Part a (2 marks)

Award one mark for each observable outcome: substitutes C(2)=12 to obtain 2Ae⁻¹=12; obtains A=6e. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to A=6e. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Substituting t=2 gives 2Ae⁻¹=12, so A=6e.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part b (2 marks)

Award one mark for each observable outcome: obtains C'(t)=6e,e^(-t / 2)(1-t / 2); uses the derivative sign to establish a maximum of 12 milligrams per litre at t=2. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to The maximum concentration is 12 milligrams per litre at t=2. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Differentiation gives C'(t)=6e,e^(-t / 2)(1-t / 2). This changes from positive to negative at t=2.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part c (3 marks)

Award one mark for each observable outcome: uses an antiderivative equivalent to -12e(t+2)e^(-t / 2); evaluates the antiderivative at t=0 and t=8; obtains 24e-120e⁻³. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to int_0⁸ C(t),dt=24e-120e⁻³. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Using int te^(-t / 2),dt=-2(t+2)e^(-t / 2), evaluate 6e[-2(t+2)e^(-t / 2)]_0^8.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part d (1 mark)

Award the mark for the correct response 3e-15e⁻³ milligrams per litre. Working is not required because the prompt does not ask for proof, verification, explanation or justification.

Acceptable alternatives: Accept a mathematically equivalent response to 3e-15e⁻³ milligrams per litre. that preserves every stated condition.

Part e (2 marks)

Award one mark for each observable outcome: forms 6et e^(-t / 2)=6 and restricts the solution to t>2; obtains the boundary time tapprox5.36 hours. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to The concentration falls below 6 after tapprox5.36 hours. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Solve 6et e^(-t / 2)=6, restricting the numerical solution to t>2. The required boundary is t=5.35669ldots.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part a (3 marks)

Award one mark for each observable outcome: uses a+2b=3a to obtain b=a; applies int_0²(a+bx),dx=1; obtains a=b=frac14. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to a=b=frac14. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: The condition a+2b=3a gives b=a. Normalisation then gives int_0^2a(1+x),dx=4a=1.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part b (2 marks)

Award one mark for each observable outcome: uses F(x)=int_0^xfrac14(1+t),dt; obtains F(x)=((x) / (4))+((x²) / (8)) for 0leq xleq2. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to F(x)=((x) / (4))+((x²) / (8)). that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Integrate f(t)=frac14(1+t) from 0 to x.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part c (2 marks)

Award one mark for each observable outcome: sets F(m)=frac12 to obtain m²+2m-4=0; selects the valid exact solution m=sqrt5-1. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to m=sqrt5-1. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Set F(m)=1 / 2. The equation m²+2m-4=0 has the valid solution m=sqrt5-1.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part d (2 marks)

Award one mark for each observable outcome: uses Pr(X>frac32mid X>1)=((1-F(3 / 2)) / (1-F(1))); obtains ((11) / (20)). Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to ((11) / (20)). that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Using the cumulative distribution from part (b), F(1)=3 / 8 and F(3 / 2)=21 / 32. The conditional probability is (1-21 / 32) / (1-3 / 8)=11 / 20.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part e (1 mark)

Award the mark for the correct response Since F(1)=frac38<frac12, the median must be greater than 1. Working is not required because the prompt does not ask for proof, verification, explanation or justification.

Acceptable alternatives: Accept a mathematically equivalent response to Since F(1)=frac38<frac12, the median must be greater than 1. that preserves every stated condition.

Part a (2 marks)

Award one mark for each observable outcome: states the maximum temperature 24 at the first time t=4; states the minimum temperature 12 at the first time t=16. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to Maximum 24 at t=4; minimum 12 at t=16. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: The cosine term equals 1 at t=4 and -1 at t=16.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part b (2 marks)

Award one mark for each observable outcome: applies the chain rule with inner derivative ((pi) / (12)); obtains T'(t)=-((pi) / (2))sin(((pi(t-4)) / (12))). Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to T'(t)=-((pi) / (2))sin(((pi(t-4)) / (12))). that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Differentiate using the chain rule.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part c (2 marks)

Award one mark for each observable outcome: substitutes t=10 into the derivative from part (b); obtains T'(10)=-((pi) / (2)) degrees per hour. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to T'(10)=-((pi) / (2)) degrees per hour. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Substituting t=10 into the derivative from part (b) gives a sine argument of pi / 2.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part d (2 marks)

Award one mark for each observable outcome: solves cos(((pi(t-4)) / (12)))=frac12; states all solutions t=0,8,24 in the required interval. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to t=0, 8, 24. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Solve cos(pi(t-4) / 12)=1 / 2 over the stated interval.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part e (2 marks)

Award one mark for each observable outcome: recognises that the cosine contribution integrates to zero over one complete cycle; obtains the average temperature 18 degrees. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to 18 degrees. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: The cosine term completes one full cycle over 24 hours, so its average contribution is zero.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part a (1 mark)

Award the mark for the correct response hat p=((139) / (250))=0.556. Working is not required because the prompt does not ask for proof, verification, explanation or justification.

Acceptable alternatives: Accept a mathematically equivalent response to hat p=((139) / (250))=0.556. that preserves every stated condition.

Part b (2 marks)

Award one mark for each observable outcome: states E(hat p)=0.60; obtains operatorname(sd)(hat p)=√(((0.60(0.40)) / (250)))approx0.0310. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to E(hat p)=0.60 and operatorname(sd)(hat p)=√(((0.60(0.40)) / (250)))approx0.0310. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: For a sample proportion, E(hat p)=p and operatorname(sd)(hat p)=√(p(1-p) / n).

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part c (1 mark)

Award the mark when the response calculates the expected counts 150 and 100 and explains that both are sufficiently large for the normal approximation.

Acceptable alternatives: Accept a mathematically equivalent response to The expected counts are 250(0.60)=150 and 250(0.40)=100, and both are sufficiently large. that preserves every stated condition.

Do not credit by itself: Do not award the mark for an unsupported conclusion when the prompt explicitly requires reasoning.

Part d (2 marks)

Award one mark for each observable outcome: standardises to obtain zapprox-1.42; obtains Pr(hat pleq0.556)approx0.078, allowing consequential use of the standard deviation from part (b). Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to Pr(hat pleq0.556)approxPr(Zleq-1.42)approx0.078. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Standardising gives z=(0.556-0.60) / 0.030983ldotsapprox-1.42.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part e (3 marks)

Award one mark for each observable outcome: uses the sample-based standard error √(((0.556(0.444)) / (250))); obtains the margin 1.96(0.03142ldots)approx0.0616; states the approximate interval (0.494, 0.618). Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to 0.556pm1.96sqrt(((0.556(0.444)) / (250)))approx(0.494, 0.618). that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: The sample-based standard error is approximately 0.03142, so the margin is approximately 0.06159.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part f (1 mark)

Award the mark for the correct response The claim p=0.60 is consistent with the sample at the 5% level because 0.60 lies inside the approximate 95% confidence interval. Working is not required because the prompt does not ask for proof, verification, explanation or justification.

Acceptable alternatives: Accept a mathematically equivalent response to The claim p=0.60 is consistent with the sample at the 5% level because 0.60 lies inside the approximate 95% confidence interval. that preserves every stated condition.

Part a (2 marks)

Award one mark for each observable outcome: identifies the vertex time t=2; obtains the minimum flow rate 2 litres per minute. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to The minimum is 2 litres per minute at t=2. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: The quadratic has its vertex at t=2, where R(2)=2.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part b (3 marks)

Award one mark for each observable outcome: uses int_0⁴(2t²-8t+10),dt; evaluates a correct antiderivative at both limits; obtains the total volume ((56) / (3)) litres. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to ((56) / (3)) litres. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Evaluate int_0⁴(2t²-8t+10),dt=56 / 3.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part c (2 marks)

Award one mark for each observable outcome: uses the result from part (b) divided by the four-minute interval; obtains the average flow rate ((14) / (3)) litres per minute. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to ((14) / (3)) litres per minute. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Divide the total volume from part (b), 56 / 3, by the interval length 4.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part d (3 marks)

Award one mark for each observable outcome: forms 2t²-8t+10=4; factorises to an equivalent of (t-1)(t-3)=0; obtains both times t=1 and t=3. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to t=1 and t=3. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Solve 2t²-8t+10=4, which reduces to (t-1)(t-3)=0.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part a (1 mark)

Award the mark for the correct response a=frac1(10). Working is not required because the prompt does not ask for proof, verification, explanation or justification.

Acceptable alternatives: Accept a mathematically equivalent response to a=frac1(10). that preserves every stated condition.

Part b (2 marks)

Award one mark for each observable outcome: forms Pr(Ygeq2mid Y>0)=((4a+2a) / (3a+4a+2a)); obtains frac23. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to frac23. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: The required conditional probability is (4a+2a) / (3a+4a+2a)=2 / 3.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part c (2 marks)

Award one mark for each observable outcome: uses E(Y)=0(a)+1(3a)+2(4a)+3(2a); obtains E(Y)=((17) / (10)). Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to E(Y)=((17) / (10)). that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Compute 0(a)+1(3a)+2(4a)+3(2a)=17a.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part d (2 marks)

Award one mark for each observable outcome: obtains E(Y²)=((37) / (10)); uses E(C)=E(Y²)+2E(Y) to obtain ((71) / (10)) dollars. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to E(C)=((71) / (10)) dollars. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: Since C=Y²+2Y, first obtain E(Y²)=37 / 10, then use E(C)=E(Y²)+2E(Y)=71 / 10.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part e (2 marks)

Award one mark for each observable outcome: calculates expected totals of 3550 dollars for the variable arrangement and 3500 dollars for the fixed arrangement; concludes that the fixed arrangement is lower by 50 dollars. Apply consequential marking only when a clearly identified earlier result is used consistently.

Acceptable alternatives: Accept a mathematically equivalent response to The fixed arrangement has the lower expected total by 50 dollars. that preserves every stated condition.; Accept an alternative valid method that establishes the same observable steps described by: The variable arrangement has expected cost 500(7.10)=3550 dollars, compared with 500(7)=3500 dollars.

Do not credit by itself: Do not award a method mark unless its corresponding observable setup or intermediate step is present.

Part f (1 mark)

Award the mark for the correct response Yes. Since E(C)=7.1, C>E(C) when Ygeq2, which has probability 0.6>0.5. Working is not required because the prompt does not ask for proof, verification, explanation or justification.

Acceptable alternatives: Accept a mathematically equivalent response to Yes. Since E(C)=7.1, C>E(C) when Ygeq2, which has probability 0.6>0.5. that preserves every stated condition.