Skill Align ACT BSSS Specialist Methods T Booklet 2 Practice Assessment Pack 0 - 2026 Edition
Original Skill Align Specialist Methods T practice-assessment Pack 0, Booklet 2; not an official ACT BSSS examination and not endorsed by ACT BSSS or the ACT Government.
- Paper
- Booklet 2 Showcase
- Reading
- 10 minutes
- Writing
- 65 minutes
- Assessment
- 50 marks
Suggested Skill Align Booklet 2 conditions: 10 minutes reading time and 65 minutes working time. An approved graphics calculator or CAS may be used. Offline stored notes and programs and locally supplied reference material may be used only when authorised by the administering college. Internet access and external communication are prohibited. ACT BSSS assessment is school based. The administering college may vary timing, technology, stored-note, program and reference-material conditions. No public PDF download is supplied with Pack 0.
Booklet 2
Booklet 2 assesses technology-supported probability, distributions, inference, regression and non-routine mathematical modelling. Answer all questions. Show complete working, reasoning and interpretation. State exact values unless an approximation is requested, and include units and context where applicable. Suggested Skill Align Booklet 2 conditions: 10 minutes reading time and 65 minutes working time. An approved graphics calculator or CAS may be used. Offline stored notes and programs and locally supplied reference material may be used only when authorised by the administering college. Internet access and external communication are prohibited. ACT BSSS assessment is school based. The administering college may vary timing, technology, stored-note, program and reference-material conditions.
Question 6
10 marksQuestion 7
10 marksQuestion 8
10 marksQuestion 9
10 marksQuestion 10
10 marksWorked Solutions And Marking Guide
Question 6
(a) E(X)=6.
Use the binomial mean.
(b) operatorname(Var)(X)=4.2.
Use the binomial variance.
(c) P(X=6)=binom(20)(6)(0.30)⁶(0.70)¹⁴approx0.1916.
Apply the binomial probability formula and retain the exact expression.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Identifies n=20 and p=0.30.
- Uses E(X)=np.
- Obtains E(X)=20(0.30)=6.
Acceptable alternatives: For Q6(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches E(X)=6.
Do not credit by itself: Omitting or contradicting this required step: Identifies n=20 and p=0.30. Giving the final result without establishing: Obtains E(X)=20(0.30)=6.
Part b (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses operatorname(Var)(X)=np(1-p).
- Substitutes 20(0.30)(0.70).
- Obtains operatorname(Var)(X)=4.2.
Acceptable alternatives: For Q6(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches operatorname(Var)(X)=4.2.
Do not credit by itself: Omitting or contradicting this required step: Uses operatorname(Var)(X)=np(1-p). Giving the final result without establishing: Obtains operatorname(Var)(X)=4.2.
Part c (4 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 4. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, retain the requested exact form; carry unrounded values until the final stated accuracy.
Mark-by-mark evidence:
- Selects the binomial coefficient binom(20)(6).
- Uses the success factor (0.30)^6.
- Uses the failure factor (0.70)¹⁴.
- Combines the factors and rounds the probability to 0.1916.
Acceptable alternatives: For Q6(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches P(X=6)=binom(20)(6)(0.30)⁶(0.70)¹⁴approx0.1916.
Do not credit by itself: Omitting or contradicting this required step: Selects the binomial coefficient binom(20)(6). Giving the final result without establishing: Combines the factors and rounds the probability to 0.1916.
Question 7
(a) z=2.
Subtract the mean and divide by the standard deviation.
(b) P(X>130)=0.0228.
Use the upper-tail complement.
(c) P(85<X<115)=0.6826.
Standardise both endpoints and use normal symmetry.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses z=(x-mu) / sigma.
- Substitutes x=130, mu=100 and sigma=15.
- Obtains z=(130-100) / 15=2.
Acceptable alternatives: For Q7(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches z=2.
Do not credit by itself: Omitting or contradicting this required step: Uses z=(x-mu) / sigma. Giving the final result without establishing: Obtains z=(130-100) / 15=2.
Part b (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Recognises X>130 as Z>2.
- Uses P(Z>2)=1-P(Z<2).
- Obtains 1-0.9772=0.0228.
Acceptable alternatives: For Q7(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches P(X>130)=0.0228.
Do not credit by itself: Omitting or contradicting this required step: Recognises X>130 as Z>2. Giving the final result without establishing: Obtains 1-0.9772=0.0228.
Part c (4 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 4. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Standardises 85 and 115 to z=-1 and z=1.
- Uses symmetry to write P(-1<Z<1)=2P(Z<1)-1.
- Substitutes 2(0.8413)-1.
- Obtains 0.6826.
Acceptable alternatives: For Q7(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches P(85<X<115)=0.6826.
Do not credit by itself: Omitting or contradicting this required step: Standardises 85 and 115 to z=-1 and z=1. Giving the final result without establishing: Obtains 0.6826.
Question 8
(a) k=1 / 8.
Set the total density area equal to one.
(b) P(X>2)=1 / 4.
Integrate the normalised density over the required interval.
(c) E(X)=4 / 3.
Integrate x f(x) over the support.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Writes int_0^4k(4-x),dx=1.
- Evaluates the integral as 8k=1.
- Obtains k=1 / 8.
Acceptable alternatives: For Q8(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches k=1 / 8.
Do not credit by itself: Omitting or contradicting this required step: Writes int_0^4k(4-x),dx=1. Giving the final result without establishing: Obtains k=1 / 8.
Part b (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Writes P(X>2)=int_2⁴(4-x) / 8,dx.
- Uses the antiderivative x / 2-x² / 16.
- Evaluates the bounds to obtain 1 / 4.
Acceptable alternatives: For Q8(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches P(X>2)=1 / 4.
Do not credit by itself: Omitting or contradicting this required step: Writes P(X>2)=int_2⁴(4-x) / 8,dx. Giving the final result without establishing: Evaluates the bounds to obtain 1 / 4.
Part c (4 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 4. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Uses E(X)=int_0^4x f(x),dx.
- Substitutes f(x)=(4-x) / 8.
- Evaluates frac18int_0⁴(4x-x²),dx.
- Obtains E(X)=4 / 3.
Acceptable alternatives: For Q8(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches E(X)=4 / 3.
Do not credit by itself: Omitting or contradicting this required step: Uses E(X)=int_0^4x f(x),dx. Giving the final result without establishing: Obtains E(X)=4 / 3.
Question 9
(a) hat p=0.60.
Average the two endpoints.
(b) The margin of error is 0.06.
Use half the interval width.
(c) The interval does not establish the claim because it contains plausible values from 0.54 to 0.55, which are not above 0.55.
Compare the whole interval with the claimed threshold and use appropriately cautious confidence language.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Adds the endpoints 0.54+0.66.
- Divides the sum by 2.
- Obtains the centre hat p=0.60.
Acceptable alternatives: For Q9(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches hat p=0.60.
Do not credit by itself: Omitting or contradicting this required step: Adds the endpoints 0.54+0.66. Giving the final result without establishing: Obtains the centre hat p=0.60.
Part b (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Calculates the interval width 0.66-0.54=0.12.
- Divides the width by 2.
- Obtains a margin of error of 0.06.
Acceptable alternatives: For Q9(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches The margin of error is 0.06.
Do not credit by itself: Omitting or contradicting this required step: Calculates the interval width 0.66-0.54=0.12. Giving the final result without establishing: Obtains a margin of error of 0.06.
Part c (4 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 4. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Identifies the threshold proportion 0.55.
- Observes that the lower endpoint 0.54 is below 0.55.
- States that the interval therefore contains plausible values not exceeding 0.55.
- Concludes that the 95% interval does not establish the claim.
Acceptable alternatives: For Q9(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches The interval does not establish the claim because it contains plausible values from 0.54 to 0.55, which are not above 0.55.
Do not credit by itself: Omitting or contradicting this required step: Identifies the threshold proportion 0.55. Giving the final result without establishing: Concludes that the 95% interval does not establish the claim.
Question 10
(a) P(0)=100 and lim_(ttoinfty)P(t)=1000.
Evaluate the model at zero and use exponential limiting behaviour.
(b) t=ln9 / 0.4approx5.49.
Set the logistic model to half its limiting value and solve.
(c) Growth accelerates before P=500, then slows; 1000 is the carrying capacity. A changed environment or population mechanism may invalidate long-term extrapolation.
Connect the S-shape to carrying capacity and qualify the model.
Detailed marking criteria
Part a (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Do not award the final-result mark when the stated mathematical form is incomplete.
Mark-by-mark evidence:
- Substitutes t=0 to obtain P(0)=1000 / (1+9)=100.
- Uses e^(-0.4t)to0 as ttoinfty.
- Obtains the limiting population 1000.
Acceptable alternatives: For Q10(a), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches P(0)=100 and lim_(ttoinfty)P(t)=1000.
Do not credit by itself: Omitting or contradicting this required step: Substitutes t=0 to obtain P(0)=1000 / (1+9)=100. Giving the final result without establishing: Obtains the limiting population 1000.
Part b (3 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 3. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, carry unrounded values until the final stated accuracy.
Mark-by-mark evidence:
- Sets 1000 / (1+9e^(-0.4t))=500.
- Rearranges to e^(-0.4t)=1 / 9.
- Obtains t=ln9 / 0.4approx5.49.
Acceptable alternatives: For Q10(b), credit an equivalent mathematically correct method only when it establishes all 3 required checkpoints and reaches t=ln9 / 0.4approx5.49.
Do not credit by itself: Omitting or contradicting this required step: Sets 1000 / (1+9e^(-0.4t))=500. Giving the final result without establishing: Obtains t=ln9 / 0.4approx5.49.
Part c (4 marks)
Award one mark for each numbered, task-specific statement below, to a maximum of 4. Apply consequential marking when a correct later method consistently uses an earlier incorrect value, unless the error materially simplifies the task. For the final-result mark, include the requested units, context or justification.
Mark-by-mark evidence:
- Identifies the initial accelerating phase of the S-shaped model.
- Identifies the slowing phase after approximately half capacity.
- Interprets 1000 as the carrying capacity.
- Explains that environmental or mechanism changes can invalidate long-term extrapolation.
Acceptable alternatives: For Q10(c), credit an equivalent mathematically correct method only when it establishes all 4 required checkpoints and reaches Growth accelerates before P=500, then slows; 1000 is the carrying capacity. A changed environment or population mechanism may invalidate long-term extrapolation.
Do not credit by itself: Omitting or contradicting this required step: Identifies the initial accelerating phase of the S-shaped model. Giving the final result without establishing: Explains that environmental or mechanism changes can invalidate long-term extrapolation.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Unit 4 — Discrete random variables — Calculate a binomial expectation | Q6(a) | 3 | ___ | Q6(a): reproduce this exact process without the solution: Identifies n=20 and p=0.30; then Uses E(X)=np; then Obtains E(X)=20(0.30)=6. |
| Unit 4 — Discrete random variables — Calculate a binomial variance | Q6(b) | 3 | ___ | Q6(b): reproduce this exact process without the solution: Uses operatorname(Var)(X)=np(1-p); then Substitutes 20(0.30)(0.70); then Obtains operatorname(Var)(X)=4.2. |
| Unit 4 — Discrete random variables — Evaluate a binomial point probability | Q6(c) | 4 | ___ | Q6(c): reproduce this exact process without the solution: Selects the binomial coefficient binom(20)(6); then Uses the success factor (0.30)⁶; then Uses the failure factor (0.70)¹⁴; then Combines the factors and rounds the probability to 0.1916. |
| Unit 4 — Continuous random variables and normal distribution — Standardise a normal observation | Q7(a) | 3 | ___ | Q7(a): reproduce this exact process without the solution: Uses z=(x-mu) / sigma; then Substitutes x=130, mu=100 and sigma=15; then Obtains z=(130-100) / 15=2. |
| Unit 4 — Continuous random variables and normal distribution — Convert a standard score to an upper-tail probability | Q7(b) | 3 | ___ | Q7(b): reproduce this exact process without the solution: Recognises X>130 as Z>2; then Uses P(Z>2)=1-P(Z<2); then Obtains 1-0.9772=0.0228. |
| Unit 4 — Continuous random variables and normal distribution — Find a symmetric central normal probability | Q7(c) | 4 | ___ | Q7(c): reproduce this exact process without the solution: Standardises 85 and 115 to z=-1 and z=1; then Uses symmetry to write P(-1<Z<1)=2P(Z<1)-1; then Substitutes 2(0.8413)-1; then Obtains 0.6826. |
| Unit 4 — Continuous random variables and normal distribution — Normalise a continuous density | Q8(a) | 3 | ___ | Q8(a): reproduce this exact process without the solution: Writes int_0^4k(4-x),dx=1; then Evaluates the integral as 8k=1; then Obtains k=1 / 8. |
| Unit 4 — Continuous random variables and normal distribution — Integrate a density over a tail interval | Q8(b) | 3 | ___ | Q8(b): reproduce this exact process without the solution: Writes P(X>2)=int_2⁴(4-x) / 8,dx; then Uses the antiderivative x / 2-x² / 16; then Evaluates the bounds to obtain 1 / 4. |
| Unit 4 — Continuous random variables and normal distribution — Calculate the expectation of a continuous random variable | Q8(c) | 4 | ___ | Q8(c): reproduce this exact process without the solution: Uses E(X)=int_0^4x f(x),dx; then Substitutes f(x)=(4-x) / 8; then Evaluates frac18int_0⁴(4x-x²),dx; then Obtains E(X)=4 / 3. |
| Unit 4 — Interval estimates for proportions — Recover a sample proportion from interval endpoints | Q9(a) | 3 | ___ | Q9(a): reproduce this exact process without the solution: Adds the endpoints 0.54+0.66; then Divides the sum by 2; then Obtains the centre hat p=0.60. |
| Unit 4 — Interval estimates for proportions — Recover a margin of error from an interval | Q9(b) | 3 | ___ | Q9(b): reproduce this exact process without the solution: Calculates the interval width 0.66-0.54=0.12; then Divides the width by 2; then Obtains a margin of error of 0.06. |
| Unit 4 — Interval estimates for proportions — Use a confidence interval to assess a threshold claim | Q9(c) | 4 | ___ | Q9(c): reproduce this exact process without the solution: Identifies the threshold proportion 0.55; then Observes that the lower endpoint 0.54 is below 0.55; then States that the interval therefore contains plausible values not exceeding 0.55; then Concludes that the 95% interval does not establish the claim. |
| Unit 2 — Exponential functions — Identify initial value and carrying capacity | Q10(a) | 3 | ___ | Q10(a): reproduce this exact process without the solution: Substitutes t=0 to obtain P(0)=1000 / (1+9)=100; then Uses e^(-0.4t)to0 as ttoinfty; then Obtains the limiting population 1000. |
| Unit 3 — The logarithmic function — Solve a logistic threshold equation | Q10(b) | 3 | ___ | Q10(b): reproduce this exact process without the solution: Sets 1000 / (1+9e^(-0.4t))=500; then Rearranges to e^(-0.4t)=1 / 9; then Obtains t=ln9 / 0.4approx5.49. |
| Unit 2 — Exponential functions — Interpret and critique a logistic model | Q10(c) | 4 | ___ | Q10(c): reproduce this exact process without the solution: Identifies the initial accelerating phase of the S-shaped model; then Identifies the slowing phase after approximately half capacity; then Interprets 1000 as the carrying capacity; then Explains that environmental or mechanism changes can invalidate long-term extrapolation. |