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pdf kx³(4−x)²: k, mean, variance

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.

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.

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.

✓ Answer checked by Claude Sonnet 5.5

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