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pdf Kx(2−x): K, mean, variance

Short answer

Integrate Kx(2−x) from 0 to 2 equals 1 gives K=3/4. Mean = 1; E(X²)=6/5, so variance = 0.2.

In a continuous distribution, whose probability density function f(x) = Kx(2 − x), 0 ≤ x ≤ 2, = 0 otherwise. Find: a) The value of K b) Mean of the distribution c) Variance of the distribution

a) Value of K

∫₀² Kx(2 − x) dx = K [x² − x³/3]₀² = K (4 − 8/3) = 4K/3 = 1, so K = 3/4.

b) Mean

E(X) = (3/4) ∫₀² x·x(2 − x) dx = (3/4) ∫₀² (2x² − x³) dx = (3/4) [2x³/3 − x⁴/4]₀²

= (3/4)(16/3 − 4) = (3/4)(4/3) = 1

(The pdf is symmetric about x = 1, which confirms the mean.)

c) Variance

E(X²) = (3/4) ∫₀² (2x³ − x⁴) dx = (3/4) [x⁴/2 − x⁵/5]₀² = (3/4)(8 − 32/5) = (3/4)(8/5) = 6/5

Variance = E(X²) − [E(X)]² = 6/5 − 1 = 1/5 = 0.2

Answer: K = 3/4, mean = 1, variance = 0.2.

✓ Answer checked by Claude Sonnet 5.5

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