The exponential probability density function (PDF) is given by:

$f(x; \lambda) = \lambda e^{-\lambda x} \quad \text{for } x \geq 0$

where $\lambda$ is the rate parameter. The cumulative distribution function (CDF), which gives $P(X \leq x)$, is:

$P(X \leq x) = F(x) = 1 - e^{-\lambda x}$

Using this formula, we can solve each part of your question.

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1. Formula for $P(X \leq x_0)$

For any $x_0$, the probability $P(X \leq x_0)$ is:

$P(X \leq x_0) = 1 - e^{-\lambda x_0}$

2. Find $P(X \leq 2)$

Using $x = 2$ in the CDF formula:

$P(X \leq 2) = 1 - e^{-\lambda \cdot 2}$

This depends on the value of $\lambda$. If you have $\lambda$, we can substitute it to find the numerical answer.

3. Find $P(X \geq 3)$

To find $P(X \geq 3)$, we use the complement rule:

$P(X \geq 3) = 1 - P(X \leq 3) = e^{-\lambda \cdot 3}$

Again, this requires the value of $\lambda$ for a numerical result.

4. Find $P(X \leq 5)$

For $x = 5$:

$P(X \leq 5) = 1 - e^{-\lambda \cdot 5}$

This will also depend on the rate parameter $\lambda$.

5. Find $P(2 \leq X \leq 5)$

The probability $P(2 \leq X \leq 5)$ can be calculated as:

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

Using the CDF values:

$P(2 \leq X \leq 5) = \left(1 - e^{-\lambda \cdot 5}\right) - \left(1 - e^{-\lambda \cdot 2}\right)$ $= e^{-\lambda \cdot 2} - e^{-\lambda \cdot 5}$

Again, with $\lambda$, this can be evaluated numerically.

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If you provide the value of $\lambda$, I can calculate exact numerical results for each part. Let me know if you'd like more details on any step!

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Would you like to proceed with further calculations, or do you have any questions?

Here are five related questions to explore exponential distributions further:

1. How does changing the rate parameter $\lambda$ affect the shape of the exponential distribution?
2. How can we interpret the mean and variance of an exponential distribution in real-world applications?
3. What are the memoryless properties of an exponential distribution, and how are they useful?
4. Can we find the median of an exponential distribution, and how does it compare to the mean?
5. How is the exponential distribution used in modeling time-to-failure scenarios in reliability analysis?

Tip: In exponential distributions, the rate parameter $\lambda$ is the reciprocal of the mean (i.e., $\lambda = \frac{1}{\mu}$). This relationship is often helpful for interpreting the parameter in practical contexts.