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Lesson 5 of 6

Statistics · Lesson 5 of 6

How Data Changes: Scaling and Corrections

“Predict the effects of shifting and scaling, recover missing values, and repair statistics after data errors.”

Learning Objectives

• Prove how adding a constant changes the mean while leaving variance and standard deviation unchanged. • Prove the effects of a positive or negative multiplier on mean, variance and standard deviation. • Recover total sums and sums of squares from a known count, mean and spread. • Find missing observations from their combined sum and sum of squares. • Correct replacement and omission errors, using the updated observation count where necessary.

Changes to observations have predictable effects

You do not always need to recalculate a statistic from every observation. If every value is shifted or scaled in the same way, the definitions tell us how the mean and spread change. When one or a few observations are mistaken, we can update the original totals instead. Both kinds of problem depend on understanding what the formulas represent.

Let the original observations be xᵢ and the changed observations be yᵢ. A shift adds a constant to every observation; a scale change multiplies every observation by a constant. In this lesson a will denote a multiplier and b an added constant. These roles are different from the chosen centre symbol used earlier for mean deviation.

Adding a constant moves the centre without changing distances

Suppose yᵢ = xᵢ + b. The total of the new observations is the old total plus nb, so the new mean is ȳ = x̄ + b. Each observation and its mean move together by the same amount. Their difference therefore stays unchanged, even if b is negative.

Example — Prove invariance under a shift

Problem
Show that adding the same number b to every observation leaves variance unchanged.

  1. 1.Start with yᵢ = xᵢ + b. Averaging gives ȳ = Σ(xᵢ + b)/n = x̄ + b.
  2. 2.Subtract the new mean: yᵢ − ȳ = xᵢ + b − (x̄ + b) = xᵢ − x̄.
  3. 3.Hence each squared deviation is unchanged: (yᵢ − ȳ)² = (xᵢ − x̄)².
  4. 4.Averaging these squares gives σᵧ² = σₓ². Taking the nonnegative square root also gives σᵧ = σₓ.
  5. 5.The centre changes, but the actual distances from that centre do not. The argument holds for positive, negative or zero b.
Example — A shift in a small dataset

Problem
Compare 2, 4, 6 with the list obtained by adding 10 to each value.

  1. 1.The original mean is 4, and deviations are −2, 0, 2. Variance is (4 + 0 + 4)/3 = 8/3.
  2. 2.The shifted observations are 12, 14, 16 and their mean is 14.
  3. 3.Their deviations from 14 are again −2, 0, 2, so variance remains 8/3 and standard deviation remains √(8/3).
  4. 4.A larger mean does not automatically imply larger spread. In a shift, the centre changes while the dispersion stays fixed.
Original: 2, 4, 6048121620Mean 4Add 10: 12, 14, 16048121620Mean 14Multiply by 2: 4, 8, 12048121620Mean 8
A shift and a scale change— All three lines use the same scale. Adding 10 moves the cluster; multiplying by 2 increases its width.

Multiplying rescales deviations

Now suppose yᵢ = axᵢ. Averaging gives ȳ = ax̄, and the new deviations are yᵢ − ȳ = a(xᵢ − x̄). Squaring introduces a², so variance is multiplied by a². Standard deviation involves a square root and is therefore multiplied by |a|, not by a negative a.

Example — A doubled dataset

Problem
Twenty observations have variance 5. Find the variance after multiplying every observation by 2.

  1. 1.The original squared-deviation total is nσ² = 20 × 5 = 100.
  2. 2.Doubling all observations also doubles their mean, so every deviation doubles.
  3. 3.Every squared deviation becomes four times as large. The new total is 4 × 100 = 400.
  4. 4.The count is still 20, so new variance is 400/20 = 20. Standard deviation doubles from √5 to 2√5.
Example — Prove the general scale rule

Problem
For yᵢ = axᵢ, prove the new mean, variance and standard deviation.

  1. 1.The new mean is ȳ = Σaxᵢ/n = ax̄.
  2. 2.Each centred deviation is yᵢ − ȳ = a(xᵢ − x̄).
  3. 3.Average the squares: σᵧ² = Σa²(xᵢ − x̄)²/n = a²σₓ².
  4. 4.Taking the nonnegative square root gives σᵧ = √(a²σₓ²) = |a|σₓ.
  5. 5.If a = 0, all values become zero and spread becomes zero. If a < 0, the ordering is reversed but spread remains nonnegative.
Example — A negative multiplier reverses order

Problem
Multiply 2, 4, 6 by −3. Find the new mean, variance and standard deviation.

  1. 1.The original mean is 4 and variance is 8/3.
  2. 2.The new observations are −6, −12, −18, whose mean is −12 = −3 × 4.
  3. 3.Variance becomes (−3)² × 8/3 = 24. Directly, deviations 6, 0, −6 have squared average 72/3 = 24.
  4. 4.Standard deviation is √24 = 2√6 = 3√(8/3), not a negative number. Use the multiplier’s magnitude.

A combined change yᵢ = axᵢ + b can be understood as scaling first and shifting afterwards. The mean becomes ax̄ + b. The shift cancels from centred deviations, leaving a(xᵢ − x̄). This combines the two proofs without requiring a fresh calculation for every dataset.

Shift and scale relationshipsLaTeX
a is the scale multiplier and b the shift. The absolute value is necessary for standard deviation when a is negative.
Variance and standard deviation scale differently

Doubling every value multiplies variance by 4 but standard deviation by 2. Adding the same constant changes neither. These are changes to the whole dataset, not rules for modifying just one observation.

Recover the totals behind the summary statistics

A known mean determines the sum of observations. A known variance then determines the sum of their squares through the second-moment identity. These two totals contain enough information to solve some missing-value and correction problems even when the full original list is unavailable. Keep the observation count explicit.

Write S = Σxᵢ for the original-value total and Q = Σxᵢ² for the squared-value total. From x̄ = S/n we get S = nx̄. Rearranging σ² = Q/n − x̄² gives Q = n(σ² + x̄²). If standard deviation is supplied, square it first to obtain σ².

Recovering totalsLaTeX
S is the sum of observations and Q the sum of their squares. Q is not S²; squaring after addition creates cross terms that are absent from Q.
Example — Recover two missing observations

Problem
Five observations have mean 4.4 and variance 8.24. Three values are 1, 2 and 6. Find the other two.

  1. 1.Total S = 5 × 4.4 = 22. The known values total 9, so missing values u and v satisfy u + v = 13.
  2. 2.Squared total Q = 5(8.24 + 4.4²) = 5(8.24 + 19.36) = 138.
  3. 3.Known squares total 1² + 2² + 6² = 41, so u² + v² = 97.
  4. 4.Expand (u + v)²: 169 = 97 + 2uv, hence uv = 36.
  5. 5.Thus u and v are roots of t² − 13t + 36 = 0, or (t − 4)(t − 9) = 0.
  6. 6.The missing observations are 4 and 9, in either order. Their sum and squares meet both conditions; finding one pair does not determine which position each occupied.
Example — A second missing-value reconstruction

Problem
Eight observations have mean 9 and variance 9.25. Six are 6, 7, 10, 12, 12 and 13. Find the remaining two.

  1. 1.S = 8 × 9 = 72. The known values sum to 60, so u + v = 12.
  2. 2.Q = 8(9.25 + 81) = 722. Known squares sum to 642, so u² + v² = 80.
  3. 3.From (u + v)² = u² + v² + 2uv, obtain 144 = 80 + 2uv, giving uv = 32.
  4. 4.The equation t² − 12t + 32 = 0 factors as (t − 4)(t − 8) = 0. The missing observations are 4 and 8.
  5. 5.The count remains eight; only the values were unknown. Both the first-power total and squared total are needed.

Replacing an incorrect observation keeps the count fixed

Suppose a recorded value w should have been v. Remove w from S and add v. For Q, remove w² and add v². The number of observations stays the same because one observation is being replaced by another. Recalculate both the mean and variance using the corrected totals, rather than assuming the old mean still applies.

Replacement correctionLaTeX
w is the incorrectly recorded value and v its correct replacement. For several replacements, subtract and add each corresponding value and square.
Example — Correct a mistaken value

Problem
A dataset of 100 observations was reported with mean 40 and standard deviation 5.1 after recording 50 instead of 40 once. Find the correct mean and standard deviation.

  1. 1.Incorrect totals are S = 100 × 40 = 4000 and Q = 100(5.1² + 40²) = 100(26.01 + 1600) = 162601.
  2. 2.Replace 50 by 40: S′ = 4000 − 50 + 40 = 3990, while Q′ = 162601 − 2500 + 1600 = 161701.
  3. 3.The corrected mean is 3990/100 = 39.9.
  4. 4.Corrected variance is 161701/100 − 39.9² = 1617.01 − 1592.01 = 25.
  5. 5.Corrected standard deviation is √25 = 5. Updating only the total S would repair the mean but leave the spread calculation incomplete.

Omitting observations changes the count as well

An omitted observation is removed entirely, with no replacement. Remove its value from S, its square from Q, and one from the count. For several omissions subtract all their values and squares and reduce the count by how many were omitted. The denominator therefore differs from a replacement problem.

Omission correctionLaTeX
r is the number of omitted observations, and wⱼ are their recorded values. Then use x̄′ = S′/n′ and (σ′)² = Q′/n′ − (x̄′)², provided n′ > 0.
Example — Omission versus replacement

Problem
Twenty observations have mean 10 and standard deviation 2. An entry 8 is incorrect. Find corrected statistics if it is omitted, and if it is replaced by 12.

  1. 1.Original totals are S = 20 × 10 = 200 and Q = 20(2² + 10²) = 2080.
  2. 2.For omission, n′ = 19, S′ = 200 − 8 = 192 and Q′ = 2080 − 64 = 2016.
  3. 3.The new mean is 192/19 ≈ 10.1053. Variance is 2016/19 − (192/19)² = 1440/361; standard deviation is √1440/19 ≈ 1.9972.
  4. 4.For replacement by 12, the count remains 20. S′ = 200 − 8 + 12 = 204 and Q′ = 2080 − 64 + 144 = 2160.
  5. 5.The mean is 204/20 = 10.2. Variance is 2160/20 − 10.2² = 3.96; standard deviation is √3.96 ≈ 1.99.
  6. 6.The two operations create different datasets. Do not reuse denominator 20 for the omitted-item case.
Example — Omit three incorrect entries

Problem
One hundred observations have mean 20 and standard deviation 3. Recorded values 21, 21 and 18 are removed. Find the new mean and standard deviation.

  1. 1.Original totals are S = 2000 and Q = 100(9 + 400) = 40900.
  2. 2.Removed values sum to 60; their squares sum to 441 + 441 + 324 = 1206.
  3. 3.New count is 97, with S′ = 1940 and Q′ = 39694. Thus the new mean is 1940/97 = 20.
  4. 4.Variance is 39694/97 − 400 = 894/97 ≈ 9.2165.
  5. 5.Standard deviation is √(894/97) ≈ 3.04. Removing observations close to the mean does not necessarily reduce the remaining dataset’s spread.
Correct squares separately and update the denominator

Removing w from S does not mean removing w from Q; Q needs w². Replacing one entry leaves the count fixed, whereas omitting it reduces the count. Use the corrected mean inside the corrected variance formula.

Check your understanding

Choose the type of operation first: a uniform transformation, a replacement or an omission. Recover totals from the old statistics when needed, then update count, sum and squared sum consistently.

Quiz

Quick check

Adding 7 to every observation does what to variance?

Quick check

Multiplying every observation by −3 does what to standard deviation?

Quick check

Twenty observations have variance 5. What is their squared-deviation total?

Quick check

Five observations have mean 4.4 and variance 8.24. What is Σxᵢ²?

Quick check

Replacing a recorded 8 by 12 changes Q = Σxᵢ² by how much?

Quick check

Omitting one observation from a list of 20 changes the count to what?

Quick check

After yᵢ = 2xᵢ − 5, an original mean 10 becomes what?

Quick check

If two missing observations have sum 13 and squared sum 97, what is their product?

Practice Problems

Practice Problems
  1. Prove the mean and variance rules for yᵢ = axᵢ + b, then explain why standard deviation requires |a|.
  2. Six observations have mean 8 and standard deviation 4. Find both statistics after multiplying every value by 3.
  3. Seven observations have mean 8 and variance 16. Five values are 2, 4, 10, 12 and 14. Recover the remaining two.
  4. A dataset has mean 12 and standard deviation 3. Find the mean, variance and standard deviation after the change yᵢ = −2xᵢ + 5.
  5. Explain why correcting a single value cannot generally be handled by the uniform-shift or uniform-scale rules.
  6. Recalculate the omission and replacement cases for the twenty-observation example without rounding the intermediate means.
  7. For the three-item omission example, explain why the mean stays 20 while the standard deviation increases slightly.
  8. Describe a correction procedure for two wrong entries, stating how count, total and squared total change.

Key Takeaways

Key Takeaways

• Adding a constant shifts the mean but leaves centred deviations and spread unchanged. • A scale multiplier a changes variance by a² and standard deviation by |a|. • Known mean and variance determine S = nx̄ and Q = n(σ² + x̄²). • Missing values can be reconstructed from their required sum and sum of squares when enough information is available. • Replacement updates totals but leaves count fixed; omission updates totals and reduces count. • Recalculate corrected variance using the corrected mean and the correct denominator.

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Grouped Variance and Shortcut Calculations

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Chapter Summary and Practice