JC2 H2 Maths Bukit Timah | Why Variance Uses n−1

JC2 H2 Mathematics tutor Bukit Timah — Did you know? Two students can calculate the same set of squared deviations, yet produce different values of “variance” because one has divided by n and the other by n−1. Which denominator is correct depends on the task. If the numbers are the entire population, dividing by n describes that population. If they are a random sample used to estimate an unknown population variance, H2 Mathematics uses the unbiased sample-variance estimator with n−1. The difference is not a mathematical preference: it is a statement about what information you have.

This Singapore–Cambridge JC2 H2 Mathematics 9758 study guide is written for Bukit Timah parents, tutors and A-Level learners who need to interpret a sample summary correctly. We will reconstruct the formula, use Σx and Σx² from a question, practise a shifted origin, examine frequency data and distinguish a sample standard deviation from an unbiased estimate of population variance. The examples use deliberately fictional scores and measurements. They do not present actual student records or examination questions.

The quick answer is immediately below. Afterwards, try the one-minute diagnostic, see the fully worked dataset, explore 52 differentiated mini-clinics, and finish with parents’ frequently asked questions. Each clinic identifies not only a number but the reason a denominator belongs in the calculation. The underlying curriculum check is the official 2027 H2 Mathematics syllabus 9758, sampling section.

The 45-second answer: divide by n−1 when estimating population variance

For observations x₁, …, xₙ from a simple random sample, calculate the sample mean x̄ = (Σx)/n. The corrected sum of squares is S = Σ(x−x̄)² = Σx² − (Σx)²/n. The unbiased estimator of the population variance is s² = S/(n−1), provided n is at least 2. A descriptive variance of those n observed values, when treated as the complete population of interest, is instead S/n. These are two different questions answered using the same deviations.

Sample estimating a population: s² = [Σx² − (Σx)²/n] / (n−1).
Complete population of n measured objects: σ² = [Σx² − (Σx)²/n] / n.
One observed value only: n−1 = 0, so unbiased sample variance cannot be computed from one observation.
Standard deviation: after finding the relevant variance, take its positive square root and state the original unit, not a squared unit.
Reality check: squared deviations and variance are non-negative; a negative result means a data or arithmetic error.

Why the missing degree of freedom matters

Imagine you know that four observations have sample mean 5. The four deviations must add to zero: (x₁−5)+(x₂−5)+(x₃−5)+(x₄−5)=0. Once you specify any three deviations, the fourth is forced to be their negative sum. When the sample mean has been estimated from the same observations, not all n deviations are independently free to vary. In this setting the loss of one degree of freedom is reflected in the correction n−1.

The correction is specifically chosen so that, under standard independent identically distributed sampling with a finite population variance, the expected value of s² equals the population variance. This is what unbiased means. It does not say your particular sample variance must equal the true population variance, or that the sample standard deviation s is itself exactly unbiased for the population standard deviation σ. Distinguishing the expectation of an estimator from its observed value prevents careless statistical claims.

For an exact population of only four fully observed items, there is no unknown larger population variance to estimate. If the question explicitly calls those four measurements the entire population, their descriptive variance is S/4. If the same four values are labelled a sample from a larger process, the usual unbiased variance estimator is S/3. A good H2 answer reads the role of the data before selecting the denominator.

The first complete worked problem

A fictional random sample contains the values 2, 4, 6, 8. There are n = 4 readings. Their total is Σx = 20, so x̄ = 20/4 = 5. Deviations from the sample mean are −3, −1, 1 and 3; their squares are 9, 1, 1 and 9. The corrected sum of squares is therefore S = 20. For an unbiased estimate of a larger population’s variance, divide by n−1=3: s² = 20/3 ≈ 6.6667 squared units.

The same result follows from summary statistics. Here Σx² = 4 + 16 + 36 + 64 = 120. Hence S = 120 − 20²/4 = 120 − 100 = 20, and s² = 20/3. The corresponding sample standard deviation is s = √(20/3) ≈ 2.5820 in the original measurement units. By contrast, if these four items constituted the entire population, their variance would be 20/4 = 5. There is no contradiction: we have changed the quantity the question asks us to report.

For a visual sense check, the sample is balanced around 5. The endpoints 2 and 8 are three units away, while 4 and 6 are one unit away. Their squared distances average 5 over the four observed points, but the unbiased estimate has a correction because we used the same data to estimate the mean. The two interpretations are both legitimate if labelled correctly.

Summarised data: the calculator’s Σx and Σx² are not the answer

In an H2 examination, the raw list may be replaced with n, Σx and Σx². Consider a new fictional summary: n = 5, Σx = 40 and Σx² = 346. The mean is 8. Compute the corrected total S = 346 − 40²/5 = 346 − 320 = 26. The unbiased variance estimate is 26/4 = 6.5. The standard deviation estimate is √6.5 ≈ 2.5495. Do not simply enter 346/4: Σx² measures squares about zero, not squares about the estimated mean.

If the data came from observations 4, 7, 9, 10 and 10, their sums indeed match this summary. Their deviations from mean 8 are −4, −1, 1, 2 and 2. The squared deviations 16+1+1+4+4 total 26, independently confirming the shortcut. Building two routes to the same answer is particularly useful for students prone to dropping brackets in (Σx)²/n.

A common transcription error is to treat (Σx)² as Σx². The first means square the sum; the second means sum the individual squares. They are generally different. For x values 1 and 3, Σx² = 1²+3² = 10, whereas (Σx)² = (1+3)² = 16. When questions display both symbols, they are offering different pieces of information, not two spellings of the same quantity.

Shifting the origin to avoid awkward arithmetic

Suppose four readings are 105, 107, 110, 112. Subtract 100 from each to work with t = 5, 7, 10, 12. This shift changes the mean but not any distance between two readings, nor their deviations from their own mean. With t values, Σt = 34 and Σt² = 25 + 49 + 100 + 144 = 318. Thus S = 318 − 34²/4 = 318 − 289 = 29 and the unbiased variance estimate is 29/3 ≈ 9.6667. The sample mean in original units is 100 + 34/4 = 108.5.

The invariance can be checked directly: subtracting the same 100 from every observation also subtracts 100 from its mean, leaving each deviation unchanged. Be careful with scaling, however. If you divide every original value by 10, the standard deviation divides by 10 and the variance divides by 100. Shifting and rescaling have different consequences. H2 learners should recognise both rather than automatically assume all transformations preserve variance.

Frequency tables: n counts observations, not rows

Imagine a fictional six-observation dataset with value 2 occurring twice, value 4 occurring once, and value 6 occurring three times. There are only three distinct values, but n = 2 + 1 + 3 = 6. The sum of the six observations is Σfx = 2×2 + 1×4 + 3×6 = 26; the sum of their squares is Σfx² = 2×4 + 1×16 + 3×36 = 132. Thus S = 132 − 26²/6 = 58/3, and s² = (58/3)/5 = 58/15 ≈ 3.8667.

The row count three is not the sample size. Counting only distinct values would distort the mean and the denominator. When reading a frequency table, multiply each value and its square by the frequency before summing. This principle also explains why a table of repeated readings and a fully expanded raw list must produce the same answer if the data mean the same thing.

Value x	Frequency f	Contribution to Σfx	Contribution to Σfx²
2	2	4	8
4	1	4	16
6	3	18	108
Total	6	26	132

A tiny counterexample: when n = 1

Suppose the sample is a single reading 17. Its sample mean is also 17, and its sum of squared deviations from that mean is zero. But this does not tell you the unknown population variance is zero. A single point has no measured within-sample spread. The unbiased sample-variance formula has denominator n−1 = 0, so it is undefined. Writing zero as the population-variance estimate would confuse a calculation about this one observed point with information about the larger process.

The same caution applies when every observed value happens to be identical. With n = 5 and observations all equal to 7, the computed unbiased sample variance is zero. That is a valid numerical estimate for that particular sample; it is not definitive proof that no variability exists in the larger population. Precision in interpretation matters just as much as precision in arithmetic.

Which part is examined in 2027 H2 Mathematics?

In the 2027 Singapore–Cambridge H2 Mathematics 9758 outline, the sampling topic includes unbiased estimates of the population mean and variance, including summarised datasets containing Σx, Σx² and shifted sums such as Σ(x−a) and Σ(x−a)². That makes the calculation here directly relevant. The syllabus does not ask students to treat a single lucky estimate as the true population parameter, and careful interpretation is an examination skill alongside formula use.

This tutorial is not a licence to use the n−1 correction for every variance printed on a calculator. Different calculator modes and question wordings may refer to population and sample measures. The correct answer is the one matched to the statistical quantity described. Label it first; then calculate. If the questions use a frequency distribution of the entire stated population, that is conceptually different from using a random sample to estimate an unseen population.

The one-minute diagnostic

Give the learner this sentence: “A random sample of four readings, 2, 4, 6, 8, is used to estimate the population variance.” Ask for the formula and an answer. The expected response is 20/(4−1) = 20/3. Now change only the wording to: “These four readings are every observation in the population of interest.” The response becomes 20/4 = 5. If they change the numbers but not their interpretation, the underlying conceptual error has not been repaired.

Then present n = 5, Σx = 40, Σx² = 346. Ask which quantity is squared in the correction. The answer is Σx = 40, so subtract 40²/5 from 346. Finally give an n = 1 case and see whether the student can explain why dividing by n−1 is impossible. These short changes discriminate real understanding from memorising a single sequence of calculator keystrokes.

Fifty-two original variance clinics, organised by the decision you must make

The following cases are fabricated learning datasets. They are not actual marks, measurements from a clinic, or claims about pupils. Work in small clusters: identify what is being estimated; decide whether n−1 belongs in the denominator; reconstruct the corrected sum of squares; and interpret the size and units. The full worked answers are deliberately visible so parents and tutors can give immediate corrective feedback. The cases vary the data shape and the presentation rather than repeating only a single formula in different font.

1. Audit the raw list — case 1

Try this original sample. The fictional observations are 9, 12, 14, 16. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 4 observations. Their sum is 51 and their sum of squares is 677, so x̄ = 51/4 = 12.75. Corrected S = 677 − 51²/4 = 26.75. Because the task requests the unbiased estimator, divide by 4−1 = 3: s² = 8.91667 squared units and s ≈ 2.98608 original units.

Interrogate the result. Compare the endpoints with the centre before calculating. The farther readings should contribute disproportionately because deviations are squared. For a complete population consisting of exactly these 4 readings, its descriptive variance would be 6.6875, not 8.91667. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

2. Compare two denominators — case 2

Try this original sample. The fictional observations are 13, 16, 19, 18, 22. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 5 observations. Their sum is 88 and their sum of squares is 1594, so x̄ = 88/5 = 17.6. Corrected S = 1594 − 88²/5 = 45.2. Because the task requests the unbiased estimator, divide by 5−1 = 4: s² = 11.3 squared units and s ≈ 3.36155 original units.

Interrogate the result. Check the total from two routes: add the list directly, and independently sum the squared entries. These operations answer different questions. For a complete population consisting of exactly these 5 readings, its descriptive variance would be 9.04, not 11.3. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

3. Reconstruct Σx² — case 3

Try this original sample. The fictional observations are 16, 20, 20, 22, 26, 27. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 6 observations. Their sum is 131 and their sum of squares is 2945, so x̄ = 131/6 = 21.83333. Corrected S = 2945 − 131²/6 = 84.83333. Because the task requests the unbiased estimator, divide by 6−1 = 5: s² = 16.96667 squared units and s ≈ 4.11906 original units.

Interrogate the result. A shifted sample mean is still a mean of the data, not a fixed population parameter. Explain why its estimation costs one degree of freedom. For a complete population consisting of exactly these 6 readings, its descriptive variance would be 14.13889, not 16.96667. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

4. Check the units and spread — case 4

Try this original sample. The fictional observations are 19, 20, 23, 25, 27, 28, 31. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 7 observations. Their sum is 173 and their sum of squares is 4389, so x̄ = 173/7 = 24.71429. Corrected S = 4389 − 173²/7 = 113.42857. Because the task requests the unbiased estimator, divide by 7−1 = 6: s² = 18.90476 squared units and s ≈ 4.34796 original units.

Interrogate the result. Look for repeated values, then verify that every occurrence is counted in n. A repeated reading still contributes a separate squared deviation. For a complete population consisting of exactly these 7 readings, its descriptive variance would be 16.20408, not 18.90476. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

5. Audit the raw list — case 5

Try this original sample. The fictional observations are 21, 25, 28, 28, 30, 34, 34, 37. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 8 observations. Their sum is 237 and their sum of squares is 7215, so x̄ = 237/8 = 29.625. Corrected S = 7215 − 237²/8 = 193.875. Because the task requests the unbiased estimator, divide by 8−1 = 7: s² = 27.69643 squared units and s ≈ 5.26274 original units.

Interrogate the result. Try deleting the largest reading and predict whether the sample spread will necessarily decrease. Test rather than assuming that every deletion must behave alike. For a complete population consisting of exactly these 8 readings, its descriptive variance would be 24.23438, not 27.69643. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

6. Compare two denominators — case 6

Try this original sample. The fictional observations are 25, 29, 30, 30. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 4 observations. Their sum is 114 and their sum of squares is 3266, so x̄ = 114/4 = 28.5. Corrected S = 3266 − 114²/4 = 17. Because the task requests the unbiased estimator, divide by 4−1 = 3: s² = 5.66667 squared units and s ≈ 2.38048 original units.

Interrogate the result. State whether the variance is in squared measurement units and convert to standard deviation only at the last step. For a complete population consisting of exactly these 4 readings, its descriptive variance would be 4.25, not 5.66667. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

7. Reconstruct Σx² — case 7

Try this original sample. The fictional observations are 28, 30, 31, 34, 36. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 5 observations. Their sum is 159 and their sum of squares is 5097, so x̄ = 159/5 = 31.8. Corrected S = 5097 − 159²/5 = 40.8. Because the task requests the unbiased estimator, divide by 5−1 = 4: s² = 10.2 squared units and s ≈ 3.19374 original units.

Interrogate the result. Imagine calling these readings the complete group rather than a sample. Recompute the descriptive population variance using n, and describe the changed question. For a complete population consisting of exactly these 5 readings, its descriptive variance would be 8.16, not 10.2. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

8. Check the units and spread — case 8

Try this original sample. The fictional observations are 31, 33, 37, 37, 40, 42. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 6 observations. Their sum is 220 and their sum of squares is 8152, so x̄ = 220/6 = 36.66667. Corrected S = 8152 − 220²/6 = 85.33333. Because the task requests the unbiased estimator, divide by 6−1 = 5: s² = 17.06667 squared units and s ≈ 4.13118 original units.

Interrogate the result. Do the calculation after adding 100 to all readings. The variance should stay unchanged because translating the origin leaves deviations intact. For a complete population consisting of exactly these 6 readings, its descriptive variance would be 14.22222, not 17.06667. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

9. Audit the raw list — case 9

Try this original sample. The fictional observations are 33, 38, 39, 40, 43, 45, 46. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 7 observations. Their sum is 284 and their sum of squares is 11644, so x̄ = 284/7 = 40.57143. Corrected S = 11644 − 284²/7 = 121.71429. Because the task requests the unbiased estimator, divide by 7−1 = 6: s² = 20.28571 squared units and s ≈ 4.50397 original units.

Interrogate the result. Scale each observation by 2. The variance should multiply by 4 even though the count n remains the same. For a complete population consisting of exactly these 7 readings, its descriptive variance would be 17.38776, not 20.28571. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

10. Compare two denominators — case 10

Try this original sample. The fictional observations are 37, 39, 41, 42, 45, 47, 49, 50. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 8 observations. Their sum is 350 and their sum of squares is 15470, so x̄ = 350/8 = 43.75. Corrected S = 15470 − 350²/8 = 157.5. Because the task requests the unbiased estimator, divide by 8−1 = 7: s² = 22.5 squared units and s ≈ 4.74342 original units.

Interrogate the result. Do not infer that an apparently modest variance guarantees a small maximum error. A single outlier can dominate squared deviations. For a complete population consisting of exactly these 8 readings, its descriptive variance would be 19.6875, not 22.5. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

11. Reconstruct Σx² — case 11

Try this original sample. The fictional observations are 40, 43, 45, 46. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 4 observations. Their sum is 174 and their sum of squares is 7590, so x̄ = 174/4 = 43.5. Corrected S = 7590 − 174²/4 = 21. Because the task requests the unbiased estimator, divide by 4−1 = 3: s² = 7 squared units and s ≈ 2.64575 original units.

Interrogate the result. If a student reports a negative corrected sum of squares, audit the parentheses and the two summation symbols before discussing statistics. For a complete population consisting of exactly these 4 readings, its descriptive variance would be 5.25, not 7. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

12. Check the units and spread — case 12

Try this original sample. The fictional observations are 43, 46, 48, 49, 53. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 5 observations. Their sum is 239 and their sum of squares is 11479, so x̄ = 239/5 = 47.8. Corrected S = 11479 − 239²/5 = 54.8. Because the task requests the unbiased estimator, divide by 5−1 = 4: s² = 13.7 squared units and s ≈ 3.70135 original units.

Interrogate the result. Report more than a naked decimal: name it as the estimated population variance and explain the denominator being used. For a complete population consisting of exactly these 5 readings, its descriptive variance would be 10.96, not 13.7. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

13. Audit the raw list — case 13

Try this original sample. The fictional observations are 45, 48, 50, 52, 53, 56. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 6 observations. Their sum is 304 and their sum of squares is 15478, so x̄ = 304/6 = 50.66667. Corrected S = 15478 − 304²/6 = 75.33333. Because the task requests the unbiased estimator, divide by 6−1 = 5: s² = 15.06667 squared units and s ≈ 3.88158 original units.

Interrogate the result. Compare the endpoints with the centre before calculating. The farther readings should contribute disproportionately because deviations are squared. For a complete population consisting of exactly these 6 readings, its descriptive variance would be 12.55556, not 15.06667. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

14. Compare two denominators — case 14

Try this original sample. The fictional observations are 49, 52, 55, 54, 58, 61, 61. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 7 observations. Their sum is 390 and their sum of squares is 21852, so x̄ = 390/7 = 55.71429. Corrected S = 21852 − 390²/7 = 123.42857. Because the task requests the unbiased estimator, divide by 7−1 = 6: s² = 20.57143 squared units and s ≈ 4.53557 original units.

Interrogate the result. Check the total from two routes: add the list directly, and independently sum the squared entries. These operations answer different questions. For a complete population consisting of exactly these 7 readings, its descriptive variance would be 17.63265, not 20.57143. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

15. Reconstruct Σx² — case 15

Try this original sample. The fictional observations are 52, 56, 56, 58, 62, 63, 63, 68. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 8 observations. Their sum is 478 and their sum of squares is 28746, so x̄ = 478/8 = 59.75. Corrected S = 28746 − 478²/8 = 185.5. Because the task requests the unbiased estimator, divide by 8−1 = 7: s² = 26.5 squared units and s ≈ 5.14782 original units.

Interrogate the result. A shifted sample mean is still a mean of the data, not a fixed population parameter. Explain why its estimation costs one degree of freedom. For a complete population consisting of exactly these 8 readings, its descriptive variance would be 23.1875, not 26.5. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

16. Check the units and spread — case 16

Try this original sample. The fictional observations are 55, 56, 59, 61. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 4 observations. Their sum is 231 and their sum of squares is 13363, so x̄ = 231/4 = 57.75. Corrected S = 13363 − 231²/4 = 22.75. Because the task requests the unbiased estimator, divide by 4−1 = 3: s² = 7.58333 squared units and s ≈ 2.75379 original units.

Interrogate the result. Look for repeated values, then verify that every occurrence is counted in n. A repeated reading still contributes a separate squared deviation. For a complete population consisting of exactly these 4 readings, its descriptive variance would be 5.6875, not 7.58333. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

17. Audit the raw list — case 17

Try this original sample. The fictional observations are 57, 61, 64, 64, 66. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 5 observations. Their sum is 312 and their sum of squares is 19518, so x̄ = 312/5 = 62.4. Corrected S = 19518 − 312²/5 = 49.2. Because the task requests the unbiased estimator, divide by 5−1 = 4: s² = 12.3 squared units and s ≈ 3.50714 original units.

Interrogate the result. Try deleting the largest reading and predict whether the sample spread will necessarily decrease. Test rather than assuming that every deletion must behave alike. For a complete population consisting of exactly these 5 readings, its descriptive variance would be 9.84, not 12.3. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

18. Compare two denominators — case 18

Try this original sample. The fictional observations are 61, 65, 66, 66, 71, 72. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 6 observations. Their sum is 401 and their sum of squares is 26883, so x̄ = 401/6 = 66.83333. Corrected S = 26883 − 401²/6 = 82.83333. Because the task requests the unbiased estimator, divide by 6−1 = 5: s² = 16.56667 squared units and s ≈ 4.07022 original units.

Interrogate the result. State whether the variance is in squared measurement units and convert to standard deviation only at the last step. For a complete population consisting of exactly these 6 readings, its descriptive variance would be 13.80556, not 16.56667. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

19. Reconstruct Σx² — case 19

Try this original sample. The fictional observations are 64, 66, 67, 70, 72, 74, 75. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 7 observations. Their sum is 488 and their sum of squares is 34126, so x̄ = 488/7 = 69.71429. Corrected S = 34126 − 488²/7 = 105.42857. Because the task requests the unbiased estimator, divide by 7−1 = 6: s² = 17.57143 squared units and s ≈ 4.19183 original units.

Interrogate the result. Imagine calling these readings the complete group rather than a sample. Recompute the descriptive population variance using n, and describe the changed question. For a complete population consisting of exactly these 7 readings, its descriptive variance would be 15.06122, not 17.57143. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

20. Check the units and spread — case 20

Try this original sample. The fictional observations are 67, 69, 73, 73, 76, 78, 79, 82. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 8 observations. Their sum is 597 and their sum of squares is 44733, so x̄ = 597/8 = 74.625. Corrected S = 44733 − 597²/8 = 181.875. Because the task requests the unbiased estimator, divide by 8−1 = 7: s² = 25.98214 squared units and s ≈ 5.09727 original units.

Interrogate the result. Do the calculation after adding 100 to all readings. The variance should stay unchanged because translating the origin leaves deviations intact. For a complete population consisting of exactly these 8 readings, its descriptive variance would be 22.73438, not 25.98214. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

21. Audit the raw list — case 21

Try this original sample. The fictional observations are 69, 74, 75, 76. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 4 observations. Their sum is 294 and their sum of squares is 21638, so x̄ = 294/4 = 73.5. Corrected S = 21638 − 294²/4 = 29. Because the task requests the unbiased estimator, divide by 4−1 = 3: s² = 9.66667 squared units and s ≈ 3.10913 original units.

Interrogate the result. Scale each observation by 2. The variance should multiply by 4 even though the count n remains the same. For a complete population consisting of exactly these 4 readings, its descriptive variance would be 7.25, not 9.66667. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

22. Compare two denominators — case 22

Try this original sample. The fictional observations are 73, 75, 77, 78, 81. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 5 observations. Their sum is 384 and their sum of squares is 29528, so x̄ = 384/5 = 76.8. Corrected S = 29528 − 384²/5 = 36.8. Because the task requests the unbiased estimator, divide by 5−1 = 4: s² = 9.2 squared units and s ≈ 3.03315 original units.

Interrogate the result. Do not infer that an apparently modest variance guarantees a small maximum error. A single outlier can dominate squared deviations. For a complete population consisting of exactly these 5 readings, its descriptive variance would be 7.36, not 9.2. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

23. Reconstruct Σx² — case 23

Try this original sample. The fictional observations are 76, 79, 81, 82, 85, 88. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 6 observations. Their sum is 491 and their sum of squares is 40271, so x̄ = 491/6 = 81.83333. Corrected S = 40271 − 491²/6 = 90.83333. Because the task requests the unbiased estimator, divide by 6−1 = 5: s² = 18.16667 squared units and s ≈ 4.26224 original units.

Interrogate the result. If a student reports a negative corrected sum of squares, audit the parentheses and the two summation symbols before discussing statistics. For a complete population consisting of exactly these 6 readings, its descriptive variance would be 15.13889, not 18.16667. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

24. Check the units and spread — case 24

Try this original sample. The fictional observations are 79, 82, 84, 85, 89, 89, 91. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 7 observations. Their sum is 599 and their sum of squares is 51369, so x̄ = 599/7 = 85.57143. Corrected S = 51369 − 599²/7 = 111.71429. Because the task requests the unbiased estimator, divide by 7−1 = 6: s² = 18.61905 squared units and s ≈ 4.31498 original units.

Interrogate the result. Report more than a naked decimal: name it as the estimated population variance and explain the denominator being used. For a complete population consisting of exactly these 7 readings, its descriptive variance would be 15.95918, not 18.61905. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

25. Audit the raw list — case 25

Try this original sample. The fictional observations are 81, 84, 86, 88, 89, 92, 94, 96. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 8 observations. Their sum is 710 and their sum of squares is 63194, so x̄ = 710/8 = 88.75. Corrected S = 63194 − 710²/8 = 181.5. Because the task requests the unbiased estimator, divide by 8−1 = 7: s² = 25.92857 squared units and s ≈ 5.09201 original units.

Interrogate the result. Compare the endpoints with the centre before calculating. The farther readings should contribute disproportionately because deviations are squared. For a complete population consisting of exactly these 8 readings, its descriptive variance would be 22.6875, not 25.92857. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

26. Compare two denominators — case 26

Try this original sample. The fictional observations are 85, 88, 91, 90. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 4 observations. Their sum is 354 and their sum of squares is 31350, so x̄ = 354/4 = 88.5. Corrected S = 31350 − 354²/4 = 21. Because the task requests the unbiased estimator, divide by 4−1 = 3: s² = 7 squared units and s ≈ 2.64575 original units.

Interrogate the result. Check the total from two routes: add the list directly, and independently sum the squared entries. These operations answer different questions. For a complete population consisting of exactly these 4 readings, its descriptive variance would be 5.25, not 7. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

27. Reconstruct Σx² — case 27

Try this original sample. The fictional observations are 88, 92, 92, 94, 98. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 5 observations. Their sum is 464 and their sum of squares is 43112, so x̄ = 464/5 = 92.8. Corrected S = 43112 − 464²/5 = 52.8. Because the task requests the unbiased estimator, divide by 5−1 = 4: s² = 13.2 squared units and s ≈ 3.63318 original units.

Interrogate the result. A shifted sample mean is still a mean of the data, not a fixed population parameter. Explain why its estimation costs one degree of freedom. For a complete population consisting of exactly these 5 readings, its descriptive variance would be 10.56, not 13.2. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

28. Check the units and spread — case 28

Try this original sample. The fictional observations are 91, 92, 95, 97, 99, 100. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 6 observations. Their sum is 574 and their sum of squares is 54980, so x̄ = 574/6 = 95.66667. Corrected S = 54980 − 574²/6 = 67.33333. Because the task requests the unbiased estimator, divide by 6−1 = 5: s² = 13.46667 squared units and s ≈ 3.6697 original units.

Interrogate the result. Look for repeated values, then verify that every occurrence is counted in n. A repeated reading still contributes a separate squared deviation. For a complete population consisting of exactly these 6 readings, its descriptive variance would be 11.22222, not 13.46667. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

29. Audit the raw list — case 29

Try this original sample. The fictional observations are 93, 97, 100, 100, 102, 106, 106. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 7 observations. Their sum is 704 and their sum of squares is 70934, so x̄ = 704/7 = 100.57143. Corrected S = 70934 − 704²/7 = 131.71429. Because the task requests the unbiased estimator, divide by 7−1 = 6: s² = 21.95238 squared units and s ≈ 4.68534 original units.

Interrogate the result. Try deleting the largest reading and predict whether the sample spread will necessarily decrease. Test rather than assuming that every deletion must behave alike. For a complete population consisting of exactly these 7 readings, its descriptive variance would be 18.81633, not 21.95238. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

30. Compare two denominators — case 30

Try this original sample. The fictional observations are 97, 101, 102, 102, 107, 108, 109, 112. Treat them as a random sample from a larger population, not as the complete population. First identify n, then calculate Σx and Σx², then estimate the population variance without prematurely rounding the mean. Before looking below, predict whether squared deviations should be dominated by values far from the centre.

Show the arithmetic. There are 8 observations. Their sum is 838 and their sum of squares is 87956, so x̄ = 838/8 = 104.75. Corrected S = 87956 − 838²/8 = 175.5. Because the task requests the unbiased estimator, divide by 8−1 = 7: s² = 25.07143 squared units and s ≈ 5.00714 original units.

Interrogate the result. State whether the variance is in squared measurement units and convert to standard deviation only at the last step. For a complete population consisting of exactly these 8 readings, its descriptive variance would be 21.9375, not 25.07143. The two numbers represent different estimands, so neither should be silently substituted for the other. Explain in one sentence why this particular question selects n−1, then verify that the standard deviation you reported is the positive square root of the sample variance.

Summarised-statistics clinics: Σx and Σx² without the full list

31. Reconstruct spread from a printed summary

Data card. An invented study gives n=5, Σx=60, Σx²=732. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 60/5 = 12. The correction for the fitted sample mean is (60)²/5 = 720. Subtract this from Σx² to obtain corrected sum 12. Divide by 4, producing s² = 3 and s ≈ 1.73205. No attempt has been made to invent the individual observations that led to this summary.

What to notice. The number of summands was given, not guessed from the number of statistics in the question. Changing the sample mean by adding the same constant to each observation would change Σx and Σx² but would leave corrected S unchanged. Describe which algebraic terms must offset that change, and explain why s is measured in the original units while s² uses squared units. A calculator is useful only after this bookkeeping is correct.

32. Reconstruct spread from a printed summary

Data card. An invented study gives n=6, Σx=90, Σx²=1372.5. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 90/6 = 15. The correction for the fitted sample mean is (90)²/6 = 1350. Subtract this from Σx² to obtain corrected sum 22.5. Divide by 5, producing s² = 4.5 and s ≈ 2.12132. No attempt has been made to invent the individual observations that led to this summary.

33. Reconstruct spread from a printed summary

Data card. An invented study gives n=7, Σx=126, Σx²=2304. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 126/7 = 18. The correction for the fitted sample mean is (126)²/7 = 2268. Subtract this from Σx² to obtain corrected sum 36. Divide by 6, producing s² = 6 and s ≈ 2.44949. No attempt has been made to invent the individual observations that led to this summary.

34. Reconstruct spread from a printed summary

Data card. An invented study gives n=8, Σx=168, Σx²=3580.5. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 168/8 = 21. The correction for the fitted sample mean is (168)²/8 = 3528. Subtract this from Σx² to obtain corrected sum 52.5. Divide by 7, producing s² = 7.5 and s ≈ 2.73861. No attempt has been made to invent the individual observations that led to this summary.

35. Reconstruct spread from a printed summary

Data card. An invented study gives n=9, Σx=216, Σx²=5256. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 216/9 = 24. The correction for the fitted sample mean is (216)²/9 = 5184. Subtract this from Σx² to obtain corrected sum 72. Divide by 8, producing s² = 9 and s ≈ 3. No attempt has been made to invent the individual observations that led to this summary.

36. Reconstruct spread from a printed summary

Data card. An invented study gives n=10, Σx=270, Σx²=7317. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 270/10 = 27. The correction for the fitted sample mean is (270)²/10 = 7290. Subtract this from Σx² to obtain corrected sum 27. Divide by 9, producing s² = 3 and s ≈ 1.73205. No attempt has been made to invent the individual observations that led to this summary.

37. Reconstruct spread from a printed summary

Data card. An invented study gives n=11, Σx=330, Σx²=9945. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 330/11 = 30. The correction for the fitted sample mean is (330)²/11 = 9900. Subtract this from Σx² to obtain corrected sum 45. Divide by 10, producing s² = 4.5 and s ≈ 2.12132. No attempt has been made to invent the individual observations that led to this summary.

38. Reconstruct spread from a printed summary

Data card. An invented study gives n=5, Σx=165, Σx²=5469. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 165/5 = 33. The correction for the fitted sample mean is (165)²/5 = 5445. Subtract this from Σx² to obtain corrected sum 24. Divide by 4, producing s² = 6 and s ≈ 2.44949. No attempt has been made to invent the individual observations that led to this summary.

39. Reconstruct spread from a printed summary

Data card. An invented study gives n=6, Σx=216, Σx²=7813.5. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 216/6 = 36. The correction for the fitted sample mean is (216)²/6 = 7776. Subtract this from Σx² to obtain corrected sum 37.5. Divide by 5, producing s² = 7.5 and s ≈ 2.73861. No attempt has been made to invent the individual observations that led to this summary.

40. Reconstruct spread from a printed summary

Data card. An invented study gives n=7, Σx=273, Σx²=10701. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 273/7 = 39. The correction for the fitted sample mean is (273)²/7 = 10647. Subtract this from Σx² to obtain corrected sum 54. Divide by 6, producing s² = 9 and s ≈ 3. No attempt has been made to invent the individual observations that led to this summary.

41. Reconstruct spread from a printed summary

Data card. An invented study gives n=8, Σx=336, Σx²=14133. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 336/8 = 42. The correction for the fitted sample mean is (336)²/8 = 14112. Subtract this from Σx² to obtain corrected sum 21. Divide by 7, producing s² = 3 and s ≈ 1.73205. No attempt has been made to invent the individual observations that led to this summary.

42. Reconstruct spread from a printed summary

Data card. An invented study gives n=9, Σx=405, Σx²=18261. You have no raw list, so you cannot reconstruct which observation took which value. What you can compute is the sample mean and unbiased population-variance estimate. This question exercises the exact sort of summary information used in the H2 sampling section.

Calculation. x̄ = 405/9 = 45. The correction for the fitted sample mean is (405)²/9 = 18225. Subtract this from Σx² to obtain corrected sum 36. Divide by 8, producing s² = 4.5 and s ≈ 2.12132. No attempt has been made to invent the individual observations that led to this summary.

Frequency-table clinics: repeated values are observations too

43. Three-value frequency audit

Fictional frequency card. A value 2 appears 2 times, value 4 appears 1 times, and value 9 appears 4 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=2+1+4 = 7. Weighted Σfx = 44 and Σfx² = 348. The corrected squared total is 348 − 44²/7 = 71.42857, so s² = 11.90476. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

Transfer task. Increase the frequency of the middle value by one and decide which quantities change before recomputing: the sample size, sum, sum of squares and sample mean all generally change. A shortcut that holds x̄ fixed after adding an observation is wrong. Describe why the denominator becomes the new sample size minus one, not the old size minus one. Parent or tutor feedback should focus on whether the learner rebuilt the whole summary consistently.

44. Three-value frequency audit

Fictional frequency card. A value 3 appears 3 times, value 5 appears 2 times, and value 10 appears 5 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=3+2+5 = 10. Weighted Σfx = 69 and Σfx² = 577. The corrected squared total is 577 − 69²/10 = 100.9, so s² = 11.21111. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

45. Three-value frequency audit

Fictional frequency card. A value 4 appears 4 times, value 6 appears 3 times, and value 11 appears 2 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=4+3+2 = 9. Weighted Σfx = 56 and Σfx² = 414. The corrected squared total is 414 − 56²/9 = 65.55556, so s² = 8.19444. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

46. Three-value frequency audit

Fictional frequency card. A value 2 appears 2 times, value 7 appears 4 times, and value 12 appears 3 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=2+4+3 = 9. Weighted Σfx = 68 and Σfx² = 636. The corrected squared total is 636 − 68²/9 = 122.22222, so s² = 15.27778. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

47. Three-value frequency audit

Fictional frequency card. A value 3 appears 3 times, value 4 appears 1 times, and value 13 appears 4 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=3+1+4 = 8. Weighted Σfx = 65 and Σfx² = 719. The corrected squared total is 719 − 65²/8 = 190.875, so s² = 27.26786. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

48. Three-value frequency audit

Fictional frequency card. A value 4 appears 4 times, value 5 appears 2 times, and value 9 appears 5 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=4+2+5 = 11. Weighted Σfx = 71 and Σfx² = 519. The corrected squared total is 519 − 71²/11 = 60.72727, so s² = 6.07273. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

49. Three-value frequency audit

Fictional frequency card. A value 2 appears 2 times, value 6 appears 3 times, and value 10 appears 2 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=2+3+2 = 7. Weighted Σfx = 42 and Σfx² = 316. The corrected squared total is 316 − 42²/7 = 64, so s² = 10.66667. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

50. Three-value frequency audit

Fictional frequency card. A value 3 appears 3 times, value 7 appears 4 times, and value 11 appears 3 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=3+4+3 = 10. Weighted Σfx = 70 and Σfx² = 586. The corrected squared total is 586 − 70²/10 = 96, so s² = 10.66667. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

51. Three-value frequency audit

Fictional frequency card. A value 4 appears 4 times, value 4 appears 1 times, and value 12 appears 4 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=4+1+4 = 9. Weighted Σfx = 68 and Σfx² = 656. The corrected squared total is 656 − 68²/9 = 142.22222, so s² = 17.77778. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

52. Three-value frequency audit

Fictional frequency card. A value 2 appears 2 times, value 5 appears 2 times, and value 13 appears 5 times. The question describes a sample from a larger population. Do not treat the three rows as n=3; first expand the frequencies mentally and decide how many independent recorded observations the table contains.

Correct calculation. n=2+2+5 = 9. Weighted Σfx = 79 and Σfx² = 903. The corrected squared total is 903 − 79²/9 = 209.55556, so s² = 26.19444. The answer is non-negative, as required for a variance. If the values were merely three labelled categories rather than numerical measurements, this calculation would not be meaningful.

Eight short decisions before you leave this lesson

Sample or population? If the wording says a random sample estimates population variance, choose n−1. If it describes the complete finite population, describe that population with n.
One observation? The unbiased variance formula is undefined when n=1; do not interpret the zero deviation as proof of zero population variability.
Translation? Adding a fixed number to every reading leaves both the corrected sum of squares and variance unchanged.
Scaling? Multiplying every reading by k multiplies its variance by k² and standard deviation by |k|.
Summarised data? Use Σx²−(Σx)²/n. Squaring Σx² or subtracting Σx²/n are different operations.
Frequency table? The sum of frequencies is n, including repetitions. Distinct values alone do not determine n.
Units? Variance has squared measurement units; standard deviation has the original units.
Unbiased? It describes the estimator’s long-run mean under appropriate sampling assumptions, not a guarantee for a particular sample.

A usable five-day home practice plan

On Monday, solve four small raw-data samples by hand, then verify Σ(x−x̄)² from the shortcut formula. On Tuesday, take summary cards with only n, Σx and Σx²; audit brackets until they are automatic. On Wednesday, practise translated readings and compare the two corrected sums of squares before and after the shift. On Thursday, complete frequency tables and explain why the total frequency, not the number of printed rows, determines n. On Friday, mix population-variance and sample-estimation questions without revealing the intended denominator. The quality of the explanation matters more than the count of repeated calculations.

A three-person small-group tutorial can allocate roles: one pupil states the estimand, another performs the calculations, and the third challenges the interpretation and units. Rotate the roles on the next question. That structure can make silent formula memorisation visible without assuming anything about a real pupil’s previous marks. A learner who can independently justify n−1, demonstrate a shift-invariance check, and reconcile raw and summary forms has a transferable skill that supports later sampling and hypothesis-testing questions.

Frequently asked questions

Why is the unbiased variance denominator n−1?

Because estimating the sample mean from the observations imposes one relationship among their deviations. The n−1 correction gives an unbiased estimate of the population variance under standard independent sampling assumptions.

Is n−1 used for every form of variance?

No. It is used for this sample-based unbiased population-variance estimate. The variance of a completely observed population of n items is calculated with denominator n.

Is the sample standard deviation exactly unbiased?

Not generally. The unbiased statement applies to the sample variance estimator s², not automatically to its square root s.

What if the question gives only Σx and Σx²?

Use x̄=Σx/n and S=Σx²−(Σx)²/n. You do not need to reconstruct the unseen raw observations to estimate variance.

Can every observation be identical and s² still be zero?

Yes, for that realised sample. A zero estimate is not proof of no possible population variability, especially if the sample is small.

Does translating data change the variance?

No. Adding or subtracting one fixed constant changes the mean but preserves deviations about the mean. Multiplication by a constant is different: it scales variance by the square of that constant.

Are grouped intervals as informative as raw frequencies?

No. An exact-value frequency table preserves each observed value through its counts, whereas broad grouped intervals can lose information. Do not treat a midpoint-based approximation as the exact variance of unknown observations.

Which examination syllabus does this guide use?

It references the 2027 H2 Mathematics syllabus 9758, particularly the sampling section and its requirement for unbiased estimates from raw and summarised data. Learners should use the document for their actual examination year.

Series links, study ownership and official source

Why Sample Variance Uses n−1: current guide
Top 10% and Inverse Normal Cutoffs
Why Variances Add When Values Subtract
Angle Between Two Planes and Normals

For the complete parent-level progression, see the Bukit Timah JC2 tuition pathway, H2 normal distribution and sampling subject guide, and immutable eduKateSG small-group mathematics tuition reference from Clementi. This page explains the n−1 estimator; it does not duplicate the wider sampling curriculum or assert current class availability.

Official source: SEAB, Singapore–Cambridge H2 Mathematics 9758, 2027 examination syllabus. All datasets, calculations, teaching sequences and diagnostics above are original instructional constructions. The guide is not an official SEAB examination paper, does not contain identifiable learner data, and makes no guarantees of grades. The explanations and examples are intended to be used alongside the learner’s actual examination-year syllabus and school guidance.

JC2 H2 Mathematics Tutor Bukit Timah | Why Sample Variance Uses n−1