Suppose that f(x) = e
−x
for x ≥ 0 is a PDF of a distribution. Determine the following:
(a) P(X > 1)
(b) P(1 < X < 2.5)
(c) P(X = 3)
(d) P(X < 4)
(e) a > 0 such that P(X ≥ a) = 0.10

Given the probability density function (PDF) \( f(x) = e^{-x} \) for \( x \geq 0 \), we are dealing with an exponential distribution with parameter \( \lambda = 1 \), since \( f(x) = \lambda e^{-\lambda x} \) is the general form of an exponential distribution. Here, \( \lambda = 1 \).

For an exponential distribution with parameter \( \lambda = 1 \), the cumulative distribution function (CDF) is:

\[
F(x) = 1 - e^{-x}
\]

Now, let's calculate each part.

---

### (a) \( P(X > 1) \)

We can compute this as:

\[
P(X > 1) = 1 - P(X \leq 1) = 1 - F(1)
\]

\[
F(1) = 1 - e^{-1}
\]

So,

\[
P(X > 1) = e^{-1}
\]

Thus:

\[
P(X > 1) = \frac{1}{e} \approx 0.3679
\]

---

### (b) \( P(1 < X < 2.5) \)

This is the probability that \( X \) lies between 1 and 2.5:

\[
P(1 < X < 2.5) = F(2.5) - F(1)
\]

Using the CDF formula \( F(x) = 1 - e^{-x} \):

\[
F(2.5) = 1 - e^{-2.5}, \quad F(1) = 1 - e^{-1}
\]

Thus:

\[
P(1 < X < 2.5) = (1 - e^{-2.5}) - (1 - e^{-1}) = e^{-1} - e^{-2.5}
\]

So,

\[
P(1 < X < 2.5) \approx 0.3679 - 0.0821 = 0.2858
\]

---

### (c) \( P(X = 3) \)

For a continuous random variable, the probability that \( X \) takes any specific value is zero:

\[
P(X = 3) = 0
\]

---

### (d) \( P(X < 4) \)

We calculate this directly using the CDF:

\[
P(X < 4) = F(4) = 1 - e^{-4}
\]

Thus:

\[
P(X < 4) = 1 - \frac{1}{e^4} \approx 1 - 0.0183 = 0.9817
\]

---

### (e) Find \( a > 0 \) such that \( P(X \geq a) = 0.10 \)

This is equivalent to solving:

\[
P(X \geq a) = 1 - P(X < a) = 1 - F(a) = 0.10
\]

So,

\[
F(a) = 0.90 \quad \Rightarrow \quad 1 - e^{-a} = 0.90
\]

Solving for \( a \):

\[
e^{-a} = 0.10 \quad \Rightarrow \quad -a = \ln(0.10) \quad \Rightarrow \quad a = -\ln(0.10)
\]

Thus:

\[
a = \ln(10) \approx 2.3026
\]

---

### Summary:

(a) \( P(X > 1) \approx 0.3679 \)

(b) \( P(1 < X < 2.5) \approx 0.2858 \)

(c) \( P(X = 3) = 0 \)

(d) \( P(X < 4) \approx 0.9817 \)

(e) \( a \approx 2.3026 \)

---

Would you like any further details or clarifications? Here are some related questions you might find interesting:

1. How would the results change if \( \lambda \) were not 1?
2. How do you find the mean and variance of an exponential distribution?
3. What is the relationship between the exponential distribution and the Poisson process?
4. How do you compute the moment generating function for an exponential distribution?
5. Can you derive the memoryless property of the exponential distribution?

**Tip:** The exponential distribution is known for its memoryless property, meaning the future probability does not depend on the past.

Suppose that f(x) = e^−x for x ≥ 0 is a PDF of a distribution. Determine the following: (a) P(X > 1) (b) P(1 < X < 2.5) (c) P(X = 3) (d) P(X < 4) (e) a > 0 such that P(X ≥ a) = 0.10

Explore the solution to a probability problem involving the exponential distribution PDF f(x) = e^−x for x ≥ 0. Calculate various probabilities, including P(X > 1), P(1 < X < 2.5), and P(X < 4), and determine the threshold a such that P(X ≥ a) = 0.10. This problem is suitable for undergraduate students studying probability.

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