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A-LevelMathematicsProbability and statistics9709/62 · May/June 2024 · Q710 Marks

The probability density function, f, of a random variable X is given by [k(1+cosx) f(x) = 0 0 ≤ x ≤ π, otherwise, where k is a constant.

  1. (a)Show that k = 1/π.[3]
  2. (b)Verify that the median of X lies between 0.83 and 0.84 .[3]
  3. (c)Find the exact value of E(X).[4]

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

  1. [M1] ∫₀^π k(1+cosx)dx = 1 Attempt integrate f(x) with correct limits and equate to 1.
  2. [A1] k[x+sinx]₀^π = 1 Correct integration.
  3. [A1] [e.g. k(π + sin π - (0 + 0)) = 1], kπ = 1, k = 1/π AG Some evidence of substitution of limits, i.e. at least one interim step (e.g. kπ = 1) as minimum requirement. Convincingly obtained; no errors seen.
  4. [M1] (1/π) * [x+sinx]₀^0.83 or (1/π) * (0.83 + sin 0.83) (and for 0.84) Substitute correct limits into their integral. OR1: integrate 0 to 0.83 and 0.84 to π. OR2: use g(m) = m + sin m - (π/2) and find g(0.83) and g(0.84). OR3: use h(m) = m + sin m and find h(0.83) and h(0.84). Both attempted.
  5. [A1] = 0.499 (3sf) (and 0.504 (3sf)) OR1: 0.499 and 0.496. OR2: g(0.83) = -0.00286/7 and g(0.84) = 0.0138/9. OR3: h(0.83) = 1.57 and h(0.84) = 1.58 or 1.59. Both correct.
  6. [A1FT] '0.499' < 0.5 < '0.504' hence 0.83 < median < 0.84 Equivalent to -0.000912 < 0 < 0.00441 hence 0.83 < median < 0.84 FT their areas; dep 0.5 is between their areas OE. OR1: 0.499 < 0.5 and 0.496 < 0.5, so 0.83 < m < 0.84. OR2: g(0.83) > 0 g(0.84) < 0 OE, so 0.83 < m < 0.84. OR3: h(0.83) < π/2 h(0.84) > π/2, so 0.83 < m < 0.84. Both statements needed. Note: A score of M1 A0 A1FT is possible. If 0 scored, SC: (1/π) * (m + sin m) = 0.5 B1 and m = 0.831 to 0.832, so 0.83 < m < 0.84 B1.
  7. [M1*] ∫₀^π xf(x)dx Attempt integrate xf(x). Ignore limits.
  8. [DM1] (1/π) * [x²/2 + xsinx - ∫sinx dx] from 0 to π Attempt to integrate (using 'parts') with correct limits, reaching an expression of the form ax² + uv – ∫v du. OR using parts to integrate x(1+cosx) reaching an expression of the form uv – ∫v du i.e. (1/π) * (x²/2 + xsinx - ∫(x + sinx) dx))
  9. [A1] (1/π) * [x²/2 + xsinx + cosx]₀^π Integration fully correct.
  10. [A1] = (π/2) - (2/π) ISW after correct exact value seen. SC1: Unsupported answer of (π/2) - (2/π) scores B3. SC2: Unsupported answer of 0.934 scores B2.

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About this question

Topic
Probability and statistics, A-Level Mathematics (syllabus 9709). Structured question, 10 marks.
Source
A-Level Mathematics, May/June 2024 Paper 6 Variant 2, question 7.

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