Given the below cumulative distribution function, find the following probabilities.
CDF:
F(x) =
0
if x <2
0.2
if 2≤x<3
0.25 if 3 ≤x <6
0.35 if 6≤x<9
1
if 9≤x
P(X = 0)=
0
P(X< 3) =
0.25
P(X < 6) =
P(6 ≤ X <9)=
P(2 < X ≤6)=
P(2 < X< 9)=
P(X ≥3)=
P(X > 6) =
0.65
く
0.65
0.15
0.15

Let's go through the given cumulative distribution function (CDF) and compute the probabilities requested. Here's a breakdown of the given CDF \( F(x) \):

- \( F(x) = 0 \) if \( x < 2 \)
- \( F(x) = 0.2 \) if \( 2 \leq x < 3 \)
- \( F(x) = 0.25 \) if \( 3 \leq x < 6 \)
- \( F(x) = 0.35 \) if \( 6 \leq x < 9 \)
- \( F(x) = 1 \) if \( x \geq 9 \)

### Now, we'll compute the requested probabilities:

1. **\( P(X = 0) \)**:
   - Since the CDF shows \( F(x) = 0 \) for \( x < 2 \), and \( X = 0 \) falls in this range, the probability is:
   \[
   P(X = 0) = 0
   \]

2. **\( P(X < 3) \)**:
   - From the CDF, for \( x = 3 \), \( F(x) = 0.25 \). Since the CDF is non-decreasing, and for \( x < 3 \), \( F(x) = 0.2 \), the probability of \( P(X < 3) \) is:
   \[
   P(X < 3) = 0.2
   \]

3. **\( P(X < 6) \)**:
   - From the CDF, for \( x = 6 \), \( F(x) = 0.35 \). Thus, the probability of \( P(X < 6) \) is:
   \[
   P(X < 6) = 0.25
   \]

4. **\( P(6 \leq X < 9) \)**:
   - From the CDF, the probability mass between 6 and 9 can be found by subtracting the values of the CDF at these points:
   \[
   P(6 \leq X < 9) = F(9) - F(6) = 1 - 0.35 = 0.65
   \]

5. **\( P(2 < X \leq 6) \)**:
   - The probability between 2 and 6 is:
   \[
   P(2 < X \leq 6) = F(6) - F(2) = 0.25 - 0.2 = 0.05
   \]

6. **\( P(2 < X < 9) \)**:
   - The probability between 2 and 9 is:
   \[
   P(2 < X < 9) = F(9) - F(2) = 1 - 0.2 = 0.8
   \]

7. **\( P(X \geq 3) \)**:
   - This can be found as:
   \[
   P(X \geq 3) = 1 - P(X < 3) = 1 - 0.2 = 0.8
   \]

8. **\( P(X > 6) \)**:
   - The probability for \( X > 6 \) is:
   \[
   P(X > 6) = 1 - F(6) = 1 - 0.35 = 0.65
   \]

### Summary of Results:
- \( P(X = 0) = 0 \)
- \( P(X < 3) = 0.2 \)
- \( P(X < 6) = 0.25 \)
- \( P(6 \leq X < 9) = 0.65 \)
- \( P(2 < X \leq 6) = 0.05 \)
- \( P(2 < X < 9) = 0.8 \)
- \( P(X \geq 3) = 0.8 \)
- \( P(X > 6) = 0.65 \)

Do you need further details on any of these probabilities?

---

**Related Questions:**
1. What is the total probability for the entire distribution?
2. How would \( P(X = 2) \) and \( P(X = 6) \) be determined?
3. What is the probability that \( X = 9 \)?
4. How does the CDF help in finding probabilities of ranges like \( P(a \leq X \leq b) \)?
5. What is the significance of jumps in the CDF for discrete distributions?

**Tip**: The probability for a single value in a continuous distribution is always zero unless the distribution has discrete components at that point.

Given the below cumulative distribution function, find the following probabilities.
CDF:
F(x) =
0 if x <2
0.2 if 2≤x<3
0.25 if 3 ≤x <6
0.35 if 6≤x<9
1 if 9≤x
P(X = 0)= 0
P(X< 3) = 0.25
P(X < 6) = P(6 ≤ X <9)= P(2 < X ≤6)= P(2 < X< 9)= P(X ≥3)= P(X > 6) = 0.65

Solve probability problems using the Cumulative Distribution Function (CDF) with step-by-step guidance. Learn how to calculate P(X < 3), P(2 < X ≤ 6), and more. Suitable for college-level statistics students.

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