hardMultiple Choice
MLA-C01 Practice Question: A machine learning team is preparing numerical…
A machine learning team is preparing numerical features for a linear regression model. Feature 'A' ranges from 0 to 1000, feature 'B' ranges from 0 to 1, and feature 'C' ranges from -10000 to 10000. The team wants to ensure that feature scales do not affect the model's coefficients and that the features are bounded between 0 and 1. Which transformation should they apply?
⚠ Common exam trap
MLA-C01 often tests the distinction between scalers by hiding the output-range requirement in the question; candidates who fixate on 'scale does not affect coefficients' alone may wrongly pick StandardScaler, missing that only MinMaxScaler guarantees the [0, 1] bound.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
MinMaxScaler
MinMaxScaler applies the formula (x - min) / (max - min), which linearly rescales each feature to the [0, 1] range regardless of its original bounds. This directly satisfies both stated requirements: eliminating scale-driven coefficient distortion in linear regression and bounding all features between 0 and 1. It is the only option that guarantees the [0, 1] output range.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
RobustScaler (based on median and IQR)
Why it's wrong here
RobustScaler centres on the median and scales by the interquartile range, so output values are not bounded to 0 and 1 as the team requires. It is tempting because it resists outliers well, and it is the right choice when extreme values must not distort scaling and no fixed output range is needed.
- ✗
Log transformation
Why it's wrong here
Log transformation compresses skewed distributions and stabilises variance, but it does not bound values to 0–1; feature C's negatives are undefined under logarithms. It is tempting because it reduces the influence of extreme magnitudes, and it would be correct for right-skewed positive data where proportionality, not fixed range, matters.
- ✗
StandardScaler (z-score normalization)
Why it's wrong here
StandardScaler centres features on mean zero with unit variance, so outputs span negative and positive values, violating the 0–1 bound requirement. It is tempting because it equalises coefficient scales, and it would be correct when the model needs standardised inputs and no bounded range is specified.
- ✓
MinMaxScaler
Why this is correct
MinMaxScaler applies x' = (x - min) / (max - min), mapping each feature linearly onto the [0, 1] range. This directly satisfies the bounded-scale constraint and prevents the wide-ranging features A and C from dominating coefficient magnitudes in the linear regression.
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