Let's calculate each probability based on the cumulative distribution function (CDF) $F(x)$.

(a) $p(2) = P(X = 2)$

To find $P(X = 2)$, we need to find the probability that $X = 2$. We can calculate this by finding the difference in the CDF values just before and just after 2:

$p(2) = P(X = 2) = F(2) - F(1)$

From the given CDF:

• $F(2) = 0.32$
• $F(1) = 0.22$

Thus, $p(2) = 0.32 - 0.22 = 0.10$

(b) $P(X > 3)$

To find $P(X > 3)$, we calculate it as $1 - F(3)$ because $F(3)$ represents $P(X \leq 3)$:

$P(X > 3) = 1 - F(3)$

From the CDF:

• $F(3) = 0.54$

Therefore, $P(X > 3) = 1 - 0.54 = 0.46$

(c) $P(2 \leq X \leq 5)$

To calculate $P(2 \leq X \leq 5)$, we find the difference between $F(5)$ and $F(2)$:

$P(2 \leq X \leq 5) = F(5) - F(2)$

From the CDF:

• $F(5) = 0.96$
• $F(2) = 0.32$

Thus, $P(2 \leq X \leq 5) = 0.96 - 0.32 = 0.64$

(d) $P(2 < X < 5)$

To find $P(2 < X < 5)$, we need to find the probability between 2 and 5, excluding both endpoints. This can be calculated as:

$P(2 < X < 5) = F(4) - F(2)$

From the CDF:

• $F(4) = 0.89$
• $F(2) = 0.32$

Therefore, $P(2 < X < 5) = 0.89 - 0.32 = 0.57$

Summary of Results

• (a) $p(2) = 0.10$
• (b) $P(X > 3) = 0.46$
• (c) $P(2 \leq X \leq 5) = 0.64$
• (d) $P(2 < X < 5) = 0.57$

Would you like more details on these calculations?

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1. How is a cumulative distribution function (CDF) used to calculate probabilities?
2. What is the difference between $P(X \geq x)$ and $P(X > x)$ when using a CDF?
3. Why does the probability of a specific value, $P(X = x)$, depend on the CDF increments?
4. How does a CDF differ from a probability density function (PDF)?
5. Can the probability mass function (PMF) be derived from the CDF directly?

Tip: Remember that for discrete random variables, probabilities at specific values are found by the difference in CDF values at neighboring points.