Continuous variable with density 6x(1−x)

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.