---
title: "pdf cx²: constant, P(1<x<2), cdf"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:pdf-cx2-0-3"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-cx2-0-3
site: BCT Engineering
---

# pdf cx²: constant, P(1<x<2), cdf

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2070 Chaitra, 4 marks
- Video: [To find CDF when PDF of continuous random variable is given](https://www.youtube.com/watch?v=fafKWPXPNY0) — Dr. Ravinder Kaur
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-cx2-0-3
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

The probability density function given by f(x) = cx², 0 < x < 3; 0, otherwise.
i) Find the value constant C
ii) Compute P(1 < x < 2)
iii) Find the distribution function

## Short answer

c=1/9 from ∫₀³ cx² dx=1. P(1<X<2)=7/27≈0.2593. Distribution function: F(x)=0 for x<0, F(x)=x³/27 for 0≤x<3, F(x)=1 for x≥3.

## Answer

#### i) Value of c
∫₀³ cx² dx = c [x³/3]₀³ = 9c = 1, so **c = 1/9**.

#### ii) P(1 < X < 2)
P = ∫₁² (x²/9) dx = [x³/27]₁² = (8 − 1)/27 = 7/27 = **0.2593**

#### iii) Distribution function
For x < 0: F(x) = 0.

For 0 ≤ x < 3: F(x) = ∫₀ˣ (t²/9) dt = x³/27.

For x ≥ 3: F(x) = 1.

```text
F(x) = 0,        x < 0
     = x³/27,    0 ≤ x < 3
     = 1,        x ≥ 3
```
Check: F(3) = 27/27 = 1 and P(1<X<2) = F(2) − F(1) = 8/27 − 1/27 = 7/27.

## Check yourself

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

1. For f(x) = cx^2 on 0 < x < 3, c equals:
   A. 1/3
   B. 1/9
   C. 1/27
   D. 3
   Answer: B. ∫0^3 cx^2 dx = 9c = 1, so c = 1/9. (Key point: Constant c)

2. P(1 < X < 2) with f(x) = x^2/9 is:
   A. 1/9
   B. 7/27
   C. 8/27
   D. 1/3
   Answer: B. ∫1^2 x^2/9 dx = (8 − 1)/27 = 7/27 = 0.2593. (Key point: Interval probability)

3. The distribution function F(x) is:
   A. 0 for x < 0; x^3/27 for 0 ≤ x < 3; 1 for x ≥ 3
   B. x^2/9 for all x
   C. 0 for x < 0; x^3/9 for 0 ≤ x < 3; 1 beyond
   D. x^3/27 for all x
   Answer: A. F integrates f up to x, so F = x^3/27 inside the range, 0 below and 1 above (F(3) = 1). (Key point: Piecewise distribution function)

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