01 · Shift a mean
A dataset has mean 18 and SD 4. Add 7 to every observation. Give the new mean and SD.
Hint
Only the centre changes.
Worked solution
New mean = 25; SD = 4.
Understand · explore · practise
Coding and transformations of data
Transform means, variances and standard deviations under shifts and signed scales, then decode coded summaries. Explore fifteen transformations and original worked practice.
Before you startMeans, variance, standard deviation and solving linear equations.
01 / Transform every observation
A coding rule changes the variable, not just the mean.
If y = ax + b, then mean(y) = a mean(x) + b.
Variance(y) = a² variance(x); SD(y) = |a| SD(x).
The added constant shifts all values equally, so it changes the centre without changing distances between observations. Multiplication changes distances by its absolute magnitude. A negative multiplier also reverses their order.
Move the centre; scale the spreadExplore
Original values: 2, 4, 6. Mean 4; descriptive variance 8/3.
y = x; values: 2, 4, 6.
Mean = 4.
Variance = 8/3; SD ≈ 1.632993.
A positive multiplier preserves order.
02 / Add or subtract a constant
Differences from the mean stay unchanged.
Every recorded journey time is reduced by two minutes.Worked example
New mean = old mean − 2
The centre shifts with every observation.
(x − 2) − (x̄ − 2) = x − x̄
The deviation from the mean is unchanged.
Variance and SD are unchanged
The same squared deviations are being averaged.
02 · Shift a variance
Subtract 100 from every value in a dataset with variance 9. Find the new variance.
Hint
The subtraction cancels in each deviation.
Worked solution
The variance remains 9; it is not 9 − 100.
03 / Multiply distances
Variance and SD scale differently.
Watch: shifts cancel but scales remain
Pause, replay or seek freely. The notes explain the same idea and stay in view.
If every observation doubles, its deviation from the new mean doubles. Its squared deviation is multiplied by four. Thus SD doubles while variance quadruples.
03 · Positive scale
A dataset has mean 5, variance 9 and SD 3. Multiply every value by 4. Give all three summaries.
Hint
Mean and SD multiply by four; variance by sixteen.
Worked solution
Mean = 20; variance = 144; SD = 12.
04 · Change units
A duration dataset in hours has mean 1.5 and SD 0.2. Convert the mean, SD and variance to minutes.
Hint
Multiply observations by 60.
Worked solution
Mean = 90 minutes; SD = 12 minutes; variance = 144 minutes². The original variance 0.04 hours² is multiplied by 60².
04 / Handle negative and zero multipliers
Standard deviation cannot be negative.
05 · Reverse the scale
Mean(x) = 6 and SD(x) = 2. For y = 10 − 3x, find mean(y), variance(y) and SD(y).
Hint
The multiplier is −3.
Worked solution
Mean(y) = 10 − 18 = −8. SD(y) = 3×2 = 6; variance(y) = 36.
06 · Zero multiplier
For y = 0x + 7, find the mean and SD of y for any non-empty dataset x. Can you recover x?
Hint
Every transformed observation is seven.
Worked solution
Mean(y) = 7 and SD(y) = 0. No: the transformation discards all information about individual x values.
05 / Invert the coding equation
Undo the entire transformation.
Coded values satisfy y = (x − 100)/5, with mean(y) = 2 and SD(y) = 3.Worked example
x = 5y + 100
Rearrange before translating summaries.
Mean(x) = 5×2 + 100 = 110
Restore both scale and location.
SD(x) = 5×3 = 15; variance(x) = 225
No 100 is added to the spread.
07 · Decode a mean and variance
y = (x − 50)/10 has mean(y) = −0.4 and variance(y) = 2. Find mean(x), variance(x) and SD(x).
Hint
x = 10y + 50.
Worked solution
Mean(x) = 46; variance(x) = 100×2 = 200; SD(x) = 10√2 ≈ 14.142.
08 · Decode a negative scale
y = (20 − x)/4 has mean(y) = 3 and SD(y) = 1.5. Find mean(x) and SD(x).
Hint
x = 20 − 4y.
Worked solution
Mean(x) = 8; SD(x) = 4×1.5 = 6. The negative multiplier does not make the SD negative.
06 / Use coded sums
Compute coded statistics before decoding.
09 · Coded totals
Ten coded values y = (x − 80)/4 have Σy = 15 and Σy² = 45. Find the original mean and descriptive variance.
Hint
Mean(y) = 1.5; variance(y) = 45/10 − 1.5².
Worked solution
Coded variance = 2.25. Mean(x) = 4×1.5 + 80 = 86. Variance(x) = 16×2.25 = 36, so SD(x) = 6.
10 · Recover an original total
For the same ten observations, find Σx directly from Σy.
Hint
Sum x = 4y + 80 across ten observations.
Worked solution
Σx = 4Σy + 10×80 = 60 + 800 = 860. Add the offset once per observation, not once for the whole sum.
07 / Track order as well as scale
A negative transformation swaps lower and upper positions.
For a positive multiplier, quartile values transform with the same formula as observations under the stated rank convention. For a negative multiplier, ordering reverses: the new lower quartile comes from the old upper quartile. Range and IQR scale by |a|; adding b does not change them.
11 · Reverse quartiles
Using the same quartile convention, Q₁(x) = 4 and Q₃(x) = 10. For y = 20 − 2x, find the new quartiles and IQR.
Hint
The largest original values become the smallest transformed values.
Worked solution
Q₁(y) = 20 − 2×10 = 0; Q₃(y) = 20 − 2×4 = 12; IQR(y) = 12 = 2×(10 − 4).
12 · Diagnose an SD rule
A learner uses SD(3x + 5) = 3 SD(x) + 5. Explain the error.
Hint
An equal shift cannot change distances.
Worked solution
Only the scale affects SD. The correct result is 3 SD(x); adding five shifts the mean and each observation equally.
08 / Decode centre and spread separately
Write the inverse rule first.
Means transform with both scale and offset. Standard deviations use the absolute scale alone; variances use its square. Keep track of units and do not use an inverse transformation when the multiplier is zero.
Section 1 of 8 · Transform every observation