A-LevelMathematicsProbability and statistics9709/62 · Oct/Nov 2025 · Q59 Marks
A random variable X has probability density function given by f(x) = { k(2x²-x³) 0 ≤ x ≤ 2, { 0 otherwise.
- (a)Show that k = 3/4.[3]
- (b)(i)Write down the value of P(X ≤ m).[1]
- (b)(ii)Hence find P(E(X) ≤ X ≤ m).[5]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.
- [1] (a) ∫k(2x²-x³) dx = 1 (limits 0 to 2)
- [1] (a) k[2x³/3 - x⁴/4] (limits 0 to 2) = 1
- [1] (a) k × 4/3 = 1 k = 3/4
- [1] (b)(i) 0.5
- [1] (b)(ii) E(X) = ∫ (3/4)x(2x² - x³) dx (limits 0 to 2)
- [1] (b)(ii) = (3/4)[2x⁴/4 - x⁵/5] (limits 0 to 2) = 8/5 or 1.6
- [1] (b)(ii) P(X < E(X)) = ∫ (3/4)(2x² - x³) dx (limits 0 to 1.6)
- [1] (b)(ii) = (3/4)[2x³/3 - x⁴/4] (limits 0 to 1.6) = 594/1250 or 297/625 or 0.4752
- [1] (b)(ii) P(E(X) ≤ X ≤ m) = 0.5 - ‘0.4752’
- [1] (b)(ii) = 0.0248 or 31/1250
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →