---
title: "Continuous variable with density 6x(1−x)"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:pdf-6x-1-minus-x"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-6x-1-minus-x
site: BCT Engineering
---

# Continuous variable with density 6x(1−x)

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2068 Jestha (old course), 6 marks
- Video: [pdf is f(x)=6x(1-x),determine  b# Mathematical statistics -1 # SM3AC # Nov 2022 # Tamil](https://www.youtube.com/watch?v=-A1KYjVzVM4) — கந்தழி infinity
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/pdf-6x-1-minus-x
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

A continuous random variable X has the density function f(x) = 6x(1 − x), 0 < x < 1; 0 elsewhere. [The task asked of this density is not visible in the scan.]

## Short answer

f(x)=6x(1−x) for 0<x<1 is a valid pdf (integral=1). Mean = 0.5, variance = 0.05. For example, P(0.25<X<0.75)=0.6875.

## Answer

*The task asked of this density is not legible in the question. The standard parts of such a question (check that it is a pdf, find the mean and variance, find a probability) are answered below.*

f(x) = 6x(1 − x) = 6x − 6x², 0 < x < 1

#### Check that f(x) is a pdf
- f(x) ≥ 0 for 0 < x < 1.
- ∫₀¹ (6x − 6x²) dx = [3x² − 2x³]₀¹ = 3 − 2 = 1.

#### Distribution function
F(x) = ∫₀ˣ (6t − 6t²) dt = 3x² − 2x³, 0 < x < 1

#### Mean
E(X) = ∫₀¹ x(6x − 6x²) dx = ∫₀¹ (6x² − 6x³) dx = [2x³ − 1.5x⁴]₀¹ = 0.5

#### Variance
E(X²) = ∫₀¹ (6x³ − 6x⁴) dx = [1.5x⁴ − 1.2x⁵]₀¹ = 0.3

Variance = 0.3 − (0.5)² = 0.05, standard deviation = 0.2236

#### Sample probabilities
- P(X < 0.5) = F(0.5) = 3(0.25) − 2(0.125) = 0.5
- P(0.25 < X < 0.75) = F(0.75) − F(0.25) = 0.84375 − 0.15625 = 0.6875

**Answer: valid pdf; mean = 0.5, variance = 0.05; P(0.25 < X < 0.75) = 0.6875.**

## Check yourself

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

1. Why is f(x) = 6x(1 − x) on (0, 1) a valid pdf?
   A. It is non-negative and integrates to 1
   B. It is symmetric
   C. Its maximum is 1
   D. Its mean is 1
   Answer: A. ∫0^1 (6x − 6x^2) dx = 3 − 2 = 1, with f ≥ 0 on the range. (Key point: pdf check)

2. The cdf for 0 < x < 1 is:
   A. 3x^2 − 2x^3
   B. 6x^2 − 6x^3
   C. 3x − 2x^2
   D. x^2 − x^3
   Answer: A. ∫0^x (6t − 6t^2) dt = 3x^2 − 2x^3. (Key point: Distribution function)

3. With E(X) = 0.5 and E(X^2) = 0.3, the variance is:
   A. 0.05
   B. 0.25
   C. 0.3
   D. 0.55
   Answer: A. 0.3 − 0.25 = 0.05. (Key point: Mean and variance)

4. P(0.25 < X < 0.75) is found as:
   A. F(0.75) − F(0.25) = 0.6875
   B. F(0.75) + F(0.25)
   C. 1 − F(0.75)
   D. F(0.5)
   Answer: A. 0.84375 − 0.15625 = 0.6875. (Key point: Probability from cdf)

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