A4 Machine learning (HL-heavy) – IB Computer Science exam questions with mark scheme

1Core3 marks

Describe the difference between supervised, unsupervised and reinforcement learning. [3]

Supervised: trained on labelled data (inputs with known correct outputs)(1)
Unsupervised: finds patterns/clusters in unlabelled data(1)
Reinforcement: an agent learns by trial and error from rewards/penalties for its actions(1)
Answer: Labelled data; unlabelled patterns; reward-based trial and error.
Examiner tip: Give the type of data (labelled/unlabelled) or feedback (reward) that defines each type.

2HardHL only4 marks

A k-nearest neighbours classifier (k = 3, Euclidean distance) is trained on the data below.

Point	x	y	Class
P1	1	2	A
P2	2	3	A
P3	3	1	B
P4	6	5	B
P5	5	6	B
P6	2	0	A

Determine the class that this classifier assigns to the new point (3, 3). Show the distances you calculate. [4]

Uses Euclidean distance √((x₂−x₁)² + (y₂−y₁)²) (allow working shown for at least one point)(1)
All six distances correct: P1 2.24, P2 1.00, P3 2.00, P4 3.61, P5 3.61, P6 3.16 (accept √5, √13, √10 etc.; accept squared distances 5, 1, 4, 13, 13, 10 used consistently for ranking)(1)
Identifies the three nearest neighbours as P2 (A), P3 (B) and P1 (A)(1)
Majority vote (2 A : 1 B) gives class A (allow follow-through from their three nearest neighbours)(1)
Answer: Class A
Examiner tip: Compute all distances, then pick the k smallest; do not stop when a pattern looks obvious.

3StandardHL only4 marks

A machine learning model achieves 99% accuracy on its training data but only 71% accuracy on new, unseen data.

(a) Identify the problem with this model. [1]

(b) Explain why this problem occurs. [1]

(a) Overfitting (accept 'high variance'; reject underfitting)(1)
(b) The model has learned the noise/specific details of the training data rather than the general pattern, so it does not generalise to new data (accept 'memorised the training data'; accept a cause linked to poor generalisation, e.g. the model is too complex for the amount of training data / was trained for too long, so it fits the training data too closely)(1)
(c) One valid remedy, e.g. use more (or more varied) training data / data augmentation(1)
(c) A second, different valid remedy, e.g. reduce model complexity (fewer features, shallower tree, fewer layers), regularisation, early stopping, dropout, tune hyperparameters using a validation set / cross-validation (do not award the same idea twice)(1)
Answer: Overfitting; remedies: more data, simpler model/regularisation.
Examiner tip: Say the model fails to generalise; 'it is not accurate' is too vague.

4HardHL only5 marks

A classifier for a disease was tested on 200 patients. 60 patients had the disease: of these, 40 were correctly identified (TP) and 20 were missed (FN). Of the 140 healthy patients, 10 were wrongly flagged (FP) and 130 were correctly cleared (TN).

(a) Calculate accuracy, precision and recall. [3]

(b) A classifier that predicts 'healthy' for everyone achieves 70% accuracy on this sample. Explain why accuracy alone is a poor measure here and which measure is most important for this application. [2]

(a) Accuracy = (40 + 130) / 200 = 0.85 (accept 85%, 17/20)(1)
(a) Precision = 40 / (40 + 10) = 0.80 (accept 80%, 4/5)(1)
(a) Recall = 40 / (40 + 20) = 0.667 (accept 2/3, 0.67, 66.7%, 67%)(1)
(b) The classes are imbalanced (most patients are healthy), so a high accuracy can hide poor performance: the all-'healthy' classifier scores 70% yet detects none of the 60 sick patients (recall 0)(1)
(b) Recall (accept sensitivity / true positive rate) is most important, because a false negative (missing a patient who has the disease) is the most costly error(1)
Answer: (a) 0.85, 0.80, 0.667 (b) accuracy is misleading with imbalanced classes; recall is key
Examiner tip: Write the formula, then substitute; confusion-matrix terms are easily swapped.

5StandardHL only4 marks

A single artificial neuron has weights w₁ = 0.5, w₂ = −0.4, w₃ = 0.3 and bias b = −0.6. It uses a step activation: output 1 if the total is > 0, otherwise 0.

(a) Calculate the output for inputs (1, 0, 1). [3]

(b) State the output for inputs (0, 1, 1). [1]

(a) Weighted sum = 1(0.5) + 0(−0.4) + 1(0.3) = 0.8 (allow 0.5 + 0.3 − 0.6 = 0.2 written in one line for this mark and the next)(1)
(a) Adds the bias: 0.8 − 0.6 = 0.2(1)
(a) 0.2 > 0, so the output is 1 (follow-through from their total)(1)
(b) 0 (working not required; total = −0.4 + 0.3 − 0.6 = −0.7)(1)
Answer: (a) 1 (b) 0
Examiner tip: Do not forget to add the bias before applying the activation function.

6HardHL only5 marks

A school uses linear regression to predict a student's final exam score, y (%), from the number of hours of revision, x. The training data came from students who revised for between 5 and 40 hours. The fitted model is

$$\hat{y} = 1.6x + 28 \qquad r^2 = 0.72$$

(a) Calculate the predicted score for a student who revises for 25 hours. [1]

(b) Explain what the slope and the intercept of this model mean in this context. [2]

(d) The model predicts a score of 124% for a student who revises for 60 hours. Explain why this prediction is unreliable. [1]

(a) 1.6 × 25 + 28 = 68 (%)(1)
(b) Slope: each extra hour of revision increases the predicted score by 1.6 percentage points (accept '1.6%' or '1.6 marks' per hour; must be in context; 'gradient of the line' alone is not enough)(1)
(b) Intercept: the predicted score for a student who does no revision is 28% (ignore any comment that x = 0 is outside the training range)(1)
(c) 72% of the variation in exam scores is explained by hours of revision (by the model); accept 'the line is a fairly good fit, with 28% of the variation due to other factors'. Reject '72% of predictions are correct'(1)
(d) 60 hours is outside the range of the training data (5–40 hours), so this is extrapolation: the linear relationship may not continue (accept: a score cannot exceed 100%)(1)
Answer: (a) 68% (b) +1.6 percentage points per extra hour; 28% predicted with no revision (c) 72% of the variation in scores is explained by the model (d) extrapolation beyond the 5–40 hour training range
Examiner tip: Always interpret slope, intercept and r² in the context of the question, and check whether a prediction lies inside the range of the training data.

7Standard3 marks

A company trains a recruitment-screening model on ten years of its past hiring decisions. Explain one ethical concern with this approach and suggest one way to reduce it. [3]

Identifies a valid concern: the past hiring decisions may contain human bias (e.g. against a gender, ethnic group or age group). Allow another valid concern such as lack of transparency/explainability or privacy/consent over applicants' data(1)
Explains the effect: the model learns and repeats (or amplifies) that bias, so some applicants are treated unfairly / discriminated against (or, for an alternative concern, a matching consequence, e.g. rejected applicants cannot be told why, so no one is accountable)(1)
A suggestion that addresses the stated concern, e.g. audit outcomes for bias, use more representative/balanced training data, remove protected and proxy features (e.g. postcode), use an explainable model, keep a human reviewing decisions(1)
Answer: Bias in historical data is reproduced; audit/balance data and keep human oversight.
Examiner tip: Link the cause (biased training data) to the effect (unfair decisions) and then give a specific remedy.

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