---
title: "pdf kx: library reserve time, find k"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:pdf-kx-reserve-book-find-k"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-kx-reserve-book-find-k
site: BCT Engineering
---

# pdf kx: library reserve time, find k

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2070 Asar, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-kx-reserve-book-find-k
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

Let X denotes the amount of time for which a book on two-hour reserve at a college library is checked out by a randomly selected students, and suppose that X has density function f(x) = kx, 0 ≤ x ≤ 2, 0 otherwise.
a) Find the value of k
b) Calculate P(X ≤ 1)
c) Calculate P(0.5 ≤ X ≤ 1.5)
d) Calculate P(1.5 < X)

## Short answer

Integrating kx from 0 to 2 gives 2k=1 so k=0.5. Then P(X≤1)=0.25, P(0.5≤X≤1.5)=0.50, and P(X>1.5)=0.4375.

## Answer

#### a) Value of k
∫₀² kx dx = k [x²/2]₀² = 2k = 1, so **k = 1/2**. Hence f(x) = x/2 and the cdf is F(x) = x²/4 for 0 ≤ x ≤ 2.

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

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

#### d) P(X > 1.5)
1 − F(1.5) = 1 − 0.5625 = **0.4375**

**Answer: k = 0.5; 0.25; 0.50; 0.4375.**

## Check yourself

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

1. For f(x) = kx on [0, 2], which equation gives k?
   A. k × 2 = 1
   B. k × 4 = 1
   C. k × 2 = 2
   D. k/2 = 1
   Answer: A. ∫0^2 kx dx = k × 2 = 1, so k = 1/2. (Key point: Finding k)

2. P(X > 1.5) is computed as:
   A. F(1.5) = 0.5625
   B. 1 − F(1.5) = 0.4375
   C. F(1.5) − F(1) = 0.3125
   D. 1 − F(0.5)
   Answer: B. The upper tail is 1 − F(1.5) = 1 − 0.5625 = 0.4375. (Key point: Complement probability)

3. P(X ≤ 1) with F(x) = x^2/4 equals:
   A. 0.5
   B. 0.25
   C. 0.75
   D. 1
   Answer: B. F(1) = 1/4 = 0.25. (Key point: Probability P(X ≤ 1))

## Close variants set in other exams

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

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