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pdf kx²: lecture ending after bell

Short answer

Normalization gives k=3. P(X ≤ 0.5) = (0.5)^3 = 0.125. For 15–30 s (0.25–0.5 min), P = (0.5)^3 − (0.25)^3 = 0.109375 (7/64).

A college professor never finishes his lecture before the bell rings to end the period and always finishes his lectures within one minute after the bell rings. Let X = the time which elapses between the bell and the end of the lecture. Suppose that the p.d.f of X is f(x) = kx², 0 ≤ x ≤ 1 = 0, otherwise i) Find the value of k ii) What is the probability that the lecture ends within ½ minute of the bell ringing? iii) What is the probability that the lecture continues beyond the bell for between 15 and 30 seconds?

For a valid pdf, the total area under f(x) must be 1.

i) Value of k

∫₀¹ kx² dx = k [x³/3]₀¹ = k/3 = 1, so k = 3.

So f(x) = 3x², 0 ≤ x ≤ 1.

ii) Lecture ends within ½ minute

P(X ≤ ½) = ∫₀^½ 3x² dx = [x³]₀^½ = (½)³ = 1/8 = 0.125

iii) Lecture continues between 15 and 30 seconds

15 s = 0.25 minute and 30 s = 0.5 minute (X is in minutes).

P(0.25 ≤ X ≤ 0.5) = [x³]₀.₂₅^₀.₅ = (0.5)³ − (0.25)³ = 0.125 − 0.015625 = 0.109375 (= 7/64)

✓ Answer checked by Claude Sonnet 5.5

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