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Adding five to every observation moves the whole data set five units along the number line. The mean increases by five, but each observation stays the same distance from the new mean. Standard deviation therefore stays unchanged. It measures spread around the mean, not the height of the mean above zero.
This guide develops that result with small invented data sets, exact calculations, transformation comparisons and answered practice. It concerns descriptive statistics for a finite data set, using the standard deviation formula with division by the number of observations.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–29
Start with the data 2, 4, 6. Their mean is four. Their deviations from the mean are negative two, zero and positive two. Add five to every observation to obtain 7, 9, 11. The mean becomes nine, and the deviations are still negative two, zero and positive two.
Since those deviations are unchanged, their squares are unchanged. The mean of those squares, called the variance here, is unchanged. Taking its square root gives the same standard deviation. In this example the variance is 8/3 and the standard deviation is √(8/3), approximately 1.633.
The distinction is between adding a constant to every value and multiplying every value by a constant. Adding moves the data without stretching their spacing. Multiplying can stretch or compress the spacing. Keep those operations separate before applying any memorised transformation rule.
The SEAB 2027 G3 Mathematics syllabus includes statistical measures of central tendency and spread. Understanding what these measures describe helps students interpret a result instead of treating a calculator output as an unexplained number.
For a nonempty data set x₁, x₂, …, xₙ, write its mean as x̄ = (x₁ + x₂ + … + xₙ)/n. The variance used throughout this guide is the sum of the squared deviations from x̄ divided by n. Standard deviation is the nonnegative square root of that variance.
Some statistical settings use a sample standard deviation with an n − 1 denominator. We are not switching between those conventions here. The shift-invariance idea also holds with that other denominator when it is defined, but mixing calculator settings can obscure an otherwise simple school-level comparison.
For the broader concept, read eduKate’s guide to how variance works. This article focuses on one common interpretation error: assuming that a higher mean automatically creates a higher standard deviation.
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The mean of 2, 4, 6 is four because their total is twelve and there are three observations. The mean of 7, 9, 11 is nine because their total is twenty-seven and there are still three observations. Both calculations describe a central value.
Knowing that one mean exceeds another by five does not tell us that the second group is more spread out. To assess spread, look at how far the observations sit from their own group’s centre, or use an appropriate measure such as standard deviation.
In the first group, the outer values lie two units below and above the mean. In the shifted group, the outer values also lie two units below and above their new mean. Both groups occupy an interval four units wide.
A higher position on the number line and a wider distribution are different properties. This distinction is especially important when interpreting tables that report only a mean and a standard deviation: one number cannot automatically stand in for the meaning of the other.
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For the original data 2, 4, 6, subtract the mean four from each observation. The deviations are 2 − 4 = −2, 4 − 4 = 0 and 6 − 4 = 2. A deviation records both direction and distance relative to the mean.
The negative and positive deviations add to zero. That is not evidence of zero spread. It is why simply averaging signed deviations cannot provide the intended measure of dispersion around the arithmetic mean.
Instead, square the deviations. The squared values are four, zero and four, whose total is eight. Divide by the number of observations, three, to obtain variance 8/3. The value is positive, correctly reflecting that the observations are not all identical.
Finally take the square root: standard deviation = √(8/3). Keep this exact form until a question requests rounding. If we give a decimal to three decimal places, it is 1.633. Rounding early is unnecessary and can make comparisons harder to inspect.
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Adding five to each original value gives 7, 9, 11. Because there are three observations, the total increases by three lots of five, which is fifteen. The original total twelve becomes twenty-seven. Dividing by three makes the mean increase by exactly five.
Now compare each new value with the new mean nine: 7 − 9 = −2, 9 − 9 = 0 and 11 − 9 = 2. These are exactly the deviations from the original data set, in the same observation order.
Nothing in the standard deviation calculation has changed after this subtraction. The squared deviations still total eight, and the denominator is still three. Both variance and standard deviation are therefore the same as before.
The essential cancellation is easy to see in ordinary words. Each observation moves up by five, and the reference mean also moves up by five. Their relative separation remains fixed. It would be wrong to move the observations but keep measuring from the old mean.
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| Quantity | Data 2, 4, 6 | Data 7, 9, 11 |
|---|---|---|
| Number of observations | 3 | 3 |
| Total | 12 | 27 |
| Mean | 4 | 9 |
| Deviations from the respective mean | −2, 0, 2 | −2, 0, 2 |
| Sum of squared deviations | 8 | 8 |
| Variance | 8/3 | 8/3 |
| Standard deviation | √(8/3) | √(8/3) |
| Range | 4 | 4 |
The total and mean change, but several measures of separation remain fixed. Reading across the rows shows why a transformation does not affect every statistic in the same way. Each statistic is built from a different operation on the data.
Use this table as a comparison, not a substitute for reasoning. If a later question changes multiplication, rounding, frequency or sample size, revisit the calculation. The unchanged-spread conclusion belongs to an exact common shift applied to every observation in an otherwise unchanged data set.
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Let the new observation be yᵢ = xᵢ + c, where c is the same constant for every observation. The new total is the old total plus nc. Dividing by n gives the new mean ȳ = x̄ + c.
Subtract the new mean from a new observation: yᵢ − ȳ = (xᵢ + c) − (x̄ + c) = xᵢ − x̄. The added c and subtracted c cancel. Every new deviation equals its matching old deviation.
Squaring preserves those equalities. Adding all the squared deviations and dividing by the same n therefore gives equal variances. Since both standard deviations are defined as nonnegative square roots, they are equal too.
This is a general argument rather than a pattern guessed from three convenient numbers. It works for negative constants, positive constants, decimal constants and zero, provided the data are finite numerical observations and the same constant is applied to each one.
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Subtract ten from every value of 2, 4, 6. The new data are −8, −6, −4. Their mean is negative six. Their deviations from that mean are −8 − (−6) = −2, −6 − (−6) = 0 and −4 − (−6) = 2.
The squared deviations are again four, zero and four. Variance remains 8/3 and standard deviation remains √(8/3). Moving below zero does not make standard deviation negative or smaller by itself.
Standard deviation cannot be negative because it is a nonnegative square root. Negative observations are perfectly compatible with a positive standard deviation. The statistic describes distances from the mean, not whether the observations are positive or negative.
This is a useful place to check signed subtraction. The expression −8 − (−6) means negative eight plus six, giving negative two. A sign error here can create a false difference between original and shifted deviations and hide the cancellation the transformation should produce.
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The same variance can be calculated as the mean of the squared observations minus the square of the mean. For 2, 4, 6, the sum of squares is 4 + 16 + 36 = 56. Thus variance = 56/3 − 4² = 56/3 − 16 = 8/3.
For 7, 9, 11, the sum of squares is 49 + 81 + 121 = 251. The new mean is nine, so variance = 251/3 − 9² = 251/3 − 81 = 8/3. The larger squared observations are balanced by the larger squared mean.
This calculation explains a common partial observation. Someone may notice that adding five makes the observations’ squares much larger and conclude that variance must increase. That ignores the subtraction of the square of the new mean in this formula.
Both parts of the expression must be updated consistently. Using the new sum of squares together with the old mean would calculate neither the old variance nor the new variance. It is a mixed calculation with no valid interpretation as the requested statistic.
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If yᵢ = xᵢ + c, then yᵢ² = xᵢ² + 2cxᵢ + c². Averaging across the n observations gives mean of y² = mean of x² + 2cx̄ + c². The middle term uses the mean of the original observations.
The new mean is x̄ + c, whose square is x̄² + 2cx̄ + c². Subtracting this square from the new mean of squares cancels the two additional terms, leaving mean of x² − x̄².
That is the original variance. This route reaches the same conclusion as comparing deviations directly, although it needs more algebra. It is useful when a question provides a total and a sum of squares instead of listing the observations.
Choose the simpler route for the information available. If the question asks only what happens to standard deviation after a common shift, a clear transformation explanation is enough. Recalculating a large data table is not necessary unless the question explicitly requests the detailed numerical working.
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Multiply 2, 4, 6 by two to obtain 4, 8, 12. The mean becomes eight. Relative to that new mean, the deviations are negative four, zero and positive four. Each deviation has doubled compared with the original data.
Their squares are sixteen, zero and sixteen, so the variance is 32/3. That is four times the original variance 8/3. The standard deviation is √(32/3) = 2√(8/3), twice the original standard deviation.
Multiplication stretches the spacing between observations and their mean. Addition merely translates those positions together. This is why the rule for multiplication cannot be borrowed for addition, despite both operations producing larger observations in this particular example.
For a general multiplier a, variance is multiplied by a² and standard deviation by |a|. The absolute value matters because standard deviation must remain nonnegative. A multiplier of zero collapses all observations to the same value and gives zero spread.
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Multiply the original values by negative two: −4, −8, −12. Their mean is negative eight. In the original observation order, their deviations are positive four, zero and negative four. The signs have reversed and the magnitudes have doubled.
Squaring removes the signs, giving sixteen, zero and sixteen. The variance is 32/3, and the standard deviation is 2√(8/3). It is not negative two times the original standard deviation; it is positive two times that value.
The order on the number line has reversed, but the distances are stretched by a factor of two. This is a clean way to understand the absolute value in the transformation rule without treating it as an isolated symbol to memorise.
If a transformation combines scaling and shifting, y = ax + b, the new mean is a times the old mean plus b. The new standard deviation is |a| times the old standard deviation. The added b does not affect the spread around the transformed mean.
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Consider 4, 4, 4. Their mean is four and every deviation is zero. The variance and standard deviation are both zero. Add five to every observation and the data become 9, 9, 9, with mean nine and the same zero spread.
This gives a particularly direct counterexample to the idea that standard deviation must increase by five. If that rule were true, identical shifted observations would somehow have standard deviation five, even though there is no separation between them at all.
For the descriptive formula used here, a single-observation data set also has zero standard deviation: its mean equals its only value. After a shift the same reasoning applies. The separate sample formula with n − 1 is not defined for one observation, reinforcing why conventions should be stated.
Zero standard deviation has a precise implication for a finite nonempty numerical data set under this formula: all observations are equal. Adding a constant to every one preserves that equality. It changes their common value, not the fact that they coincide.
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Range is maximum minus minimum. Adding c to every observation moves the maximum to maximum plus c and the minimum to minimum plus c. Subtraction cancels the two added constants, leaving the same range.
Quartiles calculated consistently from the ordered data also move by the same constant under a common shift. The difference between the upper and lower quartiles, the interquartile range, therefore stays unchanged. The choice of a consistent quartile convention does not alter this translation property.
The median increases by c because the middle ordered value, or the average of the two middle values, moves by c. Any modal values also move by c; their frequencies do not change. These measures describe location rather than separation.
Do not turn this into the claim that every possible statistic remains unchanged. The mean, median, total, minimum and maximum all move or change in particular ways. Ask what operation defines each statistic before predicting the transformation’s effect on it.
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Question: a data set has mean twelve and standard deviation three. Every observation is increased by five. Find the new mean and standard deviation, and explain why the same addition does not apply to both statistics.
The new mean is 12 + 5 = 17. Each observation and the mean have moved by five, so each deviation from the mean is unchanged. The new standard deviation is therefore three, not eight. No original observation list is needed to establish this result.
If the question instead states that every original observation is doubled and then increased by five, the new mean is 2(12) + 5 = 29. The new standard deviation is 2(3) = 6. Doubling stretches deviations; the later shift does not stretch them further.
A complete answer labels the statistics and states which transformation rule applies. Writing only seventeen and three leaves the reasoning hidden. A short explanation about deviations shows that the unchanged standard deviation is understood rather than guessed from a remembered answer pattern.
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Use the descriptive standard deviation convention stated at the beginning. For transformation-only questions, explain your rule briefly. For listed data, calculate exact variance before giving a rounded standard deviation if needed.
Question A. The data 1, 3, 5 have mean three. Add four to each observation. Find the new mean and compare the old and new standard deviations. Question B. A data set has mean twenty and standard deviation six. Subtract seven from each observation.
Question C. A data set has mean eight and standard deviation two. Multiply every observation by three. Find the new mean, variance and standard deviation. Question D. Use the same original statistics but multiply each observation by negative three and then add ten.
Question E. Calculate the variance of 2, 4, 6 using the mean-of-squares formula. Repeat for 7, 9, 11. Question F. Three identical observations each equal nine. What happens to their standard deviation if all three increase by five?
Question G. Starting with 2, 4, 6, append one additional observation equal to five. Is this the same transformation as adding five to every observation? Find the new mean. Question H. Explain the error in using the old mean when calculating deviations for shifted data.
Question I. A data set has mean fifteen, standard deviation four and range twelve. Add two to every observation. State the three new statistics. Question J. A data set is transformed by y = −x + 7. Its original mean is ten and standard deviation is five. Find the transformed statistics.
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Answer A. The new data are 5, 7, 9 and the new mean is seven. Both data sets have deviations −2, 0, 2, so both standard deviations are √(8/3). Answer B. The new mean is thirteen and the standard deviation remains six.
Answer C. The mean becomes twenty-four and the standard deviation becomes six. The original variance is four, so the new variance is nine times four, which is thirty-six. Answer D. The new mean is −3(8) + 10 = −14. Standard deviation is |−3|(2) = 6 and variance is thirty-six.
Answer E. The original variance is 56/3 − 16 = 8/3. The shifted variance is 251/3 − 81 = 8/3. Both parts of the formula change, but their difference is equal. Answer F. The observations become fourteen, fourteen and fourteen, so standard deviation remains zero.
Answer G. Appending a value changes the number of observations from three to four. The new mean is (2 + 4 + 6 + 5)/4 = 17/4, not nine. It is not a common shift of the original three observations.
Answer H. The shifted observations must be compared with their own new mean. Keeping the old mean measures distances from a different reference point and does not calculate the shifted data’s variance. Answer I. Mean becomes seventeen, standard deviation remains four and range remains twelve.
Answer J. The transformed mean is −10 + 7 = −3. Standard deviation remains five because the multiplier has absolute value one. Reflection reverses signed deviations and the shift moves their centre, but neither changes their magnitudes in this particular transformation.
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Start with 2, 4, 6 and append five to obtain 2, 4, 6, 5. The word add can describe this operation in ordinary language, but it is not the same mathematical instruction as increasing every existing observation by five.
The new total is seventeen, the count is four and the sum of squares is 4 + 16 + 36 + 25 = 81. The new variance is 81/4 − (17/4)² = 324/16 − 289/16 = 35/16.
The new standard deviation is √35/4, approximately 1.479. In this example it differs from √(8/3). Appending a value can change the mean and spread in ways that depend on the added value; there is no universal unchanged-spread rule for that operation.
Translate the instruction into an explicit new list if the wording is unclear. A shift keeps the count and changes every existing entry. An appended observation increases the count and keeps the existing entries. The two lists make the distinction immediately checkable.
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Suppose only the final observation of 2, 4, 6 increases by five. The new data are 2, 4, 11. This is not a common shift because the first two observations remain where they were. Their distances from the new mean need not stay the same.
The total is seventeen, the mean is 17/3 and the sum of squares is 4 + 16 + 121 = 141. Variance is 141/3 − (17/3)² = 47 − 289/9 = 134/9. Standard deviation is √134/3, approximately 3.859.
The spread increases in this example, but do not generalise that changing one observation always increases it. Moving an unusually distant value towards the other observations can reduce spread. The direction depends on the original data and the specific change.
The reliable rule is conditional: an exact equal addition to every observation preserves standard deviation. If the changes are unequal, that particular guarantee no longer applies. Calculate or analyse the actual transformed data instead of extending the rule beyond its stated conditions.
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Suppose a table records value two with frequency two and value six with frequency one. The underlying observations are 2, 2, 6. There are three observations, not two. The mean is ten thirds because the total is ten.
Adding five to every observation produces 7, 7, 11. In a frequency table, the values become seven and eleven, while the frequencies remain two and one. The mean increases from ten thirds to twenty-five thirds, a change of five.
The standard deviation remains unchanged because each of the three underlying observations has shifted equally. Treating the two distinct table entries as equally frequent observations would describe a different data set and could give a different mean and spread.
When shifting grouped data for an estimate, move the corresponding class boundaries and midpoints consistently while retaining the frequencies. The midpoint-based estimate of standard deviation is unchanged by a common exact shift of that representation. It remains an estimate when the original individual values are unknown.
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The shift rule concerns exact addition. If observations are shifted and then rounded, rounding can change their separations. The resulting calculation involves both a translation and a rounding procedure, so the exact invariance rule cannot simply be assumed for the final rounded list.
For instance, take 1.4 and 1.6. Their mean is 1.5 and standard deviation is 0.1. Add 0.3 exactly to get 1.7 and 1.9; their mean is 1.8 and standard deviation is still 0.1.
If those shifted observations are then rounded to the nearest whole number, both become two. The rounded data now have standard deviation zero. Rounding collapsed a distinction that was present before; the common shift itself did not do that.
Keep full precision during a calculation and round the requested final statistic as instructed. When working with recorded data that have already been rounded, describe the statistics of those recorded values rather than silently treating them as exact measurements of an unrounded underlying quantity.
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If observations are lengths measured in centimetres, deviations from their mean are also in centimetres. Squared deviations are in square centimetres, so variance has those squared units. Taking the square root returns standard deviation to centimetres.
Adding five centimetres to every recorded length raises the mean by five centimetres but leaves the spread in centimetres unchanged. It would not make sense to add five centimetres directly to a variance expressed in square centimetres.
Changing the measurement unit is a scaling transformation rather than a pure shift. Converting centimetres to millimetres multiplies each numerical value by ten, so the numerical mean and standard deviation both multiply by ten. Variance multiplies by one hundred.
The physical collection of lengths has not changed, but the numerical unit representation has. State the units when comparing statistics from different tables. Equal numerical standard deviations in different units do not automatically describe equal physical spreads.
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The original and shifted groups have equal standard deviations but different means. They are an immediate example of why standard deviation alone cannot reconstruct every observation or tell you the group’s location on the number line.
Even equal means and equal standard deviations need not determine a unique data set. This article does not require solving that broader reconstruction question, but it is important not to interpret an unchanged spread statistic as proof that nothing at all changed.
For a focused companion, read why equal standard deviations can come from different data. Together the two guides separate three ideas: location, spread and the full pattern of observations.
In a written comparison, report only what the available statistics support. You can say the mean increased by five and standard deviation remained unchanged under a stated common shift. You cannot infer a real class’s learning progress or consistency from an invented transformation example.
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If a calculator gives different standard deviations for 2, 4, 6 and 7, 9, 11, first check the entered data rather than rejecting the transformation property. A mistyped value or frequency can quietly change the data set.
Next check that the same statistic was selected both times. Some calculators display both a denominator-n standard deviation and a denominator-(n − 1) sample standard deviation. Comparing different settings does not test shift invariance using one consistent convention.
Then clear or inspect any previous data entries. An extra stored observation changes the count. A shift comparison requires matching counts and matching observation-by-observation changes, so leftover entries can invalidate the intended comparison even when the visible three new values look correct.
Finally compare exact expressions or sufficient displayed precision. Two differently rounded displays can look inconsistent if one result was copied with fewer digits. The theoretical equality is exact for the stated transformation; any discrepancy needs an explanation in the data, settings or numerical handling.
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Write the transformation in symbols before doing statistics. The phrase add five to every observation becomes yᵢ = xᵢ + 5. The phrase multiply every observation by five becomes yᵢ = 5xᵢ. That small step prevents two different operations from sharing the same vague word change.
Predict the new mean and spread. Then choose a tiny data set to verify the prediction using deviations. If your prediction says standard deviation rises after a common shift, the example 2, 4, 6 provides a quick test of that claim.
For longer questions, identify whether the information is a raw list, a frequency table, totals with sums of squares, or already calculated statistics. Choose working that matches the representation. Unnecessary data reconstruction can create more opportunities for errors without adding useful reasoning.
The mathematics learning hub links the wider statistics sequence. Choose variance, standard deviation, averages or transformations according to the first step you cannot yet explain clearly, rather than repeating a full topic indiscriminately.
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SECTION 26 OF 29
25. Shifting the reference point is not recalculating spread from zero
The squared distances of 2, 4, 6 from zero are four, sixteen and thirty-six. Their average is 56/3. That quantity is the mean square of the observations, not their variance about their own mean. Variance requires the centre specified by its definition.
After adding five, the squared distances of 7, 9, 11 from zero have average 251/3. That average has changed substantially because all observations moved farther from zero. It does not contradict the unchanged variance around the new mean nine.
The difference between these calculations identifies a hidden reference-point error. If someone measures all shifted observations from zero or from the old mean, they are answering a different distance question. Standard deviation uses deviations from the data set’s own arithmetic mean.
To make the reference explicit, write both the observed value and the centre before subtracting. For the shifted observation seven, use 7 − 9, not 7 − 4. The former gives negative two and belongs in the shifted variance calculation; the latter measures a displacement relative to the old centre.
This is a useful connection to coordinate reasoning. Location relative to a fixed origin changes under translation, while relative spacing within the translated collection does not. The apparent conflict disappears once both distance questions name their reference points.
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SECTION 27 OF 29
26. Repeating all observations and shifting them are separate ideas
Take the original data 2, 4, 6 and repeat each observation once to obtain 2, 2, 4, 4, 6, 6. The total doubles from twelve to twenty-four, and the count doubles from three to six. The mean remains four.
Each squared deviation also appears twice, so the sum of squared deviations doubles from eight to sixteen. Dividing by six gives 16/6 = 8/3. Under the denominator-n convention, repeating every observation the same number of times leaves variance and standard deviation unchanged.
This is not the same explanation as a common shift. Repetition changes the count and frequencies uniformly, while shifting changes the values and keeps their count fixed. Both can preserve descriptive standard deviation, but for different reasons that should be visible in the working.
Repeating only selected observations is different again. It changes their relative frequencies and can change the mean and spread. The condition every observation the same number of times is essential to the repetition argument, just as the condition the same constant for every observation is essential to the shift argument.
If a question combines transformations, separate them. For example, repeating all observations twice and then adding five to each preserves the original denominator-n standard deviation at both stages. The mean stays fixed during repetition and then increases by five during the shift.
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Consider 1, 1, 5, 5. Their mean is three. Their deviations are negative two, negative two, positive two and positive two. The squared deviations total sixteen; dividing by four gives variance four, so standard deviation is exactly two.
Add five to every observation to obtain 6, 6, 10, 10. Their mean is eight. The deviations from eight are again negative two, negative two, positive two and positive two, so variance remains four and standard deviation remains two.
This example uses repeated observations and an integer standard deviation, avoiding any distraction from an irrational square root. The invariance does not depend on all observations being distinct or on the original standard deviation having a particular numerical form.
As a final check, compare the ranges: five minus one and ten minus six both equal four. The data’s location moved, but the internal separation did not. A small independent example can verify your interpretation before you apply the rule to a larger table.
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The questions below summarise the conditions behind the result. Before applying a transformation rule, check what happens to every observation, whether the count stays fixed, and which standard deviation convention the calculation uses.
If every value becomes larger, why does spread not become larger?
Because the mean also becomes larger by the same amount. Spread around the mean depends on their relative distances, not their absolute positions above zero. Translating every point and its centre together preserves those distances.
Does this work when the constant is negative?
Yes. Subtracting the same constant amount from every observation also shifts the mean by that amount and preserves deviations. Negative data values do not produce negative standard deviation; the statistic remains nonnegative.
Does adding another observation preserve standard deviation?
Not in general. Appending changes the count and can change the mean and spread. It is a different operation from adding a constant to each existing observation, so calculate the new data rather than borrowing the shift rule.
What should I remember first?
Addition shifts location; multiplication scales distances. For y = ax + b, the mean becomes a times the old mean plus b, and standard deviation becomes |a| times the old standard deviation. Always check that the transformation applies to every observation.
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