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The given exponential probability density function is:

\[
f(x) = \frac{1}{4} e^{- \frac{x}{4}} \quad \text{for} \quad x \geq 0
\]

Let’s break down each part of the problem.

### Part (a)
You are asked to choose the correct formula for \( P(x \leq x_0) \), which is the cumulative distribution function (CDF) of the given exponential distribution.

The probability density function (PDF) of an exponential distribution is generally given by:

\[
f(x) = \lambda e^{-\lambda x}
\]
with \(\lambda = \frac{1}{4}\) for this case.

The CDF, \( P(x \leq x_0) \), for an exponential distribution is:

\[
P(x \leq x_0) = 1 - e^{-\lambda x_0}
\]

For \(\lambda = \frac{1}{4}\), this becomes:

\[
P(x \leq x_0) = 1 - e^{- \frac{x_0}{4}}
\]

Thus, the correct answer is option **2**.

### Part (b)
To find \( P(x \leq 2) \), we use the formula for the CDF:

\[
P(x \leq 2) = 1 - e^{- \frac{2}{4}} = 1 - e^{-0.5}
\]

Using a calculator:

\[
P(x \leq 2) = 1 - 0.6065 = 0.3935
\]

So, \( P(x \leq 2) \approx 0.3935 \).

### Part (c)
Similarly, for \( P(x \leq 3) \):

\[
P(x \leq 3) = 1 - e^{- \frac{3}{4}} = 1 - e^{-0.75}
\]

Using a calculator:

\[
P(x \leq 3) = 1 - 0.4724 = 0.5276
\]

So, \( P(x \leq 3) \approx 0.5276 \).

### Part (d)
For \( P(x \leq 5) \):

\[
P(x \leq 5) = 1 - e^{- \frac{5}{4}} = 1 - e^{-1.25}
\]

Using a calculator:

\[
P(x \leq 5) = 1 - 0.2865 = 0.7135
\]

So, \( P(x \leq 5) \approx 0.7135 \).

### Part (e)
To find \( P(2 \leq x \leq 5) \), we calculate:

\[
P(2 \leq x \leq 5) = P(x \leq 5) - P(x \leq 2)
\]

From part (b) and (d):

\[
P(2 \leq x \leq 5) = 0.7135 - 0.3935 = 0.32
\]

So, \( P(2 \leq x \leq 5) \approx 0.3200 \).

---

Do you have any questions or would you like more details on any of these parts?

Here are 5 related questions:
1. What is the general form of an exponential distribution's PDF?
2. How would the CDF formula change if \(\lambda\) were different?
3. What is the significance of the parameter \(\lambda\) in an exponential distribution?
4. How do you compute the probability for a range \(a \leq x \leq b\)?
5. What is the expected value and variance of an exponential distribution?

**Tip**: Remember that the exponential distribution is memoryless, meaning that the probability of an event happening after a certain time does not depend on how much time has already passed.

Consider the following exponential probability density function f(x) = (1/4)e^(-x/4) for x ≥ 0. Find the CDF and solve for specific probabilities: P(x ≤ 2), P(x ≤ 3), P(x ≤ 5), and P(2 ≤ x ≤ 5).

Learn how to solve exponential distribution problems, including calculating the CDF and finding probabilities for specific ranges. This example uses the exponential distribution with a given probability density function and explores calculations such as P(x ≤ 2) and P(2 ≤ x ≤ 5).

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