A-LevelMathematicsProbability and statistics9709/61 · Oct/Nov 2022 · Q67 Marks

The diagram shows the graph of the probability density function of a random variable X that takes values between –1 and 3 only. It is given that the graph is symmetrical about the line x = 1. Between x = −1 and x = 3 the graph is a quadratic curve. The random variable S is such that E(S) = 2 × E(X) and Var(S) = Var(X).

(a)On the grid below, sketch a quadratic graph for the probability density function of S.[1]
(b)The random variable T is such that E(T) = E(X) and Var(T) = ¼ Var(X). On the grid below, sketch a quadratic graph for the probability density function of T.[2]
(c)It is now given that f(x) = { (3 + 2x - x²) for -1 ≤ x ≤ 3, and 0 otherwise. Given that P(1 − a < X < 1 + a) = 0.5, show that a³ – 12a + 8 = 0.[3]
(d)Hence verify that 0.69 < a < 0.70.[1]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

[B1] Curve of similar shape, x = 0 to x = 4, with highest point (2, 0.375)
[B1] Curve of similar shape, from x = 0 to x = 2, highest point at x = 1
[B1] Highest point (1, 0.75)
[M1] ∫ from 1-a to 1+a of 3/32 (3 + 2x - x²) dx = 0.5
[A1] [3x + x² - x³/3] with correct limits
[A1] a^3 – 12a + 8 = 0
[B1] 0.69^3 - 12×0.69 + 8 = 0.049 (2 sf) > 0; 0.70^3 – 12×0.70 + 8 = -0.057 (2 sf) < 0. Hence 0.69 < a < 0.70. OR: a = 0.695 (3 sf) which is between 0.69 & 0.70.

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