A-LevelMathematicsProbability and statistics9709/62 · Oct/Nov 2022 · Q78 Marks

[Figure 7.1] The diagram shows the graph of the probability density function, f, of a random variable X which takes values between –3 and 2 only.

(a)Given that the graph is symmetrical about the line x = −0.5 and that P(X < 0) =p, find P(−1 <X<0) in terms of p.[2]
(b)(i)It is now given that the probability density function shown in the diagram is given by f(x) = { a-b(x² + x) -3 ≤ x ≤ 2, { 0 otherwise, where a and b are positive constants. Show that 30a – 55b = 6.[3]
(b)(ii)By substituting a suitable value of x into f(x), find another equation relating a and b and hence determine the values of a and b.[3]

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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.

[M1] 7(a) 1-p or p-0.5 SOI, e.g. on diagram.
[A1] 7(a) [P(−1 <X<0) =] 2p – 1 Clearly as final answer.
[M1] 7(b)(i) ∫(-3 to 2) (a−b(x² +x))dx = 1 OE Attempt integral, with correct limits and RHS.
[A1] 7(b)(i) [ax - b(x³/3 + x²/2)] with limits -3 to 2 OE Correct integration.
[A1] 7(b)(i) 2a – 8b/3 – 2b + 3a – 9b + 9b/2 = 1 or 2a – 4b – 8b/3 – 9a/2 + 81b/4 – 9b = -0.5 leading to 30a – 55b = 6_AG Correctly obtained. No errors seen.
[*M1] 7(b)(ii) a – b(9 – 3) = 0 or a – b(4 + 2) = 0 [hence a = 6b = 0] Use f(-3) = 0 or f(2) = 0. Further attempts at integration M0.
[DM1] 7(b)(ii) Attempt to solve 30a – 55b = 6 and their a = 6b = 0
[A1] 7(b)(ii) a = 36/125 or 0.288 b = 6/125 or 0.048

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