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.