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?

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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.