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pdf k(5+2x): find k, mean, variance

Short answer

k=1/22 from ∫₂⁴ k(5+2x)dx=22k=1. Mean=∫₂⁴ x(5+2x)dx/22=101/33≈3.0606; E(X²)=320/33≈9.6970, variance=9.6970−(101/33)²≈0.3297.

If we the following probability density function f(x) = k(5 + 2x) for 2 ≤ x ≤ 4, 0 otherwise. Find the value of K and mean and variance of random variable X.

f(x) = k(5 + 2x), 2 ≤ x ≤ 4.

Value of k

∫₂⁴ k(5 + 2x) dx = k [5x + x²]₂⁴ = k [(20 + 16) − (10 + 4)] = 22k = 1

k = 1/22

Mean

E(X) = (1/22) ∫₂⁴ (5x + 2x²) dx = (1/22) [5x²/2 + 2x³/3]₂⁴

At x = 4: 5(16)/2 + 2(64)/3 = 40 + 42.667 = 82.667
At x = 2: 5(4)/2  + 2(8)/3  = 10 + 5.333  = 15.333
Difference = 67.333

E(X) = 67.333/22 = 101/33 = 3.0606

Variance

E(X²) = (1/22) ∫₂⁴ (5x² + 2x³) dx = (1/22) [5x³/3 + x⁴/2]₂⁴

At x = 4: 5(64)/3 + 256/2 = 106.667 + 128 = 234.667
At x = 2: 5(8)/3  + 16/2  = 13.333  + 8   = 21.333
Difference = 213.333

E(X²) = 213.333/22 = 320/33 = 9.6970

Var(X) = 9.6970 − (3.0606)² = 9.6970 − 9.3674 = 0.3297

Answer: k = 1/22, mean = 3.0606, variance = 0.3297

✓ Answer checked by Claude Sonnet 5.5

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