Data preprocessing: cleaning, feature selection, dimensionality reduction (HL) – IB Computer Science exam questions with mark scheme

1CoreHL only6 marks

A clinic is preparing data to train a machine learning model that predicts a patient's risk of type 2 diabetes. An extract of the raw data is shown.

VisitID	Age	Mass (kg)	Smoker	Visit date	Receptionist
V101	34	72.5	Yes	2026-03-14	Ms Patel
V102	51		No	2026-03-15	Mr Okafor
V103	245	80.1	No	2026-03-15	Ms Patel
V104	29	65.0	Y	16/03/2026	Mr Okafor
V105	62	90.3	No	2026-03-17	Ms Patel
V105	62	90.3	No	2026-03-17	Ms Patel

(a) Identify four different data-quality problems in this extract. For each, refer to a specific record or field. [4]

(b) Explain why the Receptionist field should be removed before the model is trained. [2]

(a) Missing data: V102 has no value for Mass (accept blank / null field)(1)
(a) Incorrect / impossible value: V103 Age 245 (accept outlier / invalid age; must name V103 or 245)(1)
(a) Inconsistent formatting: V104 Smoker recorded as "Y" instead of "Yes" or V104 date in DD/MM/YYYY instead of YYYY-MM-DD (either for one mark; do not award twice)(1)
(a) Duplicate record: V105 appears twice (accept duplicate rows)(1)
(b) Receptionist is irrelevant data: it has no (causal) relationship with diabetes risk(1)
(b) Keeping it adds noise / could let the model learn a spurious pattern (e.g. linked to which patients a receptionist sees), reducing accuracy / generalisation (accept wastes computation / increases dimensionality)(1)
Answer: (a) missing mass (V102); impossible age 245 (V103); inconsistent format Y / date (V104); duplicate V105. (b) irrelevant to the outcome, adds noise / spurious correlations.
Examiner tip: Each problem must be tied to a record; "some data is missing" without saying where scores nothing.

2StandardHL only6 marks

An estate agent is building a K-nearest neighbours (K-NN) model to estimate house prices. Two of the features are floor area (m²) and number of bedrooms. The floor areas in the training data are:

48, 65, 72, 90, 110, 160

The data is normalized using min–max scaling: $$x' = \frac{x - x_{\min}}{x_{\max} - x_{\min}}$$

(a) Calculate the normalized value of a floor area of 90 m². [2]

(b) Explain why the features should be normalized before the K-NN model is used. [2]

(c) A new house with a floor area of 175 m² is to be valued. Using the training minimum and maximum, calculate its normalized floor area and comment on the result. [2]

(a) $\frac{90-48}{160-48}$ seen (accept $\frac{42}{112}$)(1)
(a) 0.375 (accept 0.38, 3/8)(1)
(b) K-NN uses distances between data points / feature values are combined in a distance calculation(1)
(b) Floor area has a much larger range (tens–hundreds) than bedrooms (1–5), so without scaling it would dominate the distance / bedrooms would have almost no influence; normalizing gives each feature equal weight(1)
(c) $\frac{175-48}{112}$ = 1.13 (accept 1.134, 1.1)(1)
(c) The value lies outside the range 0–1 because the new value exceeds the training maximum; the same training min/max must still be used (do not re-scale with the new value) / the model may extrapolate poorly(1)
Answer: (a) 0.375 (c) 1.13, outside 0–1
Examiner tip: In (b), say which feature dominates and why: "to make the data the same" is not enough.

3CoreHL only5 marks

A fitness app records the resting heart rate (beats per minute) of five users:

58, 62, 64, 66, 70

Before training a model, this feature is standardized using $z = \frac{x - \mu}{\sigma}$, where $\mu$ is the mean and $\sigma$ is the standard deviation $\sigma = \sqrt{\frac{\sum (x-\mu)^2}{n}}$.

(a) Calculate the mean, $\mu$. [1]

(b) Calculate the standard deviation, $\sigma$. [1]

(d) Outline two differences between standardization and min–max normalization. [2]

(a) 64(1)
(b) 4 (working: $\sqrt{(36+4+0+4+36)/5} = \sqrt{16}$)(1)
(c) $(70-64)/4 = 1.5$ (follow through from (a) and (b); allow 1.34 if sample standard deviation 4.47 used)(1)
(d) Standardization rescales to mean 0 and standard deviation 1, whereas normalization rescales to a fixed range (usually 0 to 1)(1)
(d) Standardized values are not bounded (can be negative or greater than 1) / standardization is less distorted by extreme outliers, since normalization depends on the min and max / standardization suits roughly normally distributed data(1)
Answer: (a) 64 (b) 4 (c) 1.5
Examiner tip: Normalization gives 0–1; standardization gives mean 0, SD 1: mixing these up loses (d).

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4StandardHL only5 marks

A teacher collected data on hours of revision and test score for ten students to train a linear regression model. The scatter graph shows the data.

(a) Identify the outlier, giving its coordinates. [1]

(b) Describe the effect that keeping this outlier would have on the regression line. [2]

(a) (2, 95): 2 hours, 95% (allow ±1 on score)(1)
(b) The line is pulled towards the outlier: the intercept is raised / the gradient (slope) is made shallower(1)
(b) The line no longer fits the other nine points well / predictions become less accurate / r² falls (allow the fit is weakened)(1)
(c) The outlier may be a genuine (valid) value, e.g. a student with strong prior knowledge, and removing it would lose real information / bias the data(1)
(c) Investigate / check the source of the value first: remove or correct it only if it is an error (e.g. data-entry mistake); otherwise keep it (accept: cap it, or use a method less sensitive to outliers)(1)
Answer: (a) (2, 95) (b) pulls the line up at low hours, shallower gradient, poorer fit (c) may be genuine; check the cause first
Examiner tip: Say which way the line moves (higher intercept, lower gradient); "it changes the line" earns nothing.

5StandardHL only6 marks

A bank is training a model to predict loan default. One field is annual income (£ thousand). Seven applicants have the following values, one of which is missing:

21, 24, 25, 27, 30, 180, missing

(a) Calculate the value that would replace the missing entry using (i) mean imputation and (ii) median imputation. [2]

(b) Justify which imputed value is more suitable for this data. [2]

(c) Describe one alternative way of handling missing data, other than imputation, and state one disadvantage of it. [2]

(a)(i) Mean = 307 ÷ 6 = 51.2 (£ thousand) (accept 51.17, 51; £51 200)(1)
(a)(ii) Median = (25 + 27) ÷ 2 = 26 (£ thousand)(1)
(b) The median, because the value 180 is an outlier / the data is skewed(1)
(b) The outlier pulls the mean far above the typical value (51.2 is higher than five of the six known incomes), so the median better represents a typical applicant(1)
(c) Delete the record / row with the missing value (accept: delete the whole field if many values are missing; flag missing as a separate category; ask the applicant again)(1)
(c) Disadvantage matching the method, e.g. deletion reduces the size of the training set / can introduce bias if values are not missing at random / deleting a field loses a possibly useful feature(1)
Answer: (a)(i) 51.2 (ii) 26 (b) median, because 180 is an outlier
Examiner tip: Divide by 6, not 7: the missing value is not part of the total.

6StandardHL only6 marks

Feature selection reduces the number of input features used to train a model.

Describe how each of the following methods selects features.

(a) filter methods [2]

(b) wrapper methods [2]

(a) Each feature is scored using a statistical measure of its relationship with the target (e.g. correlation, chi-squared, information gain / mutual information)(1)
(a) Done before / independently of training any model; features below a threshold (or outside the top k) are removed, so the method is fast(1)
(b) Subsets of features are chosen and a model is trained and evaluated on each subset (e.g. by validation accuracy)(1)
(b) A search strategy (e.g. forward selection, backward elimination) adds or removes features, keeping the subset that performs best; accurate but computationally expensive(1)
(c) Feature selection happens as part of the model training process itself(1)
(c) Example of the mechanism, e.g. L1 (lasso) regularization shrinks the weights of unhelpful features to zero / a decision tree only splits on the most informative features (accept feature importance from a random forest)(1)
Answer: Filter: statistical score, no model. Wrapper: train and evaluate models on feature subsets. Embedded: selection built into training (e.g. lasso, decision trees).
Examiner tip: Wrapper and embedded both use a model: the difference is whether models are retrained for each subset (wrapper) or selection happens during one training run (embedded).

7HardHL only6 marks

A filter method is used to select features for a model that predicts the sale price of a house. The correlation coefficient between each feature and the sale price is shown.

Feature	Correlation with sale price, r
Floor area	0.82
Number of rooms	0.78
Distance to station	−0.61
House number	0.03
Day of week listed	−0.05

The correlation between floor area and number of rooms is 0.93.

A feature is removed if $|r| < 0.2$.

(a) Identify the features that are removed. [1]

(b) Explain why distance to station is kept even though its correlation is negative. [1]

(d) Outline one limitation of selecting features using correlation alone. [1]

(a) House number and day of week listed (both needed)(1)
(b) The size / magnitude |r| = 0.61 shows a fairly strong (inverse) relationship; the sign only gives the direction (price falls as distance rises)(1)
(c) The two features are highly correlated with each other (0.93), so they carry largely the same / redundant information(1)
(c) Removing one loses little predictive information but reduces dimensionality / training time / risk of overfitting (accept multicollinearity makes model weights unstable)(1)
(c) Remove number of rooms, because floor area has the higher correlation with sale price (0.82 > 0.78)(1)
(d) Correlation only measures linear relationships / each feature is judged on its own, so interactions between features (useful in combination) are missed(1)
Answer: (a) house number, day of week listed (c) remove number of rooms (redundant with floor area, weaker r)
Examiner tip: A negative r is not a weak r: filter on the absolute value.

8HardHL only6 marks

A wrapper method using forward selection is applied to four features A, B, C and D. It starts with no features. In each round it trains a model for every way of adding one more feature to the current set, then adds the feature that gives the highest validation accuracy. It stops when the best new model improves accuracy by less than 0.02 on the current set.

Features used	Validation accuracy
A	0.71
B	0.64
C	0.69
D	0.58
A, B	0.74
A, C	0.79
A, D	0.72
A, B, C	0.80
A, C, D	0.78

(a) Trace the forward selection to find the final set of features. Show the accuracy after each round. [3]

(b) State the number of models trained during the search. [1]

(c) An exhaustive search would train a model for every non-empty subset of features. Calculate how many models this would need for 20 features, and hence explain why wrapper methods usually use a search strategy such as forward selection. [2]

(a) Round 1: A chosen (0.71, highest single feature)(1)
(a) Round 2: C added, {A, C} = 0.79 (improvement 0.08 ≥ 0.02)(1)
(a) Round 3: best is {A, B, C} = 0.80, improvement only 0.01 < 0.02 so stop; final set {A, C}(1)
(b) 9 (4 + 3 + 2)(1)
(c) $2^{20} - 1$ = 1 048 575 models(1)
(c) Each model must be trained and validated, so exhaustive search is far too computationally expensive / time-consuming; forward selection needs at most 20 + 19 + … + 1 = 210 models (accept any valid statement that it grows exponentially)(1)
Answer: (a) {A, C}, accuracy 0.79 (b) 9 (c) 1 048 575
Examiner tip: Check the stopping rule in each round: adding B to {A, C} gains only 0.01, so B is not added.

9HardHL only6 marks

A classifier is trained on a fixed set of 200 examples. The graph shows its validation accuracy as more features are used.

(a) State the number of features that gives the highest validation accuracy. [1]

(b) Explain why the accuracy falls when more features are added beyond this point, even though more information is available. [3]

(c) Suppose the range of every feature is divided into 5 equal intervals, giving $5^d$ cells for $d$ features. Calculate the average number of training examples per cell when (i) 3 features and (ii) 8 features are used. [2]

(a) 8(1)
(b) As dimensions increase, the volume of the feature space grows exponentially, so the fixed 200 examples become sparse (curse of dimensionality)(1)
(b) There is not enough data to cover the space / learn reliable patterns, so the model overfits the training data (fits noise / irrelevant features)(1)
(b) Distances between points become similar / less meaningful in high dimensions, so distance-based methods (e.g. K-NN) cannot distinguish near from far neighbours (accept added features may be irrelevant / noisy)(1)
(c)(i) $200 / 5^3 = 200/125 = 1.6$(1)
(c)(ii) $200 / 5^8 = 200/390\,625 = 0.000512$ (accept 5.12 × 10⁻⁴, 0.0005)(1)
Answer: (a) 8 (c)(i) 1.6 (ii) 0.000512
Examiner tip: Link the fall to sparse data and overfitting; "too many features confuse the model" earns nothing.

10StretchHL only6 marks

A school security system uses a machine learning model to recognise students' faces. Each image is a 64 × 64 pixel greyscale image, and each pixel is used as a separate input feature. The training set contains 1500 images.

Discuss the importance of applying dimensionality reduction to this data before training the model. [6]

Number of features = 64 × 64 = 4096, which is large compared with 1500 training examples (more features than examples)(1)
Curse of dimensionality: data is sparse in a 4096-dimensional space, so the model is likely to overfit and generalise poorly to new faces(1)
Many pixels are redundant / highly correlated with neighbouring pixels, or irrelevant (e.g. background), so they can be combined or removed with little loss of information(1)
Dimensionality reduction (e.g. principal component analysis) transforms the pixels into a smaller set of new features that keep most of the variation (e.g. overall face shape)(1)
Benefits: faster training and recognition, less memory / storage, distance-based comparison between faces becomes more meaningful, data can be visualised(1)
Drawbacks / evaluation: some information is lost, which may reduce accuracy if important detail is discarded; the new features are harder to interpret / explain; the number of dimensions kept must be tuned (e.g. by validation)(1)
Balanced conclusion: reduction is important here because the feature count far exceeds the sample size, provided enough variance is retained(1)

Level	Marks	What it looks like
Level 3	5–6	Discussion applied to the face-image context, quantifying the feature count against the training set size; explains the curse of dimensionality and redundancy, and weighs benefits against drawbacks to reach a supported conclusion.
Level 2	3–4	Explains some reasons dimensionality reduction helps (e.g. overfitting, speed) with some reference to the context; drawbacks missing or undeveloped.
Level 1	1–2	Lists isolated points about reducing features or speed with little explanation or link to the context.
No marks	0	Nothing worthy of credit.

Answer: Indicative: 4096 features vs 1500 images; curse of dimensionality and overfitting; redundant correlated pixels; PCA keeps most variance; faster, less memory; but information loss and less interpretable.
Examiner tip: "Discuss" needs both benefits and drawbacks: candidates who only list advantages are capped in the middle band.

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