---
title: "Exponential cdf: P(X>2), mean, variance"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:exponential-cdf-1-minus-e-2x"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/exponential-cdf-1-minus-e-2x
site: BCT Engineering
---

# Exponential cdf: P(X>2), mean, variance

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2075 Asoj, 6 marks; 2073 Chaitra, 5 marks; 2072 Kartik, 5 marks (asked 3 times)
- Video: [MA25C11 | MA3355 | MA3391 | MA3303 | Exponential Distribution | Problems on Exponential Distribution](https://www.youtube.com/watch?v=Py4S1hTJPwg) — Mathematics Kala
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/exponential-cdf-1-minus-e-2x
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

The distribution function for a random variable x is
F(x) = 1 − e^(−2x) for x ≥ 0
     = 0 for x < 0
a) Find P(X > 2)
b) Find mean and variance of the variable X.

## Short answer

The distribution is exponential with rate 2. P(X>2)=e^-4=0.0183, mean=0.5, and variance=0.25.

## Answer

The distribution function F(x) = 1 − e⁻²ˣ (x ≥ 0) is that of an **exponential distribution** with rate λ = 2. Its pdf is

f(x) = dF/dx = 2e⁻²ˣ, x ≥ 0.

#### a) P(X > 2)
P(X > 2) = 1 − F(2) = 1 − (1 − e⁻⁴) = e⁻⁴ = **0.0183**

#### b) Mean and variance
```text
E(X)  = ∫₀^∞ x · 2e⁻²ˣ dx = 1/2 = 0.5
E(X²) = ∫₀^∞ x² · 2e⁻²ˣ dx = 2/λ² = 2/4 = 0.5
Var   = E(X²) − [E(X)]² = 0.5 − 0.25 = 0.25
```
(Integration by parts, or the standard results mean = 1/λ and variance = 1/λ².)

**Answer: P(X > 2) = e⁻⁴ = 0.0183, mean = 0.5, variance = 0.25**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. F(x) = 1 - e^(-2x) is the cdf of:
   A. Normal
   B. Gamma with alpha = 2 and beta = 1
   C. Poisson
   D. Exponential with rate 2
   Answer: D. It matches the exponential cdf with lambda = 2. (Key point: Identify)

2. The pdf is obtained by:
   A. Integrating F
   B. Differentiating F: 2e^(-2x)
   C. Dividing F by x
   D. Subtracting 1 from F
   Answer: B. f(x) = dF/dx. (Key point: pdf)

3. P(X > 2) is:
   A. F(2) = 0.9817
   B. 1 - F(2) = e^-4 = 0.0183
   C. e^-2
   D. 2e^-4
   Answer: B. Survival function. (Key point: P(X>2))

4. Mean and variance are:
   A. 2 and 4
   B. 0.5 and 0.5
   C. 0.25 and 0.5
   D. 1/lambda = 0.5 and 1/lambda^2 = 0.25
   Answer: D. Var = E(X^2) - mean^2 = 0.5 - 0.25. (Key point: Mean and variance)

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