Supervised learning: linear regression, K-NN, decision trees and evaluation metrics (HL) – IB Computer Science exam questions with mark scheme

1HardHL 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.

2StandardHL 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.

3HardHL 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.

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4HardHL 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.

5CoreHL only4 marks

A music-streaming company uses the decision tree shown to predict whether a subscriber will Renew or Cancel their subscription.

Subscriber	Logins per week	Months subscribed	Reported a problem?
P	5	3	Yes
Q	1	18	No

(a) State the prediction the tree makes for subscriber P and for subscriber Q. [2]

(b) State what is represented by a leaf node in a decision tree. [1]

(c) Outline one advantage, for the company, of using a decision tree rather than a k-nearest neighbours (K-NN) classifier. [1]

(a) P: Cancel (5 ≥ 3 so Yes branch; reported a problem = Yes)(1)
(a) Q: Renew (1 < 3 so No branch; 18 ≥ 12 = Yes)(1)
(b) A final classification / predicted class (label) / output of the model; accept “the decision / outcome”; do not accept “the end of the tree” alone(1)
(c) Any one, with outline: the rules are human-readable so staff can see why a subscriber is predicted to cancel / the model is explainable (transparent); OR prediction is fast because only a few comparisons are needed rather than computing the distance to every stored record; OR the training data does not need to be stored/searched at prediction time; OR features do not need to be scaled/normalised. Do not accept “more accurate”(1)
Answer: (a) P: Cancel; Q: Renew (b) the predicted class / final decision (c) e.g. rules are interpretable, so the company can see why a customer is likely to cancel
Examiner tip: Follow the branch labels exactly: “≥ 3” means 3 itself goes down the Yes branch, and students who swap Yes/No lose both marks in (a).

6StandardHL only5 marks

A web-hosting company records the mean response time of its server, y (ms), for different numbers of concurrent users, x. The graph shows the training data and the line of best fit produced by a linear regression model.

(a) Using the graph, determine the slope and the intercept of the regression line, and hence write the equation of the model. [2]

(b) Use your equation to predict the response time when there are 55 concurrent users. [1]

(c) For this model, r² = 0.90. A second model, which predicts response time from the time of day, has r² = 0.31 on the same data.

Explain which model the company should use, with reference to the meaning of r². [2]

(a) slope = 2 (ms per user) and intercept = 40 (ms) read from graph (allow slope 1.9 to 2.1 and intercept 38 to 42 if consistent with a correct reading)(1)
(a) $\hat{y} = 2x + 40$ (follow through their slope and intercept)(1)
(b) 150 ms (follow through from their equation; units required)(1)
(c) r² is the proportion of the variation in response time explained by the model / input variable; 0.90 means about 90% of the variation is explained (compared with 31%)(1)
(c) so the number-of-users model should be used, as its predictions fit the data more closely / are more reliable (do not award for choosing the model without reasoning)(1)
Answer: (a) slope 2, intercept 40, $\hat{y} = 2x + 40$ (b) 150 ms (c) the users model: it explains about 90% of the variation in response time compared with 31%
Examiner tip: Read the intercept where the line crosses x = 0, not at the first data point, and pick two far-apart grid points on the line (not on the plotted data) to get the slope.

7CoreHL only4 marks

Supervised learning can be used for regression or for classification.

(a) For each task, state whether it is a regression task or a classification task.

(i) Predicting the remaining battery life, in hours, of a laptop. [1]

(ii) Predicting whether a product review is positive, neutral or negative. [1]

(b) Describe how a linear regression algorithm chooses the line of best fit for a set of training data. [2]

(a)(i) Regression (output is a continuous numerical value)(1)
(a)(ii) Classification (output is one of a fixed set of categories/labels)(1)
(b) For each training example the residual / error between the actual value and the value predicted by the line is found (accept “vertical distance from point to line”)(1)
(b) The slope and intercept are chosen so that the sum of the squared errors/residuals is minimised (accept “least squares”; accept “minimises the mean squared error”; do not accept “goes through as many points as possible”)(1)
Answer: (a)(i) regression (ii) classification (b) calculates the residuals and picks the slope and intercept that minimise the sum of squared residuals (least squares)
Examiner tip: “The line goes through the middle of the points” does not score: examiners want the idea of minimising the total squared error.

8HardHL only6 marks

A music app classifies songs as Workout (W) or Relax (R) using a k-nearest neighbours (K-NN) classifier with Euclidean distance. Each song is described by a tempo score and an energy score, both on a scale of 0 to 10.

Song	Tempo	Energy	Class
A	5	4	W
B	3	6	R
C	6	6	R
D	1	4	R
E	7	3	W
F	8	7	W
G	0	9	R

A new song, S, has tempo 4 and energy 4.

(a) State the class assigned to S when k = 1. [1]

(b) Determine the class assigned to S when k = 3. Show your working. [3]

(c) The developers add a third feature, song length in seconds (values range from 120 to 600), but do not rescale any of the features. Explain the effect this would have on the classifier. [2]

(a) Workout (nearest neighbour is A, distance 1)(1)
(b) At least two correct distances from S, e.g. A = 1, B = $\sqrt{5}$ ≈ 2.24, C = $\sqrt{8}$ ≈ 2.83, D = 3, E = $\sqrt{10}$ ≈ 3.16 (accept squared distances used consistently)(1)
(b) Three nearest neighbours identified as A, B and C(1)
(b) Relax, as 2 of the 3 neighbours (B and C) are R; majority vote (do not award a correct class without the correct three neighbours)(1)
(c) Song length has a much larger numerical range than tempo/energy (0–10), so differences in length dominate the distance calculation(1)
(c) so the nearest neighbours are effectively chosen by length alone and tempo and energy are almost ignored / classification becomes less accurate; features should be normalised/scaled to the same range (accept standardised)(1)
Answer: (a) Workout (b) neighbours A (1), B (2.24), C (2.83): two R, one W, so Relax (c) length values dwarf the 0–10 scores, so distance is dominated by length and the other features are ignored unless the data is normalised
Examiner tip: In (b) the answer changes from (a): always list the k nearest distances before voting, and in (c) say why the large range matters (it dominates the distance), not just “it should be normalised”.

9HardHL only6 marks

An email provider tests its spam filter on 500 emails. The results are shown in the confusion matrix, where “positive” means the email is spam.

	Actually spam	Actually not spam
Predicted spam	72	8
Predicted not spam	18	402

(a) Calculate the precision and the recall of the filter. [2]

(b) Calculate the F1 score of the filter. Give your answer to 3 significant figures. [2]

(c) The provider can change a threshold that increases recall but reduces precision. Explain why the provider is likely to prefer high precision for this application. [2]

(a) Precision = 72 / (72 + 8) = 0.9 (accept 90%)(1)
(a) Recall = 72 / (72 + 18) = 0.8 (accept 80%)(1)
(b) F1 = 2 × precision × recall / (precision + recall) with their values substituted, e.g. $\frac{2 \times 0.9 \times 0.8}{0.9 + 0.8}$ (accept $\frac{2 \times 72}{2 \times 72 + 8 + 18}$)(1)
(b) 0.847 (follow through from their precision and recall)(1)
(c) Low precision means more false positives: genuine emails are wrongly put in the spam folder(1)
(c) The user may never see an important genuine email, which is more harmful than a few spam emails reaching the inbox (false negatives), so false positives are the costlier error(1)
Answer: (a) precision 0.9, recall 0.8 (b) F1 = 0.847 (c) a false positive hides a genuine email, which costs the user more than letting some spam through
Examiner tip: Precision divides by the predicted spam row (72 + 8); recall divides by the actual spam column (72 + 18). Mixing these up costs both marks in (a).

10StretchHL only7 marks

A data scientist trains decision trees to predict whether a loan applicant will repay. The hyperparameter maximum tree depth is varied from 1 to 8. The graph shows the accuracy of each tree on the training data and on a separate validation set.

(a) State the maximum depth that should be chosen. Justify your choice. [2]

(b) Explain, with reference to the graph, why the tree with maximum depth 1 is described as underfitting. [2]

(d) After choosing the depth, the data scientist evaluates the final model on a third set of data, the test set. Explain why the validation accuracy should not be reported as the model’s final performance. [1]

(a) Depth 4(1)
(a) It gives the highest validation accuracy (0.84), i.e. best performance on unseen data (do not accept “highest training accuracy”)(1)
(b) Both training accuracy (0.68) and validation accuracy (0.66) are low / the lowest on the graph (accept values ± 0.01)(1)
(b) The tree is too simple (one split) to capture the patterns in the data, so it performs poorly even on the data it was trained on(1)
(c) Training accuracy keeps increasing (towards 1.00) while validation accuracy falls (from 0.84 to 0.73) / the gap between them widens(1)
(c) The model is overfitting: the deeper tree memorises the training data, including noise, and so generalises less well to new data(1)
(d) The depth was chosen because it scored best on the validation set, so the validation accuracy is biased/optimistic; an untouched test set gives an unbiased estimate of performance on new data (accept “the validation set has influenced the model”)(1)
Answer: (a) depth 4, highest validation accuracy (0.84) (b) training and validation accuracy are both low, so the model is too simple to learn the pattern (c) training accuracy rises to 1.00 but validation accuracy falls to 0.73: overfitting (d) the validation set was used to choose the depth, so its score is optimistically biased
Examiner tip: Choose the hyperparameter from the validation curve, not the training curve, and quote values from the graph to earn the “with reference to the graph” marks.

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