---
title: "Verify pdf 2e^(−2x); P(1<x<3); mean"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:exponential-pdf-2e-2x-verify"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/exponential-pdf-2e-2x-verify
site: BCT Engineering
---

# Verify pdf 2e^(−2x); P(1<x<3); mean

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2069 Chaitra, 6 marks
- Video: [Solved Problems: Probability Density function](https://www.youtube.com/watch?v=doiKcOGzb9o) — Rajveer Shastri
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/exponential-pdf-2e-2x-verify
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

If a random variable X has a function f(x) = 2e^(−2x) for x > 0; 0 for x ≤ 0. Find (i) Verify that the function is probability density function (ii) P(1 < x < 3) (iii) Find mean and variance

## Short answer

The function is non-negative and ∫₀^∞ 2e^(−2x) dx = 1, so it's a valid pdf. P(1<X<3)=e^(−2)−e^(−6)≈0.1329. It is exponential with λ=2, thus mean=1/λ=0.5, variance=1/λ²=0.25.

## Answer

f(x) = 2e^(−2x) for x > 0 is an exponential pdf with λ = 2.

#### (i) Verify that f(x) is a pdf
1. f(x) = 2e^(−2x) > 0 for all x > 0, and f(x) = 0 otherwise, so f(x) ≥ 0.
2. ∫₀^∞ 2e^(−2x) dx = [−e^(−2x)]₀^∞ = 0 − (−1) = 1.

Both conditions hold, so f(x) is a valid probability density function.

#### (ii) P(1 < X < 3)
P(1 < X < 3) = ∫₁³ 2e^(−2x) dx = [−e^(−2x)]₁³ = e^(−2) − e^(−6)

= 0.1353 − 0.0025 = **0.1329**

#### (iii) Mean and variance
For the exponential distribution with parameter λ:
- Mean = 1/λ = 1/2 = **0.5**
- Variance = 1/λ² = 1/4 = **0.25**

Check by integration: E(X) = ∫₀^∞ x·2e^(−2x) dx = 1/2; E(X²) = ∫₀^∞ x²·2e^(−2x) dx = 1/2; so Var = 1/2 − 1/4 = 1/4.

## Check yourself

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

1. What two conditions verify that f(x) = 2e^(−2x) is a pdf?
   A. f ≥ 0 and total integral equals 1
   B. f ≤ 1 and mean equals 1
   C. f increasing and integral equals 0
   D. f symmetric and integral equals 2
   Answer: A. Non-negativity and ∫ f dx = 1 are required; here ∫0^∞ 2e^(−2x) dx = 1. (Key point: pdf verification)

2. P(1 < X < 3) equals:
   A. e^(−2) − e^(−6) = 0.1329
   B. e^(−6) − e^(−2)
   C. 1 − e^(−6)
   D. e^(−2) + e^(−6)
   Answer: A. ∫1^3 2e^(−2x) dx = [−e^(−2x)]1^3 = e^(−2) − e^(−6). (Key point: Interval probability)

3. With λ = 2, the mean is:
   A. 2
   B. 1/2
   C. 1/4
   D. 4
   Answer: B. Mean = 1/λ = 0.5. (Key point: Mean of exponential)

4. With λ = 2, the variance is:
   A. 1/2
   B. 2
   C. 1/4
   D. 4
   Answer: C. Variance = 1/λ^2 = 1/4 = 0.25. (Key point: Variance of exponential)

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