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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?

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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.

Bir sürekli rastgele değişken X’in olasılık dağılım fonksiyonu 0 < X < 8 için f(x) = 3(8x - x^2)/256 olarak verilmiştir. Kümülatif dağılım fonksiyonunu ve aşağıdaki olasılıkları hesaplayınız: (a) P(X < 2), (b) P(X < 9), (c) P(2 < X < 4), (d) P(X > 6), (e) P(X < x) = 0.95 ise x=?

Learn how to calculate the cumulative distribution function (CDF) and specific probabilities for a continuous probability distribution with f(x) = 3(8x - x^2)/256 over the interval 0 < X < 8. This problem involves integration and understanding of continuous random variables.

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