---
title: "pdf x/2: library reserve time"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:pdf-x-over-2-reserve-time"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-x-over-2-reserve-time
site: BCT Engineering
---

# pdf x/2: library reserve time

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2073 Shrawan, 5 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/pdf-x-over-2-reserve-time
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

Let X denote the amount of time for which a book on two hour reserve at a college library is checked out by a randomly selected student and suppose that X has density function f(x) = (1/2)x for 0 ≤ x ≤ 2, 0 otherwise. Calculate P(X ≤ 1) and P(0.5 ≤ X ≤ 1.5).

## Short answer

The cdf is F(x)=x²/4 for 0≤x≤2. P(X≤1)=F(1)=0.25, and P(0.5≤X≤1.5)=F(1.5)−F(0.5)=0.50.

## Answer

Check: ∫₀² (x/2) dx = [x²/4]₀² = 1, so f(x) is a valid pdf. The cdf for 0 ≤ x ≤ 2 is F(x) = ∫₀ˣ (t/2) dt = x²/4.

#### P(X ≤ 1)
P(X ≤ 1) = F(1) = 1²/4 = **0.25**

#### P(0.5 ≤ X ≤ 1.5)
P = F(1.5) − F(0.5) = (1.5)²/4 − (0.5)²/4 = 2.25/4 − 0.25/4 = 0.5625 − 0.0625 = **0.50**

**Answer: P(X ≤ 1) = 0.25 and P(0.5 ≤ X ≤ 1.5) = 0.50.**

## Check yourself

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

1. Why is f(x) = x/2 on [0, 2] a valid pdf?
   A. It is positive and its total area is 1
   B. Its mean is 1
   C. f(2) = 1
   D. It is linear
   Answer: A. ∫0^2 x/2 dx = 1 and f ≥ 0, which are the two pdf conditions. (Key point: Validity check)

2. The cdf for 0 ≤ x ≤ 2 is:
   A. x/2
   B. x^2/4
   C. x^2/2
   D. x^3/6
   Answer: B. F(x) = ∫0^x t/2 dt = x^2/4. (Key point: CDF derivation)

3. P(0.5 ≤ X ≤ 1.5) is found as:
   A. F(1.5) + F(0.5)
   B. F(1.5) − F(0.5) = 0.50
   C. F(1.5) × F(0.5)
   D. 1 − F(1.5)
   Answer: B. An interval probability is the difference of the cdf values: 0.5625 − 0.0625 = 0.50. (Key point: Interval probability)

## Close variants set in other exams

- [pdf kx: library reserve time, find k](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apdf-kx-reserve-book-find-k) — Probability and Statistics, chapter 3, asked 2×