Transformations: What Shifting and Scaling Do

You’ll be able to

Predict the effect of adding a constant on center, spread and shape
Predict the effect of multiplying by a constant on center, spread and shape
Explain why standardizing produces a distribution with mean 0 and standard deviation 1

Adding a constant moves, it does not stretch

Add the same number c to every value and the whole distribution slides along the axis. Every measure of center — mean, median, quartiles, percentiles — increases by c. Every measure of spread — standard deviation, IQR, range — is unchanged, because the gaps between observations are exactly what they were. The shape is unchanged too. This is intuitive once you picture the histogram physically sliding: the picture is identical, the axis labels have shifted.

Multiplying stretches everything

Multiply every value by a constant a and both center and spread are multiplied. The mean and median become a times what they were; the standard deviation and IQR become |a| times what they were — the absolute value, because a spread cannot be negative. The variance, being the square of the standard deviation, is multiplied by a². Shape is again unchanged: stretching a histogram horizontally does not create or remove skew.

Linear transformation y = a·x + b

mean_y = a·mean_x + b median_y = a·median_x + b SD_y = |a|·SD_x IQR_y = |a|·IQR_x variance_y = a²·variance_x

The additive constant b appears in the centers and never in the spreads. This one line answers every transformation question on the exam.

Standardizing is a transformation, not a new idea

The z-score z = (x − μ)/σ is exactly a linear transformation with b = −μ/σ and a = 1/σ. Subtracting μ shifts the center to 0; dividing by σ scales the spread to 1. So a standardized distribution always has mean 0 and standard deviation 1 — by construction, not by coincidence. Crucially, standardizing does not make a distribution Normal. A skewed distribution standardized is a skewed distribution with mean 0, and treating a z-score as a Normal percentile without checking shape is a common and costly error.

Worked example

A set of temperatures has mean 20 °C and standard deviation 4 °C. Converting to Fahrenheit uses F = 1.8C + 32. Find the mean, standard deviation and variance in Fahrenheit.

1.This is a linear transformation with a = 1.8 and b = 32.
2.Mean: 1.8(20) + 32 = 36 + 32 = 68 °F.
3.Standard deviation: |1.8|(4) = 7.2 °F. The +32 does not appear, because shifting every value leaves the gaps unchanged.
4.Variance: (1.8)²(16) = 3.24(16) = 51.84, which is also 7.2² as it must be.

Answer: Mean 68 °F, standard deviation 7.2 °F, variance 51.84 °F². The shape of the distribution is unchanged by the conversion.

Checkpoint

Every score in a class is increased by 5 points. Which statistic is unchanged?

Watch out

Standardizing does not normalize. A z-score tells you how many standard deviations from the mean an observation lies, which is meaningful for any distribution — but converting it to a percentile with a Normal table is only valid if the distribution was approximately Normal to begin with.

Checkpoint

If every value in a data set is multiplied by 3, the variance is multiplied by —

Answer the 2 checkpoints as you read.