---
title: "pdf Kx(2−x): K, mean, variance"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:pdf-kx-2-minus-x"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-kx-2-minus-x
site: BCT Engineering
---

# pdf Kx(2−x): K, mean, variance

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2068 Chaitra, 2+2+2 marks
- Video: [PQT RANDOM VARIABLE |Find K,Mean,Variance,rth moment for f(x)=Kx(2-x) #WelcomeEngineers #Probability](https://www.youtube.com/watch?v=tmZK7Mv9GJE) — WELCOME ENGINEERS
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-kx-2-minus-x
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

In a continuous distribution, whose probability density function f(x) = Kx(2 − x), 0 ≤ x ≤ 2, = 0 otherwise. Find:
a) The value of K
b) Mean of the distribution
c) Variance of the distribution

## Short answer

Integrate Kx(2−x) from 0 to 2 equals 1 gives K=3/4. Mean = 1; E(X²)=6/5, so variance = 0.2.

## Answer

#### a) Value of K
∫₀² Kx(2 − x) dx = K [x² − x³/3]₀² = K (4 − 8/3) = 4K/3 = 1, so **K = 3/4**.

#### b) Mean
E(X) = (3/4) ∫₀² x·x(2 − x) dx = (3/4) ∫₀² (2x² − x³) dx = (3/4) [2x³/3 − x⁴/4]₀²

= (3/4)(16/3 − 4) = (3/4)(4/3) = **1**

(The pdf is symmetric about x = 1, which confirms the mean.)

#### c) Variance
E(X²) = (3/4) ∫₀² (2x³ − x⁴) dx = (3/4) [x⁴/2 − x⁵/5]₀² = (3/4)(8 − 32/5) = (3/4)(8/5) = 6/5

Variance = E(X²) − [E(X)]² = 6/5 − 1 = **1/5 = 0.2**

**Answer: K = 3/4, mean = 1, variance = 0.2.**

## Check yourself

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

1. ∫0^2 x(2 − x) dx equals 4/3, so K is:
   A. 4/3
   B. 3/4
   C. 3/2
   D. 1/2
   Answer: B. K × 4/3 = 1 gives K = 3/4. (Key point: Finding K)

2. Why is the mean exactly 1?
   A. The pdf is symmetric about x = 1
   B. K equals 3/4
   C. The variance is 1
   D. The range ends at 2
   Answer: A. x(2 − x) is symmetric about x = 1, and integration gives E(X) = 1 as well. (Key point: Mean by symmetry)

3. With E(X^2) = 6/5 and E(X) = 1, the variance is:
   A. 6/5
   B. 1/5
   C. 11/5
   D. 1
   Answer: B. Var = E(X^2) − [E(X)]^2 = 6/5 − 1 = 0.2. (Key point: Variance)

## Related in Probability and Statistics

- [pdf kx³(4−x)²: k, mean, variance](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apdf-kx3-4-minus-x-squared) — Probability and Statistics, chapter 3, asked 2×
- [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×
- [pdf definition; f(x)=Kx(1−x) probabilities](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apdf-kx-1-minus-x-probabilities) — Probability and Statistics, chapter 3
- [pdf k(5+2x): find k, mean, variance](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apdf-k-5-plus-2x-mean) — Probability and Statistics, chapter 3, asked 2×
- [pdf Ax(6−x)²: find A, mean, variance](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apdf-ax-6-minus-x-squared) — Probability and Statistics, chapter 3, asked 2×
- [Mean and variance of pdf 12x²(1−x)](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apdf-12x2-1-minus-x-mean) — Probability and Statistics, chapter 3, asked 2×

## Related in other subjects

- [Fermi-Dirac function and 1.5kT probability](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Afermi-dirac-1-5kt) — Electrical Engineering Material, chapter 1
- [Collapse value of P: continuous beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-p-continuous-beam-p-2p) — Theory of Structures II, chapter 5