A repeated measurement can look reassuring and still hide a problem
A school reading corner uses a converging lens as a magnifier. The preceding Physics guide explains why a printed object must be within the focal length for the familiar upright, enlarged virtual image. Now several learners try to set the lens at the same intended object distance. The team records the distances actually achieved. One learner says, “The average is correct, so our setup is reliable.” Another notices that individual readings can vary sharply. Which is right?
The difference between these claims is a Mathematics problem, not a new lesson about ray diagrams. A mean tells us about the centre of recorded values. It does not alone tell us how tightly those values cluster or whether every observation stayed inside a required boundary. A useful examination answer reads both centre and spread before drawing a real-world conclusion.
Our three classroom datasets are invented examples, and their small sample sizes do not establish performance across a real school. They are designed to practise accurate arithmetic, statistical interpretation and clear reasoning for G3 Mathematics. The arrangement uses a lens with a nominal focal length of 10.0 cm, and the intended object–lens separation is 6.0 cm. The Physics limit and the statistical description are related but distinct. We must not substitute one for the other.
The three batches of lens-setting data
Batch A records object–lens distances in centimetres of 5.9, 6.0, 6.1, 6.0 and 6.0. Batch B records 5.3, 5.9, 6.0, 6.1 and 6.7. Batch C records 5.9, 6.0, 6.1, 6.0 and 10.0. Each batch contains five observations, collected as a separate illustrative run. Assume the readings are recorded to the nearest 0.1 cm and refer to the same nominal point on the apparatus.
Before calculating, notice that the batches answer different questions. A is clustered near the target. B straddles the same target more widely. C has four measurements near the target and one at the nominal focal distance. The outlying reading could be a recording error, a genuine setup error or a different procedure. Statistics alone does not tell us which explanation is true.
The distance 10.0 cm in C matters for a second reason. It is at the labelled focal point of the ideal lens model. Therefore the claim “all five settings were safely within the focal length” is not supported by those five measurements. A mean below 10.0 cm cannot repair a single reading that reaches the boundary. This is the central lesson of the article: descriptive statistics and constraint checking have to work together.
Write the question before operating the calculator
If the question asks for mean, total the five distances and divide by five. If it asks which batch is more consistent, compare a measure of spread such as range or standard deviation. If it asks which settings met the condition “distance less than 10.0 cm”, check every recorded value against that threshold. These are three different operations, and each has its own answer.
Students often start by pressing the calculator keys they used in the previous question. That can lead to correct arithmetic applied to the wrong task. Slow down for one sentence: “I need to compare how variable these settings are.” That points toward range or standard deviation. “I need to estimate a typical setting” points toward mean or median. “I need to know whether every setting met a strict limit” calls for a value-by-value check, not just a single summary statistic.
Once you recognise the demand, the calculation becomes manageable. Write units beside the first result and in the final conclusion. The original readings are in centimetres, so the mean, median, range and standard deviation are also expressed in centimetres. Variance, if discussed as an intermediate step, is in square centimetres. A common mistake is to change units silently during calculation and then forget the resulting scale.
Find the mean without losing the units
Batch A totals 5.9 + 6.0 + 6.1 + 6.0 + 6.0 = 30.0 cm. Divide by five observations to obtain a mean of 6.0 cm. Batch B totals 5.3 + 5.9 + 6.0 + 6.1 + 6.7 = 30.0 cm, so its mean is also 6.0 cm. Both means match the intended setting even though one batch contains much larger deviations.
Batch C totals 5.9 + 6.0 + 6.1 + 6.0 + 10.0 = 34.0 cm. Its mean is 34.0 divided by five, or 6.8 cm. The fourth and fifth readings do not vanish because the first three are close to the target. Every recorded value belongs in the mean unless the question explicitly gives a legitimate exclusion rule.
A mean is a useful way to describe total divided evenly among observations. It is not the distance measured in every trial. The sentence “Batch B was always at 6.0 cm” is false, even though its mean is 6.0 cm. The sentence “The mean of Batch B was 6.0 cm” is supported by the calculation. This small wording distinction is worth practising because it separates a numerical summary from a physical history.
Use the median to read a typical position robustly
To find the median of five values, arrange them from least to greatest and select the middle, or third, value. Batch A becomes 5.9, 6.0, 6.0, 6.0, 6.1; its median is 6.0 cm. Batch B becomes 5.3, 5.9, 6.0, 6.1, 6.7; its median is also 6.0 cm. Batch C becomes 5.9, 6.0, 6.0, 6.1, 10.0; its median is again 6.0 cm.
All three batches have the same median. That fact does not make them equally consistent. The middle value can remain fixed while the most extreme readings move much farther away. The median is less affected by a single unusual high observation than the mean is. Compare C’s median of 6.0 cm with its mean of 6.8 cm: the difference alerts us to the influence of the 10.0 cm observation.
Do not use median as an excuse to ignore the unusual observation. If we are investigating a strict physical boundary, the unusual setting may be the most consequential reading. A median answers where the middle observation sits after sorting; it does not establish that each trial satisfied the required condition.
Range gives the first picture of spread
The range is the largest observation minus the smallest observation. For Batch A, the range is 6.1 minus 5.9, or 0.2 cm. For Batch B, the range is 6.7 minus 5.3, or 1.4 cm. For Batch C, the range is 10.0 minus 5.9, or 4.1 cm. These numbers immediately show differences that the medians conceal.
Range is easy to calculate and explain. If the only information available is the largest and smallest values, it may be the most direct measure. Its limitation is equally important: it depends on the two extremes and says little about how the other observations are distributed. Two batches can have the same range even when one has four values near the centre and the other has many values scattered through the full interval.
Here, the ranges establish that A is the tightest set and C has the widest span, but we should not stop there. A detailed G3 comparison can use standard deviation to consider every deviation from the mean. A sound explanation might say, “Batch A has a smaller range, suggesting that its recorded settings were more closely grouped.” The word suggesting is appropriate because the sample consists of just five invented readings.
Standard deviation measures variation around the mean
The standard deviation of a dataset is a measure of how far its observations typically lie from their mean. It is not simply the difference between the largest and smallest reading. The G3 Mathematics K310 syllabus includes calculating standard deviation for grouped and ungrouped data, and using mean and standard deviation to compare sets. A candidate should therefore be able to calculate it correctly and interpret what a smaller or larger value means in context.
For Batch A the mean is 6.0 cm. The deviations from the mean are −0.1, 0, +0.1, 0 and 0 cm. Square them, add them, divide by five under the syllabus’s stated standard-deviation formula for the dataset, and take the square root. The sum of squared deviations is 0.02 cm². Dividing by five gives 0.004 cm², and the square root is about 0.0632 cm.
That value is much smaller than the range because the two statistics measure different features of the same dataset. When reporting it, do not write 0.0632 cm². Standard deviation has the same units as the observations. The intermediate variance is in square centimetres; taking the square root returns centimetres.
Compare A and B by calculation, not guesswork
For Batch B the mean is also 6.0 cm. The deviations are −0.7, −0.1, 0, +0.1 and +0.7 cm. Squaring gives 0.49, 0.01, 0, 0.01 and 0.49 cm², whose total is 1.00 cm². Divide by five to obtain 0.20 cm². The standard deviation is the square root of 0.20, or approximately 0.447 cm to three significant figures.
Batch A and Batch B have equal means of 6.0 cm, but their standard deviations are about 0.0632 cm and 0.447 cm respectively. A has less spread around the intended setting in these five trials. This conclusion is stronger than saying A “looks tidier”, because it rests on an explicit measure. It is still narrower than a claim of universal reliability. Another set of trials might behave differently.
The comparison also shows why an exam answer must name both the statistic and its meaning. Writing “0.0632 is smaller” is a numerical observation. Writing “Batch A has a smaller standard deviation, so its recorded object distances are more closely grouped around their mean” is an interpretation. The latter earns its place in the real-world discussion.
The unusual C reading changes more than the average
Batch C has mean 6.8 cm. Its deviations are −0.9, −0.8, −0.7, −0.8 and +3.2 cm. Squared deviations are 0.81, 0.64, 0.49, 0.64 and 10.24 cm². Their sum is 12.82 cm². Divide by five to obtain 2.564 cm², then take the square root to get about 1.60 cm to three significant figures.
The large standard deviation tells us that C is a much less tightly clustered batch. The five values do not support a simple statement that the setup repeats the intended 6.0 cm setting consistently. Four values near 6.0 cm are not enough to erase one substantial departure. Nor is a large deviation itself evidence of carelessness. It may suggest a problem worth checking, but no statistical calculation identifies the cause.
Because C includes a measurement of 10.0 cm, the physical limit becomes especially important. With a nominal 10.0 cm focal length, that reading does not meet a strict “less than 10.0 cm” requirement. Averages and standard deviations describe the dataset; the boundary check describes compliance with a stated condition. Both are needed for an honest decision.
A box plot can show what a mean cannot
If the question provides a box-and-whisker plot, read the median, quartiles and extremes rather than assuming the length of the box represents the mean. The lower and upper quartiles bracket the middle half of the ordered data under the given convention, and the interquartile range is the upper quartile minus the lower quartile. A box plot can make differences in the central spread easy to see without plotting every observation separately.
Do not mix conventions for five-value quartiles. Different learning resources handle quartile positions in small datasets differently. A specific exam question may give a definition or a cumulative-frequency procedure. Follow the stated method; if the task is simply to compare centre and spread, choose a dataset or representation where the definition is clear. The most important reasoning is that a central interval and a full range answer different questions.
When reading a box plot, check the scale. A graph can exaggerate a small difference if the axis is truncated, or hide an important threshold if the focal boundary is not shown. State a comparison with the actual units and values. “The median is the same but the upper whisker reaches a larger distance” is far more informative than “The second box looks worse.”
A dot diagram helps uncover the pattern
Imagine a horizontal axis labelled object–lens distance in centimetres. Place one dot for every measurement. Batch A stacks three dots at 6.0 and has one dot on either side at 5.9 and 6.1. Batch B spreads dots from 5.3 to 6.7. Batch C has four dots near 6.0 and one far to the right at 10.0. A dot diagram makes the difference visible before any calculation.
Drawing the diagram does not require decorative colours or an elaborate story. Label the axis and use a consistent scale. If you change the scale between batches, a visually narrow cluster may appear wide, and the comparison becomes unreliable. Put all three on comparable axes whenever the question asks for a direct visual comparison. Keep the 10.0 cm boundary visible as a reference point.
A good interpretation mentions the pattern and a statistic. “The readings in Batch C are concentrated around 6.0 cm except for one reading at 10.0 cm, giving a larger range and standard deviation than A” makes use of both visual and numerical evidence. It stops short of calling the unusual value an error without supporting information.
What to do with an outlier
A data point that is much farther from the others deserves investigation, not automatic deletion. Check whether the observer used the same point on the lens holder, whether the ruler was read correctly, whether the printed target moved, and whether the reading was written down accurately. If a documented mistake is found, a corrected dataset may be justified. If no mistake is established, the original reading should be retained in the reported data and identified as unusual.
Suppose a learner removes 10.0 cm from Batch C and calculates the mean of the remaining four values. The reduced mean is 6.0 cm. That calculation can be useful when the question explicitly asks for the mean excluding the unusual observation. It does not prove the original batch was reliable, and it does not make the fifth trial disappear from history. Always state the rule used to exclude data.
The distinction matters outside examinations. A report that silently removes inconvenient measurements can make an apparatus appear more consistent than it was. Honest Mathematics records the original set, explains any corrections and connects the conclusion to the actual procedure. A convenient answer is not necessarily a defensible one.
A threshold is not the same as a typical value
The intended magnifier condition is that the object distance is less than the lens’s focal length in the simple model. For a nominal focal length of 10.0 cm, every measurement in A and B lies below that boundary. Four measurements in C lie below it; the fifth equals 10.0 cm. If the rule is strict, equality does not count as satisfying “less than”. It would be wrong to round a calculated mean and use it as a substitute for checking the five observations.
The story gets more demanding if the printed lens focal length is itself approximate. A nominal value of 10.0 cm is not necessarily a precisely measured threshold. The exercise has supplied the nominal value, so use it as the classroom model requires, but keep the limitation in mind when making recommendations. A real safety or accessibility claim would require appropriate measurements and a defined margin, not a fragile decision at a rounded boundary.
The mathematical habit is simple: identify the inequality, then test each relevant value. If the prompt says “at most”, equality is allowed. If it says “less than”, equality is excluded. If the data are rounded, do not invent unseen decimal places. The logic of the boundary is as important as the arithmetic.
Interpreting standard deviation responsibly
A smaller standard deviation indicates less spread in the recorded values around their mean. It does not necessarily mean the mean is closer to a desired target. Suppose another group always sets a lens exactly at 7.0 cm when the requested distance is 6.0 cm. Its standard deviation would be zero, but it would be consistently wrong. Reliability of repeated settings and accuracy relative to the target are related yet different concerns.
That counterexample explains why the two-statistic comparison is useful. The mean helps assess location relative to target, while standard deviation describes spread. Compare both; then check whether critical boundaries are satisfied. You may also need to consider whether all groups used the same instruments and procedure. Comparing statistics from different conditions without acknowledging them can produce a false sense of certainty.
A strong G3 response might say, “Batches A and B have the same mean of 6.0 cm, but A has a smaller standard deviation, so A’s recorded distances are more consistent. That does not by itself prove greater reading comfort.” Each clause performs a different job: number, comparison, interpretation and limit.
Make the calculations efficient under examination conditions
The K310 Mathematics examination has two papers, each two hours and fifteen minutes, each worth 90 marks and 50 percent of the total. Paper 1 contains about 26 shorter answers; Paper 2 has nine to ten questions and ends with a real-world problem. The official syllabus expects essential working, even where an approved calculator can be used. For a statistics question, show the mean and the relevant comparison rather than presenting a bare calculator display.
Select the calculator’s appropriate statistics mode and verify that you entered the intended values, not cumulative totals or the wrong frequency. For a five-value ungrouped dataset, the standard deviation defined in the syllabus uses division by the total number of observations. Some calculators display more than one kind of standard deviation, often one with denominator n and another with n−1. Read the label and choose the one matching the relevant course formula.
Do not round intermediate values too early. For final non-exact numerical answers, follow the K310 default of three significant figures unless a different accuracy is stated. Write 0.447 cm for Batch B, but keep sufficient precision in intermediate calculations. Contextual reasoning remains necessary after the calculator has finished.
Group frequency data carefully
Suppose another exercise reports that a group obtained 5.8 cm twice, 6.0 cm four times and 6.2 cm twice. The total frequency is eight, not three. The sum of all measured distances is 2(5.8) + 4(6.0) + 2(6.2) = 48.0 cm. The mean is 48.0 divided by eight, or 6.0 cm. When constructing the frequency table, use clear columns for the value, frequency and product of value and frequency.
The standard deviation can be worked out by expanding the repeated observations or by using the frequency formula supplied in the course. Each observation’s squared deviation is weighted by its frequency. Here four readings lie exactly at 6.0 cm, two are 0.2 cm below and two are 0.2 cm above. The total weighted squared deviation is 4(0.04) = 0.16 cm², so the variance is 0.02 cm² and the standard deviation is about 0.141 cm.
Frequency tables require careful reading because one row may represent several observations. If a student divides by the number of different values rather than the sum of frequencies, the mean and standard deviation may both be wrong. This is a simple procedural error with large consequences.
When the real-world conclusion is a recommendation
Assume the school chooses a demonstration routine based only on the five recorded distance settings and the stated 10.0 cm nominal focal threshold. Batch A provides the tightest cluster around the 6.0 cm target. Batch B has the same mean but wider variation. Batch C contains a reading at the focal boundary, and its summary statistics are strongly affected by that reading. A carefully limited choice would favour investigating the routine behind A before claiming a repeatable public demonstration.
But there is no evidence here about the actual quality of the letters seen by different readers, and no usability data. The selected routine may produce more consistent positioning without satisfying every visitor. A responsible recommendation should say exactly what was compared. It should also propose a further check: test with intended users, a controlled lighting setup, and a clear record of what each reader can see under named conditions.
A mathematical conclusion becomes stronger when it separates what has been calculated from what remains to be tested. The numbers support consistency comparisons. The Physics model supports a provisional object-position interpretation. Reader comfort is still an open question. Combining these stages without confusing them is the purpose of an EMS investigation.
Original practice clinic: solve, compare and explain
The next exercises use independently constructed numbers and questions. They are not drawn from a SEAB past paper. Work each item before reading the solution. Some require only arithmetic; others ask whether a statistical statement is justified. The final part of the practice moves towards the sort of interpretation that may appear inside a longer contextual problem.
Worked check 1: Four distances are 5.8, 5.9, 6.1 and 6.2 cm. Find the mean.
Their total is 24.0 cm. Divide by four to obtain 6.0 cm. Show the addition or a clear sum rather than presenting only the final number. This mean represents the centre of the four readings; it does not say they were identical. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 2: For the same four values, find the range.
The maximum is 6.2 cm and the minimum is 5.8 cm. Subtract 5.8 from 6.2 to obtain 0.4 cm. Range uses only the two extremes and has the same unit as the original measurements. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 3: Find the median of 5.1, 5.9, 6.0, 6.0 and 8.0 cm.
The readings are already in ascending order. With five observations the third value is the median, which is 6.0 cm. Notice that the single larger value affects the mean but does not displace the middle reading. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 4: A group reports mean 6.0 cm and SD 0.05 cm. Another reports mean 6.0 cm and SD 0.50 cm. Interpret the difference.
Both groups are centred at 6.0 cm by their reported means. The smaller standard deviation indicates that the first group’s measurements are more tightly clustered around that mean. It does not prove that their eyesight, reading speed or device satisfaction is better. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 5: Which has the larger range: 5.7, 6.0, 6.3 or 5.4, 6.0, 6.6?
The first set has range 6.3−5.7=0.6 cm. The second has range 6.6−5.4=1.2 cm. The second range is twice as large. Explain the comparison using actual spread, not merely the list that looks longer. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 6: If the nominal focal length is 10.0 cm, does a reading of 10.0 cm satisfy d < 10.0 cm?
No. A strict less-than inequality excludes equality. In the ideal lens construction, an object at the focus is not the same magnifier arrangement as an object inside the focal length. Do not replace the strict symbol with an informal “about ten”. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 7: Batch D gives 6.0, 6.0, 6.0 cm. What is its standard deviation?
Each value equals the mean of 6.0 cm, so all deviations are zero. Their squared sum is zero and the standard deviation is zero. This states consistency within the tiny batch, not proof that the measuring instrument was correctly calibrated. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 8: Batch E gives 7.0, 7.0, 7.0 cm for a target of 6.0 cm. What is the problem with choosing it for its SD?
Its standard deviation is zero, so all readings agree. However, its mean is 7.0 cm, one centimetre away from the specified target. Consistent measurements can still be inaccurate relative to the goal. Discuss location and spread together. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 9: Why is 6.8 cm not a suitable description of every trial in Batch C?
Six point eight is the mean, obtained by dividing the total of 34.0 cm by five. Four readings cluster around six centimetres and one reaches ten. The mean is a summary, not a value common to all observations. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 10: Calculate Batch B’s standard deviation using the deviations from 6.0 cm.
The squared deviations are 0.49, 0.01, 0, 0.01 and 0.49. Their sum is 1.00. Dividing by five gives variance 0.20 cm²; the square root is approximately 0.447 cm. Retain precision until the final step. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 11: Give two reasons to avoid deleting an unusual reading before investigation.
The value could reflect a genuine failure of the setup, which matters for the decision, or it could reflect a transcription or measurement error that needs documentation. Removing it without evidence makes the dataset look artificially more consistent. State any exclusion rule openly. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 12: An axis shows only 5.8 to 6.2 cm. What risk arises when comparing a new dataset with a value of 10.0 cm?
The scale hides the far-out observation or forces it off the graph, so the visual comparison becomes misleading. Extend the axis or choose a common meaningful scale, mark the target and boundary, and disclose how the full set is represented. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 13: Eight measurements are grouped as 5.8 cm twice, 6.0 cm four times, 6.2 cm twice. Find their mean.
The weighted sum is 2×5.8 + 4×6.0 + 2×6.2 = 48.0 cm. Divide by total frequency 8 to obtain 6.0 cm. Dividing by the three distinct entries is the wrong procedure for a frequency table. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 14: For that grouped set, find standard deviation to three significant figures.
There are two deviations −0.2 cm, four zero deviations and two +0.2 cm. The sum of squared deviations is 0.16 cm². Divide by eight to get 0.02 cm², then take the square root: 0.141 cm to three significant figures. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 15: What happens to range if only an interior observation changes but the extreme values stay the same?
The range stays the same because it uses only maximum minus minimum. The standard deviation may change because every observation contributes to it. This explains why range is a convenient but incomplete description of variation. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 16: Why is the statement ‘the smaller mean is always better’ false?
A mean has to be interpreted against the objective. If the target is 6.0 cm, a mean of 5.0 cm is not automatically better than 6.1 cm. Distance from the target, spread and physical constraints determine whether the result serves the task. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 17: Five readings are 5.9, 6.0, 6.1, 6.0, 6.0. How many are within 0.1 cm of 6.0, including the endpoints?
All five qualify because each lies in the closed interval from 5.9 to 6.1 cm. The endpoints count when the condition says within or gives inclusive inequalities. This is a value-by-value threshold check, not a calculation of the mean. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 18: Change the condition to strictly closer than 0.1 cm to 6.0. How many qualify?
Only the three readings equal to 6.0 cm qualify. The distances 5.9 and 6.1 are exactly 0.1 cm from the target and fail a strict less-than requirement. The change from inclusive to strict inequality changes the answer. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 19: A calculator gives two SD values, 0.447 and 0.500 for the same readings. What should you do?
Check which standard-deviation convention the calculator modes use. The K310 data-set formula divides the sum of squared deviations by the total frequency. Select the output that matches the syllabus formula and do not treat both outputs as interchangeable. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 20: A batch has the smallest range but a mean far from target. Can it be declared the best?
Not without a stated decision rule. A small range suggests consistency; the mean measures location. A batch may be consistently set at the wrong value. State the target, compare centre and spread, and check the important threshold before recommending it. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 21: The report claims one extreme trial ‘proves the device is unsafe’. Does the dataset establish that?
No. The data show a setting at a boundary in a fictional geometry model. Safety is a broader claim requiring a defined hazard, relevant conditions and suitable evidence. The defensible conclusion is that the batch did not meet the stipulated strict distance condition in all five measurements. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
Worked check 22: What is the best final line for comparing A, B and C?
A and B share a mean of 6.0 cm, but A has the smaller spread. C has a higher mean, a wider range and one value at the 10.0 cm boundary. On these recorded trials, A has the most consistent setting near the stated target; further testing is needed before making user-comfort claims. If writing this as an examination solution, show the relevant numerical work or logical comparison, preserve the unit, and make the conclusion specific to the stated readings. An answer that uses a correct statistic for the wrong question still misses the decision being asked for.
A three-pass self-correction method
First reread the task and underline the demand: mean, median, standard deviation, data display, or threshold compliance. Second audit the arithmetic: total frequency, sum of values, squared deviations and the correct calculator mode. Third audit the words in your conclusion: did you claim more than the statistics show? That third pass distinguishes competent calculation from competent mathematical reasoning.
Use one short repair exercise for the type of error you actually made. If you divided by three instead of eight in grouped data, practise frequency totals. If you found range when asked for standard deviation, practise identifying spread measures. If you claimed everyone read comfortably, revise the inference rather than repeating the calculations. A steady repair routine is better than rushing into another full paper without understanding the first failure.
For continuity, the PSLE Learning Guide builds early habits of interpreting questions and presenting relevant working. The Mathematics Hub offers broader secondary routes, while the Science Hub supplies the optical context. Volume 0480 owns the magnifier-image physics; this volume owns the statistics of repeated settings. Keeping those responsibilities separate avoids doing the same lesson twice.
Official references and the next step
The current SEAB 2027 G3 syllabus list identifies Mathematics K310, English K300 and Physics K323. Consult the Mathematics K310 PDF for the exact statistics outcomes, paper arrangements and accuracy notes. It describes two 2-hour-15-minute papers, each worth 50 percent, and includes real-world problem solving and statistical comparison. Our dataset is an original teaching case, not a national examination paper.
The previous reading-aid physics lesson is Volume 0480. For the different problem of projecting and sizing a real image, use Volume 0477. The Volume 0479 projector audit demonstrates how mathematical claims must fit scientific observations. Do not merge its screen-size decision with our repeatability decision.
The next volume asks how to explain an accessibility request in accurate English. The numbers give a strong basis for reporting a repeatable setting. They do not authorise a blanket promise about all readers. Learning to preserve that difference across subjects is a practical examination skill and a useful habit beyond school.
