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pdf cx²: constant, P(1<x<2), cdf

Short answer

c=1/9 from ∫₀³ cx² dx=1. P(1<X<2)=7/27≈0.2593. Distribution function: F(x)=0 for x<0, F(x)=x³/27 for 0≤x<3, F(x)=1 for x≥3.

The probability density function given by f(x) = cx², 0 < x < 3; 0, otherwise. i) Find the value constant C ii) Compute P(1 < x < 2) iii) Find the distribution function

i) Value of c

∫₀³ cx² dx = c [x³/3]₀³ = 9c = 1, so c = 1/9.

ii) P(1 < X < 2)

P = ∫₁² (x²/9) dx = [x³/27]₁² = (8 − 1)/27 = 7/27 = 0.2593

iii) Distribution function

For x < 0: F(x) = 0.

For 0 ≤ x < 3: F(x) = ∫₀ˣ (t²/9) dt = x³/27.

For x ≥ 3: F(x) = 1.

F(x) = 0,        x < 0
     = x³/27,    0 ≤ x < 3
     = 1,        x ≥ 3

Check: F(3) = 27/27 = 1 and P(1<X<2) = F(2) − F(1) = 8/27 − 1/27 = 7/27.

✓ Answer checked by Claude Sonnet 5.5

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