Mathematics AA: Standard Level

1. 1.

   A discrete random variable X takes the values 0, 1 and 2 with probabilities 0.25, 0.50 and 0.25. Calculate E(X) and Var(X).

   [5 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
   2. Work through this mathematical step: Calculates E(X) = 0(0.25) + 1(0.50) + 2(0.25). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   3. Work through this mathematical step: Obtains E(X) = 1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   4. Work through this mathematical step: Calculates E(X²) = 0²(0.25) + 1²(0.50) + 2²(0.25). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   5. Work through this mathematical step: Obtains E(X²) = 1.5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   6. Work through this mathematical step: Uses Var(X) = E(X²) − [E(X)]² to obtain 0.5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Variance uses E(X²), not the square of every probability.

   Marking points

   • Calculates E(X) = 0(0.25) + 1(0.50) + 2(0.25).
   • Obtains E(X) = 1.
   • Calculates E(X²) = 0²(0.25) + 1²(0.50) + 2²(0.25).
   • Obtains E(X²) = 1.5.
   • Uses Var(X) = E(X²) − [E(X)]² to obtain 0.5.

   Examiner tip: Variance uses E(X²), not the square of every probability.

2. 2.

   Marking analysis: A learner attempts the following task: “A discrete random variable X takes the values 0, 1 and 2 with probabilities 0.25, 0.50 and 0.25. Calculate E(X) and Var(X).” Their response addresses only this point: “Calculates E(X) = 0(0.25) + 1(0.50) + 2(0.25).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [5 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
   2. Requirement 1: Recognises credit for the stated point: Calculates E(X) = 0(0.25) + 1(0.50) + 2(0.25). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   3. Requirement 2: Identifies the missing requirement: Obtains E(X) = 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   4. Requirement 3: Identifies the missing requirement: Calculates E(X²) = 0²(0.25) + 1²(0.50) + 2²(0.25). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   5. Requirement 4: Identifies the missing requirement: Obtains E(X²) = 1.5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   6. Requirement 5: Identifies the missing requirement: Uses Var(X) = E(X²) − [E(X)]² to obtain 0.5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

   Marking points

   • Recognises credit for the stated point: Calculates E(X) = 0(0.25) + 1(0.50) + 2(0.25).
   • Identifies the missing requirement: Obtains E(X) = 1.
   • Identifies the missing requirement: Calculates E(X²) = 0²(0.25) + 1²(0.50) + 2²(0.25).
   • Identifies the missing requirement: Obtains E(X²) = 1.5.
   • Identifies the missing requirement: Uses Var(X) = E(X²) − [E(X)]² to obtain 0.5.

   Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

3. 3.

   A biased coin has probability 0.60 of landing heads. It is tossed 8 times. Find the probability of exactly 5 heads and state the binomial parameters used.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
   2. Work through this mathematical step: States X ~ B(8, 0.60). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   3. Work through this mathematical step: Uses P(X=5) = 8C5(0.60)⁵(0.40)³. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   4. Work through this mathematical step: Evaluates the combination 8C5 = 56. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   5. Work through this mathematical step: Obtains P(X=5) ≈ 0.279. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Exactly five heads requires both the five successes and three failures.

   Marking points

   • States X ~ B(8, 0.60).
   • Uses P(X=5) = 8C5(0.60)⁵(0.40)³.
   • Evaluates the combination 8C5 = 56.
   • Obtains P(X=5) ≈ 0.279.

   Examiner tip: Exactly five heads requires both the five successes and three failures.

4. 4.

   Marking analysis: A learner attempts the following task: “A biased coin has probability 0.60 of landing heads. It is tossed 8 times. Find the probability of exactly 5 heads and state the binomial parameters used.” Their response addresses only this point: “States X ~ B(8, 0.60).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
   2. Requirement 1: Recognises credit for the stated point: States X ~ B(8, 0.60). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   3. Requirement 2: Identifies the missing requirement: Uses P(X=5) = 8C5(0.60)⁵(0.40)³. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   4. Requirement 3: Identifies the missing requirement: Evaluates the combination 8C5 = 56. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   5. Requirement 4: Identifies the missing requirement: Obtains P(X=5) ≈ 0.279. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

   Marking points

   • Recognises credit for the stated point: States X ~ B(8, 0.60).
   • Identifies the missing requirement: Uses P(X=5) = 8C5(0.60)⁵(0.40)³.
   • Identifies the missing requirement: Evaluates the combination 8C5 = 56.
   • Identifies the missing requirement: Obtains P(X=5) ≈ 0.279.

   Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

5. 5.

   A class test has scores that are normally distributed with mean 68 and standard deviation 8. Find the probability that a randomly chosen student scores above 80.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
   2. Work through this mathematical step: Standardises using z = (80 − 68)/8 = 1.5. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   3. Work through this mathematical step: Uses the normal distribution to find P(Z > 1.5). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   4. Work through this mathematical step: Obtains P(X > 80) ≈ 0.0668. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Standardise when using a standard-normal table, or use a calculator's normal cumulative distribution with mean 68 and standard deviation 8 directly. Either method is valid.

   Marking points

   • Standardises using z = (80 − 68)/8 = 1.5.
   • Uses the normal distribution to find P(Z > 1.5).
   • Obtains P(X > 80) ≈ 0.0668.

   Examiner tip: Standardise when using a standard-normal table, or use a calculator's normal cumulative distribution with mean 68 and standard deviation 8 directly. Either method is valid.

6. 6.

   Marking analysis: A learner attempts the following task: “A class test has scores that are normally distributed with mean 68 and standard deviation 8. Find the probability that a randomly chosen student scores above 80.” Their response addresses only this point: “Standardises using z = (80 − 68)/8 = 1.5.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
   2. Requirement 1: Recognises credit for the stated point: Standardises using z = (80 − 68)/8 = 1.5. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   3. Requirement 2: Identifies the missing requirement: Uses the normal distribution to find P(Z > 1.5). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   4. Requirement 3: Identifies the missing requirement: Obtains P(X > 80) ≈ 0.0668. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

   Marking points

   • Recognises credit for the stated point: Standardises using z = (80 − 68)/8 = 1.5.
   • Identifies the missing requirement: Uses the normal distribution to find P(Z > 1.5).
   • Identifies the missing requirement: Obtains P(X > 80) ≈ 0.0668.

   Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

7. 7.

   The events A and B satisfy P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Determine whether A and B are independent, showing your reasoning.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
   2. Work through this mathematical step: States the independence condition P(A)×P(B) = P(A ∩ B). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   3. Work through this mathematical step: Calculates P(A)×P(B) = 0.4 × 0.5 = 0.20. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   4. Work through this mathematical step: Compares 0.20 to the given P(A ∩ B) = 0.2 and concludes A and B are independent. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Independence must be tested numerically against the multiplication rule; it cannot be assumed from the wording alone.

   Marking points

   • States the independence condition P(A)×P(B) = P(A ∩ B).
   • Calculates P(A)×P(B) = 0.4 × 0.5 = 0.20.
   • Compares 0.20 to the given P(A ∩ B) = 0.2 and concludes A and B are independent.

   Examiner tip: Independence must be tested numerically against the multiplication rule; it cannot be assumed from the wording alone.

8. 8.

   Marking analysis: A learner attempts the following task: “The events A and B satisfy P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Determine whether A and B are independent, showing your reasoning.” Their response addresses only this point: “States the independence condition P(A)×P(B) = P(A ∩ B).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
   2. Requirement 1: Recognises credit for the stated point: States the independence condition P(A)×P(B) = P(A ∩ B). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   3. Requirement 2: Identifies the missing requirement: Calculates P(A)×P(B) = 0.4 × 0.5 = 0.20. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   4. Requirement 3: Identifies the missing requirement: Compares 0.20 to the given P(A ∩ B) = 0.2 and concludes A and B are independent. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
   5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

   Marking points

   • Recognises credit for the stated point: States the independence condition P(A)×P(B) = P(A ∩ B).
   • Identifies the missing requirement: Calculates P(A)×P(B) = 0.4 × 0.5 = 0.20.
   • Identifies the missing requirement: Compares 0.20 to the given P(A ∩ B) = 0.2 and concludes A and B are independent.

   Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

9. 9.

   A survey of 200 students found that 120 study Mathematics and 90 study Physics, with 50 studying both. A student is chosen at random. Find the probability that the student studies Mathematics or Physics (or both).

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
   2. Work through this mathematical step: Uses the addition rule P(M ∪ P) = P(M) + P(P) − P(M ∩ P). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   3. Work through this mathematical step: Calculates each probability: P(M) = 120/200, P(P) = 90/200, P(M ∩ P) = 50/200. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   4. Work through this mathematical step: Substitutes to obtain 0.6 + 0.45 − 0.25. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   5. Work through this mathematical step: Obtains P(M ∪ P) = 0.8. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
   6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Adding P(M) and P(P) directly double-counts students who study both, so the intersection must be subtracted.

   Marking points

   • Uses the addition rule P(M ∪ P) = P(M) + P(P) − P(M ∩ P).
   • Calculates each probability: P(M) = 120/200, P(P) = 90/200, P(M ∩ P) = 50/200.
   • Substitutes to obtain 0.6 + 0.45 − 0.25.
   • Obtains P(M ∪ P) = 0.8.

   Examiner tip: Adding P(M) and P(P) directly double-counts students who study both, so the intersection must be subtracted.

10. 10.

    Marking analysis: A learner attempts the following task: “A survey of 200 students found that 120 study Mathematics and 90 study Physics, with 50 studying both. A student is chosen at random. Find the probability that the student studies Mathematics or Physics (or both).” Their response addresses only this point: “Uses the addition rule P(M ∪ P) = P(M) + P(P) − P(M ∩ P).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Uses the addition rule P(M ∪ P) = P(M) + P(P) − P(M ∩ P). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Calculates each probability: P(M) = 120/200, P(P) = 90/200, P(M ∩ P) = 50/200. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Substitutes to obtain 0.6 + 0.45 − 0.25. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Obtains P(M ∪ P) = 0.8. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Uses the addition rule P(M ∪ P) = P(M) + P(P) − P(M ∩ P).
    • Identifies the missing requirement: Calculates each probability: P(M) = 120/200, P(P) = 90/200, P(M ∩ P) = 50/200.
    • Identifies the missing requirement: Substitutes to obtain 0.6 + 0.45 − 0.25.
    • Identifies the missing requirement: Obtains P(M ∪ P) = 0.8.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

11. 11.

    A discrete random variable X has probability distribution P(X = x) = kx for x = 1, 2, 3, 4, and P(X = x) = 0 otherwise. Find the value of k and calculate E(X).

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Uses the condition that all probabilities sum to 1: k(1+2+3+4) = 1. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Solves to obtain k = 1/10. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Lists the probabilities: P(1)=0.1, P(2)=0.2, P(3)=0.3, P(4)=0.4. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Calculates E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Obtains E(X) = 3.0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Use the total-probability condition to determine k. You may calculate E(X) symbolically first and substitute k later.

    Marking points

    • Uses the condition that all probabilities sum to 1: k(1+2+3+4) = 1.
    • Solves to obtain k = 1/10.
    • Lists the probabilities: P(1)=0.1, P(2)=0.2, P(3)=0.3, P(4)=0.4.
    • Calculates E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4).
    • Obtains E(X) = 3.0.

    Examiner tip: Use the total-probability condition to determine k. You may calculate E(X) symbolically first and substitute k later.

12. 12.

    Marking analysis: A learner attempts the following task: “A discrete random variable X has probability distribution P(X = x) = kx for x = 1, 2, 3, 4, and P(X = x) = 0 otherwise. Find the value of k and calculate E(X).” Their response addresses only this point: “Uses the condition that all probabilities sum to 1: k(1+2+3+4) = 1.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Uses the condition that all probabilities sum to 1: k(1+2+3+4) = 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Solves to obtain k = 1/10. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Lists the probabilities: P(1)=0.1, P(2)=0.2, P(3)=0.3, P(4)=0.4. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Calculates E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Obtains E(X) = 3.0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Uses the condition that all probabilities sum to 1: k(1+2+3+4) = 1.
    • Identifies the missing requirement: Solves to obtain k = 1/10.
    • Identifies the missing requirement: Lists the probabilities: P(1)=0.1, P(2)=0.2, P(3)=0.3, P(4)=0.4.
    • Identifies the missing requirement: Calculates E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4).
    • Identifies the missing requirement: Obtains E(X) = 3.0.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

13. 13.

    Exam scores for a large cohort are approximately normally distributed with mean 62 and standard deviation 10. Find the score that separates the top 10% of students from the rest.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Identifies that the required z-value satisfies P(Z > z) = 0.10. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Uses the inverse normal to obtain z ≈ 1.2816. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Uses x = μ + zσ. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Obtains x ≈ 74.8. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Finding a boundary score from a percentile requires the inverse normal function, working from probability back to a z-value.

    Marking points

    • Identifies that the required z-value satisfies P(Z > z) = 0.10.
    • Uses the inverse normal to obtain z ≈ 1.2816.
    • Uses x = μ + zσ.
    • Obtains x ≈ 74.8.

    Examiner tip: Finding a boundary score from a percentile requires the inverse normal function, working from probability back to a z-value.

14. 14.

    Marking analysis: A learner attempts the following task: “Exam scores for a large cohort are approximately normally distributed with mean 62 and standard deviation 10. Find the score that separates the top 10% of students from the rest.” Their response addresses only this point: “Identifies that the required z-value satisfies P(Z > z) = 0.10.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Identifies that the required z-value satisfies P(Z > z) = 0.10. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Uses the inverse normal to obtain z ≈ 1.2816. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Uses x = μ + zσ. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Obtains x ≈ 74.8. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Identifies that the required z-value satisfies P(Z > z) = 0.10.
    • Identifies the missing requirement: Uses the inverse normal to obtain z ≈ 1.2816.
    • Identifies the missing requirement: Uses x = μ + zσ.
    • Identifies the missing requirement: Obtains x ≈ 74.8.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

15. 15.

    A fair six-sided die is rolled twice. Find the probability that the sum of the two rolls is greater than 9.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Identifies that there are 36 equally likely outcomes in total. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Lists the outcomes with sum greater than 9: sums of 10, 11 and 12. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Counts these outcomes correctly as 6 (namely (4,6),(5,5),(5,6),(6,4),(6,5),(6,6)). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: States the probability as 6/36 = 1/6. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: List outcomes as ordered pairs systematically to avoid missing or double-counting any combination.

    Marking points

    • Identifies that there are 36 equally likely outcomes in total.
    • Lists the outcomes with sum greater than 9: sums of 10, 11 and 12.
    • Counts these outcomes correctly as 6 (namely (4,6),(5,5),(5,6),(6,4),(6,5),(6,6)).
    • States the probability as 6/36 = 1/6.

    Examiner tip: List outcomes as ordered pairs systematically to avoid missing or double-counting any combination.

16. 16.

    Marking analysis: A learner attempts the following task: “A fair six-sided die is rolled twice. Find the probability that the sum of the two rolls is greater than 9.” Their response addresses only this point: “Identifies that there are 36 equally likely outcomes in total.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Identifies that there are 36 equally likely outcomes in total. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Lists the outcomes with sum greater than 9: sums of 10, 11 and 12. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Counts these outcomes correctly as 6 (namely (4,6),(5,5),(5,6),(6,4),(6,5),(6,6)). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: States the probability as 6/36 = 1/6. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Identifies that there are 36 equally likely outcomes in total.
    • Identifies the missing requirement: Lists the outcomes with sum greater than 9: sums of 10, 11 and 12.
    • Identifies the missing requirement: Counts these outcomes correctly as 6 (namely (4,6),(5,5),(5,6),(6,4),(6,5),(6,6)).
    • Identifies the missing requirement: States the probability as 6/36 = 1/6.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

17. 17.

    A test for a disease is 95% accurate for people who have the disease and 90% accurate for people who do not. In a population, 2% of people have the disease. Find the probability that a randomly selected person tests positive.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Identifies the two paths to a positive test: (disease and correctly positive) or (no disease and incorrectly positive). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Calculates P(disease and positive) = 0.02 × 0.95 = 0.019. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Calculates P(no disease and positive) = 0.98 × 0.10 = 0.098. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Adds the two mutually exclusive paths. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Obtains P(positive) = 0.117. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A tree diagram makes conditional-probability problems like this much easier to set up correctly.

    Marking points

    • Identifies the two paths to a positive test: (disease and correctly positive) or (no disease and incorrectly positive).
    • Calculates P(disease and positive) = 0.02 × 0.95 = 0.019.
    • Calculates P(no disease and positive) = 0.98 × 0.10 = 0.098.
    • Adds the two mutually exclusive paths.
    • Obtains P(positive) = 0.117.

    Examiner tip: A tree diagram makes conditional-probability problems like this much easier to set up correctly.

18. 18.

    Marking analysis: A learner attempts the following task: “A test for a disease is 95% accurate for people who have the disease and 90% accurate for people who do not. In a population, 2% of people have the disease. Find the probability that a randomly selected person tests positive.” Their response addresses only this point: “Identifies the two paths to a positive test: (disease and correctly positive) or (no disease and incorrectly positive).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Identifies the two paths to a positive test: (disease and correctly positive) or (no disease and incorrectly positive). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Calculates P(disease and positive) = 0.02 × 0.95 = 0.019. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Calculates P(no disease and positive) = 0.98 × 0.10 = 0.098. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Adds the two mutually exclusive paths. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Obtains P(positive) = 0.117. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Identifies the two paths to a positive test: (disease and correctly positive) or (no disease and incorrectly positive).
    • Identifies the missing requirement: Calculates P(disease and positive) = 0.02 × 0.95 = 0.019.
    • Identifies the missing requirement: Calculates P(no disease and positive) = 0.98 × 0.10 = 0.098.
    • Identifies the missing requirement: Adds the two mutually exclusive paths.
    • Identifies the missing requirement: Obtains P(positive) = 0.117.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

19. 19.

    Each of 10 independent calls to a service desk is resolved on the first call with probability 0.6. Find the probability that exactly 4 calls are resolved on the first call.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Models the number resolved as X ~ B(10, 0.6). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Uses P(X = 4) = 10C4 × 0.6⁴ × 0.4⁶. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Obtains P(X = 4) = 0.111476736 ≈ 0.1115. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A binomial model uses a fixed number of independent trials, two outcomes per trial and the same success probability on each trial.

    Marking points

    • Models the number resolved as X ~ B(10, 0.6).
    • Uses P(X = 4) = 10C4 × 0.6⁴ × 0.4⁶.
    • Obtains P(X = 4) = 0.111476736 ≈ 0.1115.

    Examiner tip: A binomial model uses a fixed number of independent trials, two outcomes per trial and the same success probability on each trial.

20. 20.

    Marking analysis: A learner attempts the following task: “Each of 10 independent calls to a service desk is resolved on the first call with probability 0.6. Find the probability that exactly 4 calls are resolved on the first call.” Their response addresses only this point: “Models the number resolved as X ~ B(10, 0.6).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Models the number resolved as X ~ B(10, 0.6). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Uses P(X = 4) = 10C4 × 0.6⁴ × 0.4⁶. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Obtains P(X = 4) = 0.111476736 ≈ 0.1115. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Models the number resolved as X ~ B(10, 0.6).
    • Identifies the missing requirement: Uses P(X = 4) = 10C4 × 0.6⁴ × 0.4⁶.
    • Identifies the missing requirement: Obtains P(X = 4) = 0.111476736 ≈ 0.1115.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

21. 21.

    Events A and B satisfy P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.8. Find P(A ∩ B) and hence P(A | B).

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Uses P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to find P(A ∩ B) = 0.6 + 0.5 − 0.8. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Obtains P(A ∩ B) = 0.3. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: States P(A | B) = P(A ∩ B)/P(B). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Substitutes 0.3/0.5 to obtain P(A | B) = 0.6. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Whenever P(A ∩ B) is not given directly, check if it can be found first from the addition rule using P(A), P(B) and P(A ∪ B) — this is a common first step before a conditional probability calculation.

    Marking points

    • Uses P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to find P(A ∩ B) = 0.6 + 0.5 − 0.8.
    • Obtains P(A ∩ B) = 0.3.
    • States P(A | B) = P(A ∩ B)/P(B).
    • Substitutes 0.3/0.5 to obtain P(A | B) = 0.6.

    Examiner tip: Whenever P(A ∩ B) is not given directly, check if it can be found first from the addition rule using P(A), P(B) and P(A ∪ B) — this is a common first step before a conditional probability calculation.

22. 22.

    Marking analysis: A learner attempts the following task: “Events A and B satisfy P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.8. Find P(A ∩ B) and hence P(A | B).” Their response addresses only this point: “Uses P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to find P(A ∩ B) = 0.6 + 0.5 − 0.8.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Uses P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to find P(A ∩ B) = 0.6 + 0.5 − 0.8. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Obtains P(A ∩ B) = 0.3. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States P(A | B) = P(A ∩ B)/P(B). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Substitutes 0.3/0.5 to obtain P(A | B) = 0.6. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Uses P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to find P(A ∩ B) = 0.6 + 0.5 − 0.8.
    • Identifies the missing requirement: Obtains P(A ∩ B) = 0.3.
    • Identifies the missing requirement: States P(A | B) = P(A ∩ B)/P(B).
    • Identifies the missing requirement: Substitutes 0.3/0.5 to obtain P(A | B) = 0.6.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

23. 23.

    X ~ B(15, 0.4). Find E(X) and Var(X).

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States E(X) = np. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Obtains E(X) = 15 × 0.4 = 6. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: States Var(X) = np(1 − p), obtaining Var(X) = 15 × 0.4 × 0.6 = 3.6. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: These two shortcut formulas apply only to a binomial random variable — do not use them for any other distribution.

    Marking points

    • States E(X) = np.
    • Obtains E(X) = 15 × 0.4 = 6.
    • States Var(X) = np(1 − p), obtaining Var(X) = 15 × 0.4 × 0.6 = 3.6.

    Examiner tip: These two shortcut formulas apply only to a binomial random variable — do not use them for any other distribution.

24. 24.

    Marking analysis: A learner attempts the following task: “X ~ B(15, 0.4). Find E(X) and Var(X).” Their response addresses only this point: “States E(X) = np.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States E(X) = np. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Obtains E(X) = 15 × 0.4 = 6. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States Var(X) = np(1 − p), obtaining Var(X) = 15 × 0.4 × 0.6 = 3.6. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States E(X) = np.
    • Identifies the missing requirement: Obtains E(X) = 15 × 0.4 = 6.
    • Identifies the missing requirement: States Var(X) = np(1 − p), obtaining Var(X) = 15 × 0.4 × 0.6 = 3.6.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

25. 25.

    The table shows the number of siblings for 20 students: 0 siblings (5 students), 1 sibling (8 students), 2 siblings (5 students), 3 siblings (2 students). Calculate the mean and standard deviation of the number of siblings.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Calculates Σfx = 0(5) + 1(8) + 2(5) + 3(2) = 24. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Calculates the mean = 24/20 = 1.2. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Calculates Σfx² = 0²(5) + 1²(8) + 2²(5) + 3²(2) = 46. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Uses variance = Σfx²/n − mean² = 46/20 − 1.2² = 0.86. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Obtains standard deviation = √0.86 ≈ 0.927. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: For frequency-table data, always weight each value by its frequency (Σfx and Σfx²) rather than treating each distinct value as occurring only once.

    Marking points

    • Calculates Σfx = 0(5) + 1(8) + 2(5) + 3(2) = 24.
    • Calculates the mean = 24/20 = 1.2.
    • Calculates Σfx² = 0²(5) + 1²(8) + 2²(5) + 3²(2) = 46.
    • Uses variance = Σfx²/n − mean² = 46/20 − 1.2² = 0.86.
    • Obtains standard deviation = √0.86 ≈ 0.927.

    Examiner tip: For frequency-table data, always weight each value by its frequency (Σfx and Σfx²) rather than treating each distinct value as occurring only once.

26. 26.

    Marking analysis: A learner attempts the following task: “The table shows the number of siblings for 20 students: 0 siblings (5 students), 1 sibling (8 students), 2 siblings (5 students), 3 siblings (2 students). Calculate the mean and standard deviation of the number of siblings.” Their response addresses only this point: “Calculates Σfx = 0(5) + 1(8) + 2(5) + 3(2) = 24.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Calculates Σfx = 0(5) + 1(8) + 2(5) + 3(2) = 24. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Calculates the mean = 24/20 = 1.2. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Calculates Σfx² = 0²(5) + 1²(8) + 2²(5) + 3²(2) = 46. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Uses variance = Σfx²/n − mean² = 46/20 − 1.2² = 0.86. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Obtains standard deviation = √0.86 ≈ 0.927. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Calculates Σfx = 0(5) + 1(8) + 2(5) + 3(2) = 24.
    • Identifies the missing requirement: Calculates the mean = 24/20 = 1.2.
    • Identifies the missing requirement: Calculates Σfx² = 0²(5) + 1²(8) + 2²(5) + 3²(2) = 46.
    • Identifies the missing requirement: Uses variance = Σfx²/n − mean² = 46/20 − 1.2² = 0.86.
    • Identifies the missing requirement: Obtains standard deviation = √0.86 ≈ 0.927.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

27. 27.

    A scatter diagram of hours studied against exam score gives a Pearson's product-moment correlation coefficient of r = 0.85. Interpret this value in context, and state one limitation of using r to conclude that studying causes higher scores.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
    2. Work through this mathematical step: States that r = 0.85 indicates a strong, positive linear correlation between hours studied and exam score. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: States that as hours studied increases, exam score tends to increase as well. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: States the limitation that correlation does not imply causation — a strong r value alone does not prove that studying causes the increase in score, since a confounding variable or coincidence could also explain the pattern. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This 'correlation is not causation' limitation applies to every correlation coefficient, no matter how strong — a high r value only ever describes the strength and direction of a linear association, never the underlying cause.

    Marking points

    • States that r = 0.85 indicates a strong, positive linear correlation between hours studied and exam score.
    • States that as hours studied increases, exam score tends to increase as well.
    • States the limitation that correlation does not imply causation — a strong r value alone does not prove that studying causes the increase in score, since a confounding variable or coincidence could also explain the pattern.

    Examiner tip: This 'correlation is not causation' limitation applies to every correlation coefficient, no matter how strong — a high r value only ever describes the strength and direction of a linear association, never the underlying cause.

28. 28.

    Marking analysis: A learner attempts the following task: “A scatter diagram of hours studied against exam score gives a Pearson's product-moment correlation coefficient of r = 0.85. Interpret this value in context, and state one limitation of using r to conclude that studying causes higher scores.” Their response addresses only this point: “States that r = 0.85 indicates a strong, positive linear correlation between hours studied and exam score.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that r = 0.85 indicates a strong, positive linear correlation between hours studied and exam score. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States that as hours studied increases, exam score tends to increase as well. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States the limitation that correlation does not imply causation — a strong r value alone does not prove that studying causes the increase in score, since a confounding variable or coincidence could also explain the pattern. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that r = 0.85 indicates a strong, positive linear correlation between hours studied and exam score.
    • Identifies the missing requirement: States that as hours studied increases, exam score tends to increase as well.
    • Identifies the missing requirement: States the limitation that correlation does not imply causation — a strong r value alone does not prove that studying causes the increase in score, since a confounding variable or coincidence could also explain the pattern.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

29. 29.

    A field study models afternoon air temperature T (°C) using tree-canopy cover x (%) with T = 31.8 − 0.084x. The correlation coefficient is r = −0.91. (a) Predict T when x = 65. (b) Interpret the gradient in context. (c) Explain why this model alone does not prove that increasing canopy cover causes the temperature change. (d) State the effect on r if every recorded canopy-cover value were increased by 5 percentage points.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Substitutes x = 65 to obtain T = 31.8 − 0.084(65) = 26.34 °C (about 26.3 °C). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Interprets the gradient: each additional percentage point of canopy cover is associated with a predicted decrease of 0.084 °C in afternoon temperature. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: States that correlation/regression does not establish causation and identifies a possible confounding factor or the need for a controlled design. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: States that r is unchanged because adding the same constant to every x-value does not change correlation. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Keep prediction, interpretation, and causation as separate tasks; a strong correlation is still not a controlled experiment.

    Marking points

    • Substitutes x = 65 to obtain T = 31.8 − 0.084(65) = 26.34 °C (about 26.3 °C).
    • Interprets the gradient: each additional percentage point of canopy cover is associated with a predicted decrease of 0.084 °C in afternoon temperature.
    • States that correlation/regression does not establish causation and identifies a possible confounding factor or the need for a controlled design.
    • States that r is unchanged because adding the same constant to every x-value does not change correlation.

    Examiner tip: Keep prediction, interpretation, and causation as separate tasks; a strong correlation is still not a controlled experiment.

30. 30.

    Marking analysis: A learner attempts the following task: “A field study models afternoon air temperature T (°C) using tree-canopy cover x (%) with T = 31.8 − 0.084x. The correlation coefficient is r = −0.91. (a) Predict T when x = 65. (b) Interpret the gradient in context. (c) Explain why this model alone does not prove that increasing canopy cover causes the temperature change. (d) State the effect on r if every recorded canopy-cover value were increased by 5 percentage points.” Their response addresses only this point: “Substitutes x = 65 to obtain T = 31.8 − 0.084(65) = 26.34 °C (about 26.3 °C).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Substitutes x = 65 to obtain T = 31.8 − 0.084(65) = 26.34 °C (about 26.3 °C). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Interprets the gradient: each additional percentage point of canopy cover is associated with a predicted decrease of 0.084 °C in afternoon temperature. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States that correlation/regression does not establish causation and identifies a possible confounding factor or the need for a controlled design. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: States that r is unchanged because adding the same constant to every x-value does not change correlation. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Substitutes x = 65 to obtain T = 31.8 − 0.084(65) = 26.34 °C (about 26.3 °C).
    • Identifies the missing requirement: Interprets the gradient: each additional percentage point of canopy cover is associated with a predicted decrease of 0.084 °C in afternoon temperature.
    • Identifies the missing requirement: States that correlation/regression does not establish causation and identifies a possible confounding factor or the need for a controlled design.
    • Identifies the missing requirement: States that r is unchanged because adding the same constant to every x-value does not change correlation.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

31. 31.

    X takes values 0, 1 and 2 with probabilities 0.2, 0.5 and 0.3 respectively. Find E(X) and Var(X).

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Weight each value by its probability. For the second moment, square the values before weighting, rather than squaring the mean.
    2. Subtracting the square of the mean removes the location effect, leaving the spread; the standard deviation would be 0.7, not 0.49.

    Marking points

    • E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 1.1.
    • E(X^2) = 0 + 1(0.5) + 4(0.3) = 1.7.
    • Var(X) = 1.7 - 1.1^2 = 0.49.

    Examiner tip: Variance and standard deviation are not interchangeable.

32. 32.

    Marking analysis: A learner attempts the following task: “X takes values 0, 1 and 2 with probabilities 0.2, 0.5 and 0.3 respectively. Find E(X) and Var(X).” Their response addresses only this point: “E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 1.1.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 1.1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: E(X^2) = 0 + 1(0.5) + 4(0.3) = 1.7. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Var(X) = 1.7 - 1.1^2 = 0.49. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: E(X) = 0(0.2) + 1(0.5) + 2(0.3) = 1.1.
    • Identifies the missing requirement: E(X^2) = 0 + 1(0.5) + 4(0.3) = 1.7.
    • Identifies the missing requirement: Var(X) = 1.7 - 1.1^2 = 0.49.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

33. 33.

    A fitted model predicts y = 2.4x + 15 from data with 1 <= x <= 5. Predict y at x = 3 and explain why using this model at x = 20 is less reliable.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Substitution at x = 3 is interpolation because it is within the sampled interval.
    2. The equation produces 63 at x = 20, but obtaining a number does not establish that the model remains valid there.

    Marking points

    • At x = 3, y = 2.4(3) + 15 = 22.2.
    • x = 20 lies outside the observed range, so its use is extrapolation.
    • The linear relationship may not continue outside the data range; no guarantee of a reliable prediction follows.

    Examiner tip: A prediction is conditional on the model's assumptions, not proof of causation.

34. 34.

    Marking analysis: A learner attempts the following task: “A fitted model predicts y = 2.4x + 15 from data with 1 <= x <= 5. Predict y at x = 3 and explain why using this model at x = 20 is less reliable.” Their response addresses only this point: “At x = 3, y = 2.4(3) + 15 = 22.2.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: At x = 3, y = 2.4(3) + 15 = 22.2. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: x = 20 lies outside the observed range, so its use is extrapolation. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: The linear relationship may not continue outside the data range; no guarantee of a reliable prediction follows. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: At x = 3, y = 2.4(3) + 15 = 22.2.
    • Identifies the missing requirement: x = 20 lies outside the observed range, so its use is extrapolation.
    • Identifies the missing requirement: The linear relationship may not continue outside the data range; no guarantee of a reliable prediction follows.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

35. 35.

    Six independent trials each have success probability 0.3. Find the probability of at least two successes and state the distribution used.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. At least two includes 2, 3, 4, 5 and 6 successes. Its complement has only two outcomes, which is quicker and less error-prone to evaluate.
    2. The binomial coefficient 6 counts the six positions in which the single success could occur.

    Marking points

    • X follows Binomial(6,0.3).
    • P(X >= 2) = 1 - P(X = 0) - P(X = 1).
    • P(X = 0) = 0.7^6 and P(X = 1) = 6(0.3)(0.7^5).
    • Probability = 0.579825, or 0.580 to three significant figures.

    Examiner tip: Subtract both zero and one success, not just zero.

36. 36.

    Marking analysis: A learner attempts the following task: “Six independent trials each have success probability 0.3. Find the probability of at least two successes and state the distribution used.” Their response addresses only this point: “X follows Binomial(6,0.3).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: X follows Binomial(6,0.3). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: P(X >= 2) = 1 - P(X = 0) - P(X = 1). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: P(X = 0) = 0.7^6 and P(X = 1) = 6(0.3)(0.7^5). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Probability = 0.579825, or 0.580 to three significant figures. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: X follows Binomial(6,0.3).
    • Identifies the missing requirement: P(X >= 2) = 1 - P(X = 0) - P(X = 1).
    • Identifies the missing requirement: P(X = 0) = 0.7^6 and P(X = 1) = 6(0.3)(0.7^5).
    • Identifies the missing requirement: Probability = 0.579825, or 0.580 to three significant figures.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

37. 37.

    Scores are modelled by a normal distribution with mean 70 and standard deviation 8. Find P(62 < X < 78) to three significant figures and interpret the result as a model prediction.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Both limits are one standard deviation from the mean, so use the central area between z = -1 and z = 1.
    2. A normal model describes a probability distribution; a finite set of actual scores can have a different observed fraction.

    Marking points

    • Standardised limits are (62 - 70)/8 = -1 and (78 - 70)/8 = 1.
    • Probability = Phi(1) - Phi(-1) = 0.682689... = 0.683.
    • The model predicts about 68.3% of scores in this interval, not an exact proportion in every sample.

    Examiner tip: Use the standard deviation 8, not its square 64, when standardising.

38. 38.

    Marking analysis: A learner attempts the following task: “Scores are modelled by a normal distribution with mean 70 and standard deviation 8. Find P(62 < X < 78) to three significant figures and interpret the result as a model prediction.” Their response addresses only this point: “Standardised limits are (62 - 70)/8 = -1 and (78 - 70)/8 = 1.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Standardised limits are (62 - 70)/8 = -1 and (78 - 70)/8 = 1. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Probability = Phi(1) - Phi(-1) = 0.682689... = 0.683. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: The model predicts about 68.3% of scores in this interval, not an exact proportion in every sample. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Standardised limits are (62 - 70)/8 = -1 and (78 - 70)/8 = 1.
    • Identifies the missing requirement: Probability = Phi(1) - Phi(-1) = 0.682689... = 0.683.
    • Identifies the missing requirement: The model predicts about 68.3% of scores in this interval, not an exact proportion in every sample.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

39. 39.

    Factory A makes 60% of a company's items and factory B makes 40%. Defect rates are 2% at A and 5% at B. An item chosen at random is defective. Find the probability it came from B.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. The condition restricts attention to defective items, so both the factory's output share and its defect rate matter.
    2. In a model batch of 10,000 items, A contributes 120 defective items and B contributes 200, so the conditional share is 200/320.

    Marking points

    • P(A and defective) = 0.6(0.02) = 0.012.
    • P(B and defective) = 0.4(0.05) = 0.020.
    • Total defect probability = 0.012 + 0.020 = 0.032.
    • P(B | defective) = P(B and defective)/P(defective).
    • The required probability is 0.020/0.032 = 0.625.

    Examiner tip: P(defective | B) = 0.05 is not P(B | defective).

40. 40.

    Marking analysis: A learner attempts the following task: “Factory A makes 60% of a company's items and factory B makes 40%. Defect rates are 2% at A and 5% at B. An item chosen at random is defective. Find the probability it came from B.” Their response addresses only this point: “P(A and defective) = 0.6(0.02) = 0.012.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: P(A and defective) = 0.6(0.02) = 0.012. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: P(B and defective) = 0.4(0.05) = 0.020. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Total defect probability = 0.012 + 0.020 = 0.032. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: P(B | defective) = P(B and defective)/P(defective). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: The required probability is 0.020/0.032 = 0.625. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: P(A and defective) = 0.6(0.02) = 0.012.
    • Identifies the missing requirement: P(B and defective) = 0.4(0.05) = 0.020.
    • Identifies the missing requirement: Total defect probability = 0.012 + 0.020 = 0.032.
    • Identifies the missing requirement: P(B | defective) = P(B and defective)/P(defective).
    • Identifies the missing requirement: The required probability is 0.020/0.032 = 0.625.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.