Two groups can have the same average and behave very differently. One may be tightly clustered around that average while the other contains both very small and very large values. A cumulative-frequency graph, a box plot and a standard deviation describe different features of that variation. The Secondary 4 challenge is not only to calculate them. It is to choose the right feature and explain what it does, and does not, establish.
This twelfth Secondary 4 Mathematics Learning Guide develops a complete route from observations to statistical interpretation. It includes raw data, frequency tables, grouped estimates, quartiles, cumulative frequency, box-and-whisker plots, standard deviation and comparison statements. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
Scope: the principal examination references are 2026 O-Level Mathematics 4052 and 2027 SEC G3 Mathematics K310. Select material with your teacher for other subject levels. All numerical data below are invented for teaching; they are not survey findings, school results or claims about actual students.
Read the question before choosing a statistic
First identify the group being described, the variable being measured and its unit. Then decide whether the question asks about a typical value, the spread, the number beyond a threshold or a comparison between groups. A beautifully calculated standard deviation does not answer a question asking for a median. A correct median does not establish consistency.
Next inspect the representation. Raw values preserve individual observations. A frequency table groups repeated values without necessarily losing their exact numerical values. A table of class intervals usually loses the individual positions within each class. A box plot compresses a distribution into selected summaries. The information you can recover depends on what was retained.
Use the sections on quartiles, cumulative frequency, box plots, standard deviation and independent practice as separate learning sessions. Keep the distinction between an exact calculation from complete data and an estimate from grouped information visible throughout.
Centre and spread answer different questions
The mean is the total of the observations divided by their number. The median is the middle of the ordered data. The mode is a most frequent value or category. Each describes a kind of centre or typicality, but they need not agree.
Spread concerns how far apart the observations are. The range uses the largest and smallest values. The interquartile range describes the width of the middle portion. Standard deviation measures variation around the mean using every observation. A data set can have a high centre and small spread, a high centre and large spread, or either pattern around a lower centre.
The interpretation also depends on the variable. A higher score may be desirable in a particular assessment, while a lower journey time may be desirable for the same route. Do not write “better” merely because a statistic is larger. Name the measured quantity and explain the practical direction of the comparison.
Quartiles begin with ordered observations
The first quartile Q1, median and third quartile Q3 locate positions through ordered data. For the small raw-data examples in this guide, use the median-of-halves convention: find the median, then find the median of the lower half and the median of the upper half. The even-sized examples avoid the additional decision about including a central observation in both halves.
Different statistical conventions and software can produce different quartiles for small data sets. Follow any method specified by the question or taught for the task. Do not combine a raw-data convention with a cumulative-frequency reading rule as though they were interchangeable procedures. Ties also mean that exactly one quarter of the observations need not lie strictly below the first quartile.
Worked example 1: median, quartiles and interquartile range
The twelve recorded completion times, in minutes, are 2, 4, 5, 6, 7, 8, 10, 12, 13, 14, 16 and 19. Find the median, quartiles, range and interquartile range. They are already ordered. The median lies between the sixth and seventh values, so it is (8 + 10)/2 = 9 minutes.
The lower half is 2, 4, 5, 6, 7, 8. Its median is (5 + 6)/2 = 5.5 minutes, so Q1 = 5.5. The upper half is 10, 12, 13, 14, 16, 19. Its median is (13 + 14)/2 = 13.5 minutes, so Q3 = 13.5.
The range is 19 − 2 = 17 minutes. The interquartile range is Q3 − Q1 = 13.5 − 5.5 = 8 minutes. The IQR is a length of an interval, not a pair of quartile values. Writing “5.5 to 13.5” identifies the interval but does not calculate its width.
Neither the median nor a quartile has to be an observed value. Here 9, 5.5 and 13.5 arise from averaging central pairs under the stated convention. That is not a problem; a statistic is a summary calculated from the observations, not necessarily one of the observations itself.
A frequency table is not a list of equally weighted rows
Suppose the values 2, 4 and 6 have frequencies 1, 2 and 1. The actual data set is 2, 4, 4, 6. There are four observations, not three. The total is 1 × 2 + 2 × 4 + 1 × 6 = 16, so the mean is 16/4 = 4.
The notation Σfx means add the products of each value x and its frequency f. The notation Σf means add the frequencies. Thus the mean is Σfx/Σf. Dividing by the number of rows would give each distinct value equal weight even when their frequencies differ.
Worked example 2: find an unknown frequency
The values 4, 6 and 10 occur with frequencies 2, k and 3. Their mean is 7. Find k. The total number of observations is k + 5, and their total value is 2 × 4 + k × 6 + 3 × 10 = 38 + 6k.
(38 + 6k)/(k + 5) = 7
38 + 6k = 7k + 35
k = 3.
Check the result: the frequencies are now 2, 3 and 3, giving eight observations and total 56. The mean is 56/8 = 7. The frequency is a non-negative integer, which fits its role as a count. If an algebraic answer gives a negative or fractional frequency in this setting, return to the setup rather than accepting it automatically.
Cumulative frequency means a running total
A class frequency counts observations within a class. A cumulative frequency counts observations up to a boundary. Confusing them changes the data. If the first two class frequencies are 4 and 12, the second cumulative frequency is 16, not 12.
For the following table, t is a positive duration in minutes. The classes use the stated boundaries, so the cumulative frequency at 20 counts durations not greater than 20 minutes. Preserve that wording when answering questions about a threshold.
| Duration t, minutes | Frequency | Cumulative frequency |
|---|---|---|
| 0 < t ≤ 10 | 4 | 4 |
| 10 < t ≤ 20 | 12 | 16 |
| 20 < t ≤ 30 | 16 | 32 |
| 30 < t ≤ 40 | 8 | 40 |
There are 40 observations. To construct a cumulative-frequency diagram, plot the upper-boundary points (10, 4), (20, 16), (30, 32) and (40, 40), together with the starting point (0, 0). Use duration on the horizontal axis and cumulative frequency on the vertical axis. The cumulative total cannot decrease as the duration threshold increases.
Reading in two directions
A threshold-to-count question starts from the horizontal axis. For example, “How many durations are not greater than 20 minutes?” gives 20 as the duration, and the answer is cumulative frequency 16. A count-to-value question starts from the vertical axis. For the median of 40 observations on a cumulative-frequency diagram, locate cumulative frequency 20, then read the associated duration.
This difference in direction is a useful diagnostic. A student may know the word “median” yet read the vertical coordinate of a point whose horizontal coordinate is 20. The correct target is the 20th cumulative position, not a duration of 20 minutes.
Worked example 3: estimate quartiles by interpolation
Use the grouped table above to estimate the median, Q1 and Q3 by linear interpolation within each class. The original individual durations are unavailable. For this exercise, assume the cumulative frequency changes linearly between successive plotted boundary points. A supplied curve may lead to slightly different readings; use the actual supplied graph when one is part of a question.
The median position is 40/2 = 20. This lies between cumulative frequencies 16 and 32, within the 20-to-30-minute class. The position is 4 observations into a class containing 16, so the interpolated duration is 20 + (4/16) × 10 = 22.5 minutes.
For Q1, locate cumulative frequency 40/4 = 10. It lies 6 observations beyond cumulative frequency 4 in the class containing 12 observations. Thus Q1 ≈ 10 + (6/12) × 10 = 15 minutes.
For Q3, locate cumulative frequency 3 × 40/4 = 30. It lies 14 observations beyond cumulative frequency 16 in the class containing 16 observations. Therefore Q3 ≈ 20 + (14/16) × 10 = 28.75 minutes. The estimated IQR is 28.75 − 15 = 13.75 minutes.
The arithmetic may be exact for the interpolation model, but the resulting quartiles are estimates of the original data. Grouping removed the precise positions within the classes. Do not turn an estimate into an exact claim merely because the calculator produces several decimal places.
Worked example 4: count above a threshold using the complement
Using the same interpolation assumption, estimate how many durations exceed 25 minutes. At 25 minutes, we are halfway through the 20-to-30-minute class. The estimated cumulative frequency is 16 + (5/10) × 16 = 24.
That is the estimated number not greater than 25 minutes. The number greater than 25 is therefore 40 − 24 = 16 observations, or about 40% of the data set. Writing 24 would answer the opposite side of the threshold.
The count greater than 30 minutes is exactly 8 from the given class table. The count greater than 25 is estimated because 25 cuts through a class. Distinguishing these two situations shows whether the learner understands what information grouping has preserved.
Take care with “greater than”, “not greater than” and “at least”. For discrete observations or repeated values at a threshold, changing strict to inclusive wording can change the count. Use the boundaries and information actually given rather than silently assuming there are no observations equal to the threshold.
Box plots: five summaries, not the original list
A basic box-and-whisker plot can show the minimum, Q1, median, Q3 and maximum. The box runs from Q1 to Q3, with a line at the median. In the basic plots discussed here, the whiskers extend to the minimum and maximum. Some software uses an outlier rule and ends the whiskers elsewhere, so read the question’s definition or the plot’s key rather than assuming every box plot follows the same convention.
A box plot is useful for comparing centre and spread compactly, but it does not reveal every observation, the mean, the standard deviation or the exact number of observations unless that information is supplied separately. Two different data sets can share the same five-number summary.
Worked example 5: build and interpret a five-number summary
Use the twelve raw completion times from worked example 1. The five-number summary is minimum 2, Q1 5.5, median 9, Q3 13.5 and maximum 19, all in minutes. Draw a horizontal number line, mark the box from 5.5 to 13.5, draw the median at 9, and extend the whiskers to 2 and 19.
The box width is 8 minutes, which is the IQR. The full whisker-to-whisker span is 17 minutes, which is the range. The right side extends farther from the median than the left side in this summary, but the plot alone does not identify the shape or frequency of every cluster within those intervals.
A longer quarter-section of the plot does not mean that section contains more observations. The quartile divisions concern positions through the ordered data; a longer interval indicates that the corresponding observations are spread over a wider numerical span. With small samples and ties, avoid asserting exact percentages below strict thresholds without inspecting the underlying data.
Compare box plots with two separate statements
Suppose two groups perform the same task under comparable conditions. Group A has median completion time 12 minutes and IQR 4 minutes. Group B has median 15 minutes and IQR 2 minutes. Group A has the lower median time, so a typical central completion time is shorter. Group B has the smaller IQR, so its middle portion is less spread out.
Those statements can both be true. Neither group is unconditionally “better” on every criterion. A comparison should state which statistic supports which conclusion. Do not write that every member of A is faster than every member of B; medians cannot establish that claim.
Standard deviation: measure variation around the mean
For a complete data set with n observations and mean x̄, the descriptive standard deviation used here is the square root of the mean squared distance from x̄. Squaring prevents positive and negative deviations from cancelling. Taking the square root returns the measure to the original unit.
Two equivalent forms are σ = √[Σ(x − x̄)²/n] and σ = √[Σx²/n − x̄²]. For a frequency table, replace the count by Σf and use σ = √[Σfx²/Σf − (Σfx/Σf)²]. Here Σfx² means multiply f by x², not square the entire product fx.
The divisor in this guide is n, or the total frequency, because we are describing the given data set. The n − 1 formula used for a sample-based estimator in other statistical work is a different convention for a different purpose. A calculator may display both, so select the result matching the required formula rather than assuming that every standard-deviation key means the same thing.
Worked example 6: calculate standard deviation directly
Find the mean and standard deviation of 2, 4, 4 and 6. The mean is 4. The deviations from the mean are −2, 0, 0 and 2. Their squares are 4, 0, 0 and 4, with total 8.
The variance is 8/4 = 2, and the standard deviation is √2 ≈ 1.41. The alternative formula gives the same result: Σx² = 4 + 16 + 16 + 36 = 72, so σ = √(72/4 − 4²) = √2.
If these values were times in minutes, the variance would have units of square minutes and the standard deviation would have units of minutes. Do not write square units beside the final standard deviation. The square root is part of the definition, not an optional last step.
Worked example 7: equal means, different consistency
Compare A: 8, 9, 10, 11, 12 with B: 2, 6, 10, 14, 18. Both means are 10. For A, the squared deviations sum to 4 + 1 + 0 + 1 + 4 = 10, so its standard deviation is √(10/5) = √2.
For B, the squared deviations sum to 64 + 16 + 0 + 16 + 64 = 160, so its standard deviation is √(160/5) = √32 = 4√2. B is more spread out around the same mean. A is more consistent according to standard deviation.
It would be wrong to say A has the higher average or that every value in A is better. The averages are equal, and the meaning of “better” depends on the variable. A useful answer separates the unchanged centre from the changed spread.
Grouped data: why class midpoints give estimates
Return to the 40 durations in class intervals. The midpoint of the first class is 5, followed by 15, 25 and 35. To estimate the mean, represent every duration in a class by that class midpoint. This gives a workable summary, but the real durations need not equal their midpoints or balance perfectly around them.
| Midpoint x | Frequency f | fx | fx² |
|---|---|---|---|
| 5 | 4 | 20 | 100 |
| 15 | 12 | 180 | 2700 |
| 25 | 16 | 400 | 10000 |
| 35 | 8 | 280 | 9800 |
| Total | 40 | 880 | 22600 |
Worked example 8: estimated mean and standard deviation
Using the midpoint table, the estimated mean is 880/40 = 22 minutes. The estimated variance is 22600/40 − 22² = 565 − 484 = 81. Therefore the estimated standard deviation is 9 minutes.
The estimated mean 22 differs from the estimated median 22.5 obtained earlier. That is not a contradiction. They measure different features and were obtained through different summaries of the same grouped information. Neither is entitled to be called the exact original mean or median because the individual durations are unknown.
A common error is to calculate (fx)² instead of fx². In the first row, fx² is 4 × 5² = 100. Squaring fx would give 20² = 400, a different quantity. Keep the column heading and its meaning visible before entering values into a calculator.
Calculator statistics: check the data and the mode
Before using statistical functions, check that the data list is empty or contains exactly the intended observations. When entering a frequency table, enter the numerical values in the value column and the counts in the frequency column. For grouped estimates, enter the class midpoints as values. A class label such as “10 to 20” is not itself one numerical observation.
After entry, verify the displayed total count against Σf. If the table has 40 observations but the calculator reports four, the frequencies have not been used correctly. Then check the mean against a rough estimate and select the standard deviation corresponding to the divisor required by the task. Button names vary by calculator model, so this guide does not prescribe an unverified universal key sequence.
Worked example 9: shifting and scaling the data
A data set has mean 10 and standard deviation 2. What happens if 7 is added to every observation? The new mean is 17, but the standard deviation stays 2. Every value and the mean move by the same amount, so the deviations from the mean are unchanged.
What happens if every observation is multiplied by 3? The new mean is 30 and the standard deviation is 6. Every deviation is multiplied by three, so the spread in the original unit triples. More generally, multiplying by a scalar k multiplies standard deviation by |k|. A negative multiplier reverses ordering but does not create negative spread.
This reasoning is useful for unit conversion. A mean of 2 minutes and standard deviation of 0.5 minutes become 120 seconds and 30 seconds. It is also a check against the mistaken idea that adding a constant to every score must make the group more variable.
Worked example 10: combine groups by totals, not by averaging averages
Twenty observations have mean 14. Another thirty observations of the same variable have mean 18. Find the combined mean. Reconstruct each total: 20 × 14 = 280 and 30 × 18 = 540. The combined total is 820 across 50 observations.
The combined mean is 820/50 = 16.4. Simply averaging 14 and 18 would give 16 and incorrectly weight the two groups equally despite their different sizes. A combined standard deviation cannot generally be obtained from the two means and counts alone; information about within-group variation is also needed.
Worked example 11: correct a wrongly entered observation
Ten observations were reported with mean 12 and sum of squared values 1600. One value was entered as 17 but should have been 7. Find the corrected mean and standard deviation. The original total was 10 × 12 = 120. Correct it by removing 17 and adding 7: the new total is 110, so the corrected mean is 11.
The sum of squares must be corrected separately: 1600 − 17² + 7² = 1600 − 289 + 49 = 1360. Hence the corrected variance is 1360/10 − 11² = 136 − 121 = 15. The corrected standard deviation is √15 ≈ 3.87.
You cannot repair the sum of squares by merely subtracting the difference between the two values. The squared contribution changed from 289 to 49. This example tests whether the learner understands the statistics being stored, rather than treating the calculator output as an unexplained final authority.
Choose a representation for the question it can answer
A stem-and-leaf diagram can preserve individual numerical observations and their order when an appropriate key is supplied. A histogram displays grouped numerical frequencies across intervals; in the equal-width cases here, comparisons between bar heights are straightforward. A cumulative-frequency diagram is useful for percentile and threshold questions. A box plot compares selected centre-and-spread summaries compactly.
No display does every job. A box plot cannot tell you the exact frequency of one score. A grouped histogram cannot recover every original value. A cumulative-frequency curve does not by itself establish why a pattern occurred. Ask which information the representation preserves before asking it to support a conclusion.
A comparison answer should not exceed the evidence
When comparing groups, state the statistic, the numerical comparison and the interpretation. For example: “Group A has the lower median journey time, 12 minutes rather than 15 minutes, so its central journey time is shorter.” Then add a separate spread statement if required. This is stronger than “A is better” because it identifies exactly what the data supports.
Check that the measurements are comparable. Means from different test scales or journey lengths cannot automatically be interpreted as a fair performance comparison. A smaller standard deviation does not establish a cause, and statistics from one observed group do not guarantee the next individual outcome. Precision in the final sentence matters as much as precision in the calculation.
Independent practice
Use the median-of-halves convention for the small raw-data question. Use the descriptive standard-deviation formula with divisor n or total frequency. Where grouped interpolation or midpoint estimates are requested, label the result as an estimate. These instructions keep the practice internally consistent without pretending that every statistical convention is identical.
- For 2, 3, 5, 7, 8, 9, 10, 14, find the median, Q1, Q3, IQR and range.
- The values 1, 2 and 3 have frequencies 2, 4 and 4. Find the mean and standard deviation.
- A data set has mean 20 and standard deviation 3. Find the new mean and standard deviation after adding 5 to every value.
- Durations have mean 2 minutes and standard deviation 0.5 minutes. Express both statistics in seconds.
- A cumulative-frequency table has totals 5, 17, 33 and 40 at upper boundaries 10, 20, 30 and 40 minutes. Estimate the median by linear interpolation.
- A basic box plot has minimum 8, Q1 12, median 16, Q3 24 and maximum 35. Find the IQR and range.
- Ten observations have mean 12 and another fifteen have mean 18. Find the combined mean.
- Values 4, 6 and 10 have frequencies 2, k and 3. The mean is 7. Find k.
- Two comparable groups have the same mean score, 72. Their standard deviations are 8 and 5. Write one justified comparison and one claim that would not be justified.
- Durations in classes 0 < t ≤ 5, 5 < t ≤ 10 and 10 < t ≤ 15 have frequencies 2, 4 and 2. Estimate the mean and standard deviation using class midpoints.
Explained answers
1. The median is (7 + 8)/2 = 7.5. The lower-half median is (3 + 5)/2 = 4, and the upper-half median is (9 + 10)/2 = 9.5. The IQR is 5.5 and the range is 12.
2. The total frequency is 10, Σfx = 22 and Σfx² = 54. The mean is 2.2, and σ = √(54/10 − 2.2²) = √0.56 ≈ 0.748. Dividing by three would confuse distinct values with the number of observations.
3. The new mean is 25 and the standard deviation remains 3. Shifting every observation by the same amount does not change its deviation from the shifted mean.
4. Multiply both statistics by 60: the mean is 120 seconds and the standard deviation is 30 seconds. This is scaling, not adding a constant.
5. The median cumulative position is 20. It is three observations beyond 17 in the class containing 16 observations. The estimate is 20 + (3/16) × 10 = 21.875 minutes, approximately 21.9 minutes to three significant figures. The decimal is an interpolation result, not an exact recovered observation.
6. IQR = 24 − 12 = 12. Range = 35 − 8 = 27. The median 16 does not enter either subtraction.
7. The combined total is 10 × 12 + 15 × 18 = 390. Divide by 25 to obtain the combined mean 15.6. The larger group receives proportionately more weight.
8. (38 + 6k)/(5 + k) = 7 gives k = 3. Reconstructing the frequencies gives eight observations with total 56, confirming the mean.
9. The group with standard deviation 5 has scores less spread out around the same mean and is more consistent by this measure. It would not be justified to claim that every student in that group scored more highly, or that its mean is higher.
10. The midpoints are 2.5, 7.5 and 12.5. The total frequency is 8, Σfx = 60 and Σfx² = 550. The estimated mean is 7.5 minutes, and the estimated standard deviation is √(550/8 − 7.5²) = √12.5 ≈ 3.54 minutes.
A teaching routine that separates calculation from interpretation
Begin by showing the learner the same data in two forms, such as an ordered list and a frequency table. Ask how many observations there are before asking for a mean. If the student counts rows instead of frequencies, repair the representation first. More practice with a formula will not fix a mistaken interpretation of what the table contains.
Next, ask threshold questions in both directions on a cumulative-frequency diagram. Give a duration and ask for a count; then give a cumulative count and ask for a duration. Finish with a count above the threshold, so the learner must use the total as well as the graph reading.
Finally, require a two-sentence comparison: one sentence about centre and one about spread. Change the context from test scores to journey times without changing the statistics. The learner should notice that the practical interpretation of a higher centre changes, while the meaning of a smaller spread remains a statement about less variation.
Questions that prevent recurring errors
Can standard deviation be negative? No. It is the non-negative square root of a mean of squared deviations. A negative value inside the final square root can signal a wrong total, premature rounding, an incorrect formula or a data-entry mistake. Do not simply remove the minus sign.
Does a smaller IQR always mean a smaller standard deviation? Not necessarily. They use different information. IQR focuses on the middle portion, while standard deviation uses every observation and responds to distances from the mean. A data set can have a narrow middle and extreme values far away.
Can I calculate an exact mean from a box plot? Not in general. The five-number summary does not determine the total of all observations. Additional data would be needed. A plot is useful precisely because it compresses information, but compression limits what can be recovered.
Why did two methods give slightly different quartiles? Check whether one result came from a small raw-data convention and the other from grouped or graphical estimation. Also inspect the graph scale and interpolation assumption. Different information and different conventions can produce different summaries without an arithmetic mistake.
Sources, boundaries and connected learning
The assessed-topic references are the statistics sections of the official 2026 Mathematics 4052 syllabus and 2027 G3 Mathematics K310 syllabus. The examples use explicit raw-data, interpolation and standard-deviation conventions so that the reasoning can be checked. They are teaching material, not an official mark scheme or a replacement for the instructions in a particular paper.
Continue with Statistics, Probability and Real-World Problems for the wider interpretation route, Accuracy, Estimation and Calculator Discipline for numerical checks, and Error Analysis and Full-Paper Recovery when a repeated mistake needs a more precise diagnosis.
A statistical answer is complete when its meaning is as accurate as its number. Name the variable, use the information the representation genuinely contains, distinguish estimates from exact results and explain the comparison without claiming more than the data supports. That is how a list of observations becomes a defensible mathematical conclusion.
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