A random variable X has probability density function f, where the graph of y = f(x) is a semicircle with centre (0, 0) and radius √(2/π), entirely above the x-axis. Elsewhere f(x) = 0 (see diagram).

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [1] (a) Calculate the area of the semicircle using (1/2)πr², where r = √(2/π). Area = (1/2)π(√(2/π))² = 1.
2. [1] (a) State that the area is 1 AND f(x) ≥ 0, which are the conditions for a PDF.
3. [1] (b) Show that angle AOB is (1/4)π radians by using cosine: cos(AOB) = x / r = (1/√(π)) / √(2/π) = 1/√(2). (AG - convincingly obtained).
4. [1] (b) Calculate Area of sector AOB = (1/2)r²θ = (1/2)(2/π)(π/4) = 1/4.
5. [1] (b) Calculate Area of triangle AOB = (1/2) * base * height, where base = 1/√(π) and height = y_B = √(r²-x²) = √(2/π - 1/π) = 1/√(π). Or (1/2) * r * r * sin(AOB).
6. [1] (b) Area of triangle OAB = 1/(2π) or 0.1592.
7. [1] (b) Calculate P(X > 1/√(π)) by subtracting the triangle area from the sector area (Area of sector AOB - Area of triangle AOB).
8. [1] (b) P(X > 1/√(π)) = 1/4 - 1/(2π) ≈ 0.0908 (3 s.f.).