---
title: "pdf kx³(4−x)²: k, mean, variance"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:pdf-kx3-4-minus-x-squared"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-kx3-4-minus-x-squared
site: BCT Engineering
---

# pdf kx³(4−x)²: k, mean, variance

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2071 Chaitra, 5 marks
- Video: [Continuous Probability Distribution-Problem](https://www.youtube.com/watch?v=Yvntmu0XXOs) — Btech Maths Hub
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-kx3-4-minus-x-squared
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

A random variable X has the following probability density function as:
f(x) = kx³(4 − x)², 0 < x < 1; 0, otherwise
Find the value of k, using this value of k find mean and variance of distribution.

## Short answer

Normalize: ∫_0^1 k x^3(4−x)^2 dx = 1 → k = 30/77 ≈ 0.3896. Mean = 422/539 ≈ 0.7829. Variance = 1385/2156 − (422/539)^2 ≈ 0.0294.

## Answer

#### Value of k
Expand x³(4 − x)² = 16x³ − 8x⁴ + x⁵.

∫₀¹ (16x³ − 8x⁴ + x⁵) dx = [4x⁴ − 8x⁵/5 + x⁶/6]₀¹ = 4 − 1.6 + 0.16667 = 2.56667 = 77/30

k × 77/30 = 1, so **k = 30/77 = 0.3896**.

#### Mean
E(X) = k ∫₀¹ (16x⁴ − 8x⁵ + x⁶) dx = k [16x⁵/5 − 8x⁶/6 + x⁷/7]₀¹

= k (3.2 − 1.33333 + 0.14286) = k × 2.00952 = (30/77) × 2.00952 = **0.7829** (= 422/539)

#### Variance
E(X²) = k ∫₀¹ (16x⁵ − 8x⁶ + x⁷) dx = k [16x⁶/6 − 8x⁷/7 + x⁸/8]₀¹

= k (2.66667 − 1.14286 + 0.125) = k × 1.64881 = 0.6424  (= 1385/2156)

Variance = E(X²) − [E(X)]² = 0.6424 − (0.7829)² = **0.0294**

**Answer: k = 30/77 ≈ 0.3896, mean ≈ 0.7829, variance ≈ 0.0294.**

## Check yourself

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

1. x^3(4 − x)^2 expands to:
   A. 16x^3 − 8x^4 + x^5
   B. 16x^3 − 4x^4 + x^5
   C. 4x^3 − x^5
   D. 16x^3 + 8x^4 + x^5
   Answer: A. (4 − x)^2 = 16 − 8x + x^2; multiplying by x^3 gives 16x^3 − 8x^4 + x^5. (Key point: Expansion step)

2. Since the integral over 0 to 1 is 77/30, k equals:
   A. 77/30
   B. 30/77
   C. 1
   D. 30/7
   Answer: B. k times the integral must be 1, so k = 30/77 ≈ 0.3896. (Key point: Value of k)

3. How is the variance obtained once E(X) and E(X^2) are known?
   A. E(X^2) − [E(X)]^2
   B. E(X) − E(X^2)
   C. [E(X^2)]^2 − E(X)
   D. E(X^2) + [E(X)]^2
   Answer: A. Var = 0.6424 − 0.7829^2 ≈ 0.0294. (Key point: Variance from moments)

4. For the mean, which integrand is used after expansion?
   A. 16x^3 − 8x^4 + x^5
   B. 16x^4 − 8x^5 + x^6
   C. 16x^5 − 8x^6 + x^7
   D. x^3(4 − x)
   Answer: B. E(X) = k ∫ x · f-part dx, multiplying by x gives 16x^4 − 8x^5 + x^6. (Key point: Mean integrand)

## Close variants set in other exams

- [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×

## Related in Probability and Statistics

- [pdf Kx(2−x): K, mean, variance](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apdf-kx-2-minus-x) — Probability and Statistics, chapter 3
- [Continuous random variable; pdf properties](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Acontinuous-rv-pdf-properties) — Probability and Statistics, chapter 3
- [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 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

- [Moments of random variable; NEB of RC-LPF; output noise power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amoments-neb-rc-lpf) — Communication Systems, chapter 9
- [Linked list has no random access](https://www.bctengineer.com/question.md?key=semester_4%2Fdata_structure_and_algorithm%3Alinked-list-no-random-access) — Data Structure and Algorithm, chapter 4
- [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