--- title: "pdf Ax(6−x)²: find A, mean, variance" subject: "Probability and Statistics" chapter: 3 key: "semester_5/probability_and_statistics:pdf-ax-6-minus-x-squared" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-ax-6-minus-x-squared site: BCT Engineering --- # pdf Ax(6−x)²: find A, mean, variance Probability and Statistics · Chapter 3: Continuous Probability Distributions - Asked in: 2071 Shrawan, 5 marks - Video: [Finding the mean and variance of a pdf](https://www.youtube.com/watch?v=cY94ubn25SI) — Mr Camilleri - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-ax-6-minus-x-squared - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3 ## Question A probability density function is given by f(x) = Ax(6 − x)² for 0 < x < 6 i) Find the value of A ii) Find the mean and variance of this distribution ## Short answer Integrate f(x)=A x(6−x)^2 over 0–6, giving A=1/108. Mean μ = E(X) = 2.4. Variance = E(X^2)−μ^2 = 7.2−5.76 = 1.44. ## Answer #### i) Value of A Expand x(6 − x)² = 36x − 12x² + x³. ∫₀⁶ (36x − 12x² + x³) dx = [18x² − 4x³ + x⁴/4]₀⁶ = 648 − 864 + 324 = 108 A × 108 = 1, so **A = 1/108**. #### ii) Mean and variance Mean: E(X) = (1/108) ∫₀⁶ x·(36x − 12x² + x³) dx = (1/108) ∫₀⁶ (36x² − 12x³ + x⁴) dx ```text = (1/108) [12x³ − 3x⁴ + x⁵/5] from 0 to 6 = (1/108) [2592 − 3888 + 1555.2] = 259.2 / 108 = 2.4 ``` E(X²) = (1/108) ∫₀⁶ (36x³ − 12x⁴ + x⁵) dx ```text = (1/108) [9x⁴ − 2.4x⁵ + x⁶/6] from 0 to 6 = (1/108) [11664 − 18662.4 + 7776] = 777.6 / 108 = 7.2 ``` Variance = E(X²) − [E(X)]² = 7.2 − (2.4)² = 7.2 − 5.76 = 1.44 **Answer: A = 1/108, mean = 2.4, variance = 1.44 (standard deviation 1.2).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is ∫0^6 x(6 − x)^2 dx and hence A? A. 108, so A = 1/108 B. 36, so A = 1/36 C. 648, so A = 1/648 D. 72, so A = 1/72 Answer: A. Expanding gives 36x − 12x^2 + x^3, which integrates to 108 over 0 to 6, so A = 1/108. (Key point: Finding A) 2. The mean of this distribution is computed as: A. ∫ f(x) dx B. ∫ x f(x) dx C. ∫ x^2 f(x) dx D. ∫ f(x)^2 dx Answer: B. E(X) = ∫ x f(x) dx, which here equals 2.4. (Key point: Mean formula) 3. Given E(X^2) = 7.2 and E(X) = 2.4, the variance is: A. 7.2 − 2.4 = 4.8 B. 7.2 − 5.76 = 1.44 C. 7.2 + 5.76 D. sqrt(7.2) − 2.4 Answer: B. Var = E(X^2) − [E(X)]^2 = 7.2 − 5.76 = 1.44. (Key point: Variance formula) ## Close variants set in other exams - [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 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(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 - [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× - [Joint pdf a(x+y): marginals, probability](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Ajoint-pdf-a-x-plus-y) — Probability and Statistics, chapter 3 - [pdf cx²: constant, P(1