $P(X < 9) = P(X < 8) = 1$.

3. (c) $P(2 < X < 4)$: Calculate $F(4) - F(2)$ to find the probability that $X$ falls between 2 and 4.

4. (d) $P(X > 6)$: This can be found by $1 - F(6)$.

5. (e) $P(X < x) = 0.95$: Solve for $x$ by setting $F(x) = 0.95$ and solving the resulting equation.

I'll proceed with these calculations now.The cumulative distribution function (CDF) $F(x)$ is:

$F(x) = -\frac{x^3}{256} + \frac{3x^2}{64}$

Now, let's use this CDF to evaluate each probability in parts (a) through (e).

1. (a) $P(X < 2)$: Substitute $x = 2$ into $F(x)$.
2. (c) $P(2 < X < 4)$: Calculate $F(4) - F(2)$.
3. (d) $P(X > 6)$: Calculate $1 - F(6)$.
4. (e) $P(X < x) = 0.95$: Solve $F(x) = 0.95$ for $x$.

I'll compute each of these values now.Here are the results for each part:

1. (a) $P(X < 2)$: $F(2) = \frac{5}{32} \approx 0.15625$
2. (c) $P(2 < X < 4)$: $F(4) - F(2) = \frac{11}{32} \approx 0.34375$
3. (d) $P(X > 6)$: $1 - F(6) = \frac{5}{32} \approx 0.15625$
4. (e) $P(X < x) = 0.95$: The value of $x$ that satisfies $F(x) = 0.95$ is approximately $x \approx 6.917$.

Would you like further details on any part of these calculations?

Related Questions

1. How do we interpret cumulative distribution functions (CDFs) in probability?
2. What steps are involved in calculating probabilities using a PDF?
3. How can we use integration to find probabilities for continuous distributions?
4. Why is $P(X < 9)$ automatically equal to 1 in this problem?
5. What methods can be used to solve equations involving CDFs?

Tip

For continuous distributions, probabilities for specific points (like $P(X = 2)$) are always zero.