Given the cumulative distribution function (CDF) of a random variable $X$ as:

$F(x) = 1 - e^{-x^2}, \quad x > 0$

We need to find the following probabilities:

(a) $P(0 < X < 3)$

Using the CDF, the probability $P(0 < X < 3)$ can be calculated as:

$P(0 < X < 3) = F(3) - F(0)$

First, let's compute $F(3)$ and $F(0)$:

$F(3) = 1 - e^{-3^2} = 1 - e^{-9}$ $F(0) = 1 - e^{-0^2} = 1 - e^0 = 0$

Thus: $P(0 < X < 3) = (1 - e^{-9}) - 0 = 1 - e^{-9}$

(b) $P(X > 1)$

To find $P(X > 1)$, we use the complement rule:

$P(X > 1) = 1 - P(X \leq 1) = 1 - F(1)$

Compute $F(1)$:

$F(1) = 1 - e^{-1^2} = 1 - e^{-1}$

Thus: $P(X > 1) = 1 - (1 - e^{-1}) = e^{-1}$

(c) $P(X = 5)$

For continuous random variables, the probability at a single point is always 0:

$P(X = 5) = 0$

Final answers:

• (a) $P(0 < X < 3) = 1 - e^{-9}$
• (b) $P(X > 1) = e^{-1}$
• (c) $P(X = 5) = 0$

Do you have any questions or need further details?

Here are 5 related questions to explore:

1. How does the probability density function (PDF) relate to the CDF?
2. What is the expected value of the random variable $X$?
3. How would you compute the variance of $X$?
4. How does the CDF change if we shift the random variable $X$?
5. What is the median of $X$ based on its CDF?

Tip: When working with CDFs, remember that probabilities are computed as differences in CDF values over intervals.