Secondary Mathematics: Quartiles, Cumulative Frequency, Box Plots and Standard Deviation

Secondary Mathematics · Worked Repair Guide 19

Two groups can have the same average and very different distributions. One may be tightly clustered, another widely scattered. A single average cannot reveal that difference. Quartiles, cumulative frequency, box plots and standard deviation provide different ways to examine spread, position and concentration without confusing a summary with the complete data.

This guide begins with actual values, then deliberately compresses them into summaries and grouped intervals. At every stage, ask what information remains available and what has been lost. A carefully calculated statistic can still support an overconfident interpretation if the learner forgets what the statistic describes.

Use Averages, Spread and Data Interpretation first when mean, median, frequency and range are not yet secure. This companion develops the next layer through original worked examples and twenty practice tasks. Every data set here is invented for teaching; none reports results from actual students or institutions.

Choose a starting point: quartile conventions, cumulative frequency, standard deviation, or the independent practice. Follow the convention and formula specified in the learner’s actual question when they differ from this guide’s stated examples.

1. Begin with values, positions and units

Take the ordered data 2, 4, 5, 7, 8, 10, 11, 13. There are eight observations, their total is 60, their mean is 7.5 and their median is also 7.5. The range is 13 − 2 = 11. Each statistic answers a different question.

The median uses positions four and five, whose values are 7 and 8. Do not confuse the position number with the observed value. In another data set, the fourth value might be 60 rather than 4. Ordering is therefore a necessary step before locating quartiles or medians.

For an entry check, find the median of 9, 2, 6, 4, 10. Order them as 2, 4, 6, 9, 10; the median is 6. If these are times in minutes, the median is 6 minutes. Units remain attached to the statistic even after the individual observations have been compressed.

2. State the quartile convention before computing

Quartiles describe positions around one quarter, one half and three quarters of ordered data. For small finite data sets, different valid conventions can produce different numerical quartiles. The official R documentation for sample quantiles, for example, describes several algorithms rather than one universal computation rule.

For the ungrouped examples on this page, use the following explicit convention: order the data; find the median; find Q1 as the median of the lower half and Q3 as the median of the upper half. If the number of observations is odd, exclude the single overall median from both halves. For even-sized halves, average their two central values.

This declaration prevents an unnecessary argument with a calculator using a different percentile algorithm. It is not a claim that every school or examination uses this convention. When a question supplies a cumulative-frequency graph or specifies a percentile method, use that method instead of silently mixing it with a median-of-halves calculation.

3. Work through an even-sized and an odd-sized example

For 2, 4, 5, 7, 8, 10, 11, 13, the lower half is 2, 4, 5, 7, so Q1 = (4 + 5)/2 = 4.5. The upper half is 8, 10, 11, 13, so Q3 = (10 + 11)/2 = 10.5. The median is 7.5.

For 1, 3, 4, 6, 8, 9, 12, the overall median is 6. Under the declared convention, the lower half is 1, 3, 4 and the upper half is 8, 9, 12. Therefore Q1 = 3 and Q3 = 9.

Including the median 6 in both halves would produce different answers, 3.5 and 8.5. That is why the convention is not a trivial footnote. Write the two halves explicitly while learning the method; it reveals whether an error came from ordering, splitting or averaging, rather than leaving every mistake hidden inside a calculator output.

4. Interquartile range measures the spread between quartiles

The interquartile range, usually abbreviated IQR, is Q3 − Q1. For the eight-value example, IQR = 10.5 − 4.5 = 6. If the values were lengths in centimetres, the IQR would be 6 cm. It is a spread, not an average and not a percentage.

The IQR focuses on the central portion of the ordered data rather than the extreme endpoints. Replace the largest value 13 by 33. Under our convention, Q1, median and Q3 remain 4.5, 7.5 and 10.5. The IQR stays 6, while the range grows from 11 to 31.

That does not mean the changed value is unimportant. It means range and IQR deliberately measure different features. A report that includes both can distinguish a wider extreme tail from a change in the middle of the distribution. Neither statistic alone reconstructs every observation.

5. Construct a basic box plot from the five-number summary

The five-number summary consists of the minimum, Q1, median, Q3 and maximum. For our first example it is 2, 4.5, 7.5, 10.5, 13. On a consistently scaled number line, draw a box from 4.5 to 10.5, a median mark at 7.5 and whiskers to 2 and 13.

This is the basic minimum-to-maximum box-plot convention. A modified box plot may instead place whiskers at the most extreme observations not flagged as outliers and plot the outliers separately. OpenStax’s box-plot discussion explicitly distinguishes these display types.

Read the legend or instructions before interpreting a whisker as the sample minimum or maximum. The same word “box plot” is used for both displays. Neither requires the box to be centred visually on the median: unequal distances between Q1, median and Q3 may convey an important difference in concentration.

6. A long box segment does not mean more observations

Box plots are based on ranks. The four broad portions divided by the quartiles each represent approximately one quarter of the ordered observations, not a quantity proportional to their drawn lengths. A longer interval means those ranks are spread over a wider numerical range.

For small samples and tied observations, do not claim that exactly 25% of the values lie inside every closed interval formed by adjacent quartiles. Endpoints may coincide with observations, and inclusive intervals can count tied boundary values in more than one description. “Approximately a quarter by rank” is a safer interpretation unless the actual data establish an exact count.

A box plot also does not reveal the exact mean, mode or sample size unless those are given separately. Two samples can share the same five-number summary while containing different internal values. The display is useful precisely because it compresses the data, but compression inevitably removes information.

7. Compare centre and spread in separate sentences

Suppose Group A has median 60 and IQR 8, while Group B has median 65 and IQR 20, measured on the same scale. Group B has a higher median. Group A has a smaller central spread. These are two observations; one does not cancel the other.

Whether higher or lower is desirable depends on the quantity. A higher test score and a higher waiting time do not have the same practical meaning. “Group B is better” is unsupported without stating the objective and considering the wider context.

Likewise, overlapping boxes do not prove that the groups are identical, and non-overlapping boxes do not establish a cause. A descriptive graph is not an experiment. State what the summaries show, identify what remains unknown and avoid inventing explanations about effort, ability or teaching from the picture alone.

8. Cumulative frequency is a running total

Consider this invented distribution of forty completion times t, measured in minutes. The intervals include their lower boundary and exclude their upper boundary. The cumulative column counts observations below the listed upper boundary.

Time interval	Frequency	Cumulative frequency
0 ≤ t < 10	4	4
10 ≤ t < 20	12	16
20 ≤ t < 30	16	32
30 ≤ t < 40	6	38
40 ≤ t < 50	2	40

The cumulative total at 30 is 4 + 12 + 16 = 32, not just the frequency 16 in the third interval. The final cumulative frequency must equal the total sample size, 40. If it does not, either a class was omitted or a running sum is wrong.

Recover an individual frequency by subtraction. Between the cumulative values 16 and 32, there are 16 observations. This reverse operation is useful when a question gives the cumulative table rather than the original frequencies. A cumulative frequency cannot decrease as the threshold rises, because previously counted observations remain counted.

9. Plot cumulative totals at class boundaries, not midpoints

For the table, plot (0, 0), (10, 4), (20, 16), (30, 32), (40, 38) and (50, 40). The horizontal coordinate is the threshold time; the vertical coordinate is the cumulative count. Midpoints would represent the wrong threshold and shift the interpretation.

When drawing from grouped data, follow the curve or joining convention required by the question. A smooth curve and straight-line interpolation can give slightly different estimates between the same plotted points. Neither method recreates the unseen raw observations exactly.

To estimate the number below 25 minutes using linear interpolation, start at 16 below 20 and move halfway through the increase of 16 from 20 to 30. The estimate is 24. This uses an assumption of uniform accumulation within that class, not new information about the actual sixteen times.

10. Read a percentile by going from count to value

With N = 40, common grouped-curve positions for Q1, the median and Q3 are cumulative counts 10, 20 and 30. Start on the cumulative-frequency axis, move across to the curve and then down to the time axis. Reversing that route answers a different question: how many observations lie below a chosen time.

Using straight-line interpolation in our table, Q1 lies between 10 and 20 minutes. The cumulative count must rise from 4 to 10, an increase of 6 out of the class total 12. Therefore Q1 ≈ 10 + (6/12) × 10 = 15 minutes.

The median lies in 20–30: 20 + [(20 − 16)/16] × 10 = 22.5 minutes. Q3 is 20 + [(30 − 16)/16] × 10 = 28.75 minutes. The estimated IQR is 13.75 minutes. These are grouped-data estimates, not the median-of-halves quartiles of a known raw list. Keep the two procedures conceptually separate.

11. A threshold count and a percentage need the correct denominator

The table tells us exactly that 32 out of 40 recorded times are below 30 minutes. That is 80%. The remaining eight observations are at least 30 minutes because of the stated interval boundaries.

Between 20 and 40 minutes, including 20 but excluding 40, there are 38 − 16 = 22 observations. Dividing by 40 gives 55%. Using 38 as the denominator would instead describe a proportion of the observations below 40, a different question.

At an unlisted threshold such as 25, the estimate of 24 observations depends on interpolation. Therefore the resulting 60% is also an estimate. Exact arithmetic performed on an estimated count does not turn the conclusion into exact evidence. Words such as below, at least and between should be translated into interval boundaries before reading the graph.

12. Standard deviation measures spread around the mean

A deviation is an observation minus the mean. For 2, 4, 5, 7, 8, 10, 11, 13, the mean is 7.5 and the deviations are −5.5, −3.5, −2.5, −0.5, 0.5, 2.5, 3.5 and 5.5. They sum to zero, so simply averaging the signed deviations would conceal the spread.

One solution is to square the deviations, average their squares, then take a square root. Using divisor n to describe this complete listed data set, the variance is Σ(x − mean)²/n and the standard deviation is its non-negative square root. OpenStax’s discussion of spread distinguishes this formula from the sample formula using n − 1.

The square makes positive and negative deviations contribute positively. The final root restores the original measurement unit. Standard deviation is not literally the average absolute distance from the mean; that would be a different measure. It is a root-mean-square measure of deviation under the stated divisor.

13. A complete standard-deviation calculation

For the eight-value example, the squared deviations are 30.25, 12.25, 6.25, 0.25, 0.25, 6.25, 12.25 and 30.25. Their sum is 98. Divide by 8 to obtain variance 12.25, then take the square root to obtain standard deviation 3.5.

The mean 7.5 tells us where the values balance; standard deviation 3.5 describes their spread around that balance point. If the observations are in minutes, variance has units minutes squared and standard deviation has units minutes.

A useful manual check is that every squared deviation is non-negative. A negative computed variance signals an error in arithmetic, data entry or premature rounding. The final standard deviation is also non-negative. Zero standard deviation means every observation equals the mean, not merely that positive and negative deviations cancel; those signed deviations cancel for every data set.

14. Dividing by n and n − 1 answers different statistical tasks

The descriptive standard deviation used above divides by the number of observations n. The usual sample standard deviation s used in estimating a population’s spread from a sample divides the sum of squared deviations by n − 1 before taking the square root. It requires n > 1.

For the same eight values, s = √(98/7) = √14, approximately 3.742. This is different from the descriptive value 3.5, although the data entries are identical. A calculator may display both outputs.

Do not decide which result to report by choosing the more familiar-looking number. Read the required formula, the meaning of the statistic and the assessment instructions. This guide labels its divisor in worked calculations rather than claiming that one convention applies to every school question. For a single listed value, descriptive spread is zero, whereas the n − 1 sample calculation is undefined.

15. The computational formula is the same calculation rearranged

For the divisor-n variance, expanding the squared deviations gives an equivalent formula: variance = Σx²/n − mean². The equality follows because Σ(x − mean)² = Σx² − 2(mean)Σx + n(mean)² and Σx = n(mean).

For the eight values, Σx² = 548. Thus variance = 548/8 − 7.5² = 68.5 − 56.25 = 12.25, agreeing with the deviation table.

Be careful not to replace Σx² by (Σx)². The first means square each observation and then add; the second means add first and square the total. Here they are 548 and 3600, plainly different numbers. This is a structural algebra issue, not a calculator issue. Keep enough precision when subtracting two nearly equal large quantities, because premature rounding can badly distort a small variance.

16. Frequencies weight the squared values as well as the mean

Suppose values 1, 3 and 5 occur with frequencies 2, 4 and 2. There are eight observations, not three. The total Σfx is 2 + 12 + 10 = 24, so the mean is 3. The squared-value total Σfx² is 2(1²) + 4(3²) + 2(5²) = 88.

The divisor-n variance is 88/8 − 3² = 2, and the standard deviation is √2. A direct deviation check gives 2(−2)² + 4(0)² + 2(2)² = 16; divide by 8 and take the root.

For grouped intervals, replacing each unknown observation by its class midpoint gives an estimated mean and estimated standard deviation. The time table’s midpoint calculation gives Σfm = 900 and Σfm² = 24,200, so estimated mean = 22.5 and estimated standard deviation = √(605 − 506.25) = √98.75 ≈ 9.937 minutes. The midpoint assumption, not the number of decimal places, determines the result’s precision.

17. Shifting and scaling data affect spread predictably

Add 10 to every value in the original eight-value list. The mean becomes 17.5, and every quartile rises by 10, but deviations from the new mean are unchanged. Therefore the standard deviation stays 3.5 and the IQR stays 6.

Multiply every original value by 3. The mean and quartiles triple; the IQR becomes 18; the standard deviation becomes 10.5. Each deviation is multiplied by 3, each squared deviation by 9, and the final square root restores a factor of 3.

More generally, scaling by a real constant b multiplies standard deviation by |b|, not b, because spread cannot be negative. Adding a constant does not change it. Under a negative scale, the order reverses, so the lower and upper quartile roles swap. These properties are useful when converting units or checking whether a calculator list was entered on the correct scale.

18. Outlier flags are prompts to investigate, not delete

A commonly used modified-box-plot rule flags values below Q1 − 1.5 IQR or above Q3 + 1.5 IQR. With Q1 = 4.5 and Q3 = 10.5, the fences are −4.5 and 19.5. In the modified list ending in 33 rather than 13, the value 33 is flagged.

Under that display convention, the upper whisker reaches the largest unflagged observation, 11, and 33 is plotted separately. A basic minimum-to-maximum box plot would instead extend its upper whisker to 33. State which display is being used.

A flagged value may be a genuine unusual observation or a recording error. The rule alone cannot decide. Do not remove a number merely because it changes the mean or makes the graph untidy. For a mathematical exercise, follow the given data; for an investigation, check the record and report any justified exclusion transparently.

19. Standard deviations cannot simply be averaged across groups

Group A contains 0, 0 and Group B contains 10, 10. Each group has standard deviation zero. Together, the data are 0, 0, 10, 10 with mean 5. Every squared deviation is 25, so the combined descriptive standard deviation is 5.

Averaging the two group standard deviations would give zero and miss the separation between the group means. Combined spread depends on variation within groups and differences between their centres.

When totals are available, combine n, Σx and Σx² and then calculate the combined mean and variance. Do not combine just the final standard deviations. This mirrors the weighted-mean principle in the earlier data guide: reconstruct the quantities needed by the definition before recomputing the summary.

20. Avoid claims that a spread measure cannot establish

A smaller standard deviation does not automatically mean more accurate measurements. Values can be tightly clustered around a biased result. Nor does it mean higher achievement; a group can be consistently low, consistently high or consistently near any other centre.

Standard deviation also does not, by itself, guarantee that a fixed percentage lies within one standard deviation of the mean. For the data 0, 0, 10, 10, the mean is 5 and descriptive standard deviation 5. All four observations lie in the inclusive interval from 0 to 10. Familiar percentage rules require additional distributional assumptions.

Keep a comparison bounded: name the variable, unit, centre, spread measure and sample being described. A statistic calculated from a sample does not automatically establish the same value for a larger population. That distinction is part of interpreting data, even before formal statistical inference is studied.

21. Capstone: combine exact summaries with estimated ones

Return to the forty invented completion times. The grouped table supports an exact count of eight times at least 30 minutes and an exact count of twenty-two in 20 ≤ t < 40. Those counts use supplied class boundaries.

It supports estimated quartiles of 15, 22.5 and 28.75 minutes under straight-line interpolation, and an estimated mean of 22.5 minutes under midpoint substitution. The matching median estimate and mean estimate do not prove that the unseen raw distribution is perfectly symmetric.

The midpoint standard-deviation estimate is about 9.937 minutes. It describes the midpoint approximation, not an exact reconstruction of the forty times. A complete report could therefore state: “Eight of forty times were at least 30 minutes. Using the stated grouped-data methods, the estimated median was 22.5 minutes, IQR 13.75 minutes and descriptive standard deviation approximately 9.94 minutes.”

That report separates known counts from estimates. It does not pretend that an extra decimal place supplies missing raw information. The distinction is the central judgement this guide is designed to practise.

22. Independent practice: twenty tasks

Use the stated median-of-halves convention for raw lists, excluding the overall median when the list length is odd. Use divisor n for descriptive standard deviation unless the task explicitly asks for the sample formula.

1. For 1, 3, 4, 6, 8, 9, 11, 14, find the median.
2. For the same list, find Q1 and Q3.
3. Find its IQR and range.
4. Write its five-number summary for a basic box plot.
5. For 2, 5, 7, 8, 10, 12, 15, find Q1, median and Q3.
6. Frequencies 3, 7, 12, 8 occur in successive classes. Find cumulative frequencies.
7. Cumulative frequencies at successive boundaries are 5, 13, 25, 30. Recover the class frequencies.
8. A cumulative-frequency graph has N = 80. At what cumulative counts would the grouped-curve quartiles be read?
9. In the worked forty-time table, find the number below 40 minutes.
10. Using that table, find the exact percentage in 20 ≤ t < 40.
11. Using linear interpolation in that table, estimate the number below 25 minutes.
12. Find the mean and descriptive standard deviation of 2, 4, 6.
13. For the same three values, calculate the sample standard deviation using divisor n − 1.
14. Find the descriptive standard deviation of 2, 2, 4, 4.
15. Values 1, 3, 5 have frequencies 2, 4, 2. Find the descriptive variance.
16. A data set has standard deviation 4. Every value increases by 7. Find the new standard deviation.
17. A data set has standard deviation 4. Every value is multiplied by −3. Find the new standard deviation.
18. Q1 = 10 and Q3 = 18. Find the 1.5 IQR fences and decide whether 35 is flagged.
19. Combine the two groups 4, 4 and 8, 8. Find their combined descriptive standard deviation.
20. Can a five-number summary alone determine the exact mean? Explain.

23. Worked answers and interpretation checks

1. Median 7. There are eight values, so average the fourth and fifth, 6 and 8. The answer is a value on the measurement scale, not the position 4.5.

2. Q1 = 3.5, Q3 = 10. The lower half is 1, 3, 4, 6; its middle values are 3 and 4. The upper half is 8, 9, 11, 14; its middle values are 9 and 11.

3. IQR 6.5; range 13. Subtract 3.5 from 10 for the central spread and 1 from 14 for the full endpoint spread. They need not be close in size.

4. 1, 3.5, 7, 10, 14. Draw the box from 3.5 to 10, its median mark at 7, and basic whiskers to 1 and 14 on a consistently scaled axis.

5. Q1 = 5, median 8, Q3 = 12. Exclude the overall median 8 before taking medians of 2, 5, 7 and 10, 12, 15. Another declared convention could differ, but this is the one specified for the task.

6. 3, 10, 22, 30. Add each new frequency to the running total. The last value equals the sum of all four frequencies.

7. 5, 8, 12, 5. The first cumulative value is the first frequency; each later frequency is the difference between neighbouring cumulative values. Their sum is 30.

8. Counts 20, 40 and 60. These are N/4, N/2 and 3N/4. Read across from each count to the graph and down to the variable’s scale to estimate the quartile value.

9. Thirty-eight. This cumulative count comes directly from the table at the upper boundary 40, so interpolation is unnecessary.

10. Fifty-five per cent. There are 38 − 16 = 22 observations in the specified interval, out of 40. The inequalities match the table’s lower-inclusive, upper-exclusive classes.

11. About twenty-four. Twenty-five lies halfway through the 20–30 class. Under linear interpolation, add half of that class frequency 16 to the cumulative count 16 at 20. The count is an estimate because the within-class observations are unknown.

12. Mean 4; standard deviation √(8/3) ≈ 1.633. Deviations are −2, 0 and 2, whose squared sum is 8. Divide by n = 3 before taking the root.

13. Sample standard deviation 2. The squared-deviation sum is still 8, but divide by n − 1 = 2. Thus s = √4. The different result comes from the specified formula, not different observations.

14. Standard deviation 1. The mean is 3. All four deviations have magnitude 1, so the mean squared deviation is 1 and its square root is 1.

15. Variance 2. Total frequency is 8, mean is 3 and Σfx² = 88. Therefore 88/8 − 9 = 2. The task asks for variance, not its square root.

16. Still 4. The mean increases by 7 as well, leaving every deviation unchanged. A translation shifts the location without changing spread.

17. Twelve. Standard deviation is multiplied by the absolute scale factor, |−3| = 3. It cannot become −12.

18. Fences −2 and 30; 35 is flagged. IQR = 8, so 1.5 IQR = 12. Subtract 12 from Q1 and add 12 to Q3. Flagging does not by itself justify deleting the observation.

19. Standard deviation 2. Combined mean is 6 and each of the four values lies two units from it. Each separate group has zero spread, showing why group standard deviations must not simply be averaged.

20. No. The summary fixes five rank-based features but generally leaves other observations unspecified. Different internal values can preserve those five features while changing the total, and hence the mean.

24. Repair interpretation as carefully as calculation

A useful correction identifies whether the problem came from ordering, quartile convention, cumulative versus ordinary frequency, divisor choice, frequency weighting, rounding or an unsupported interpretation. “Selected the sample output when the task asked for divisor n” is more actionable than “calculator mistake”.

After correcting a calculation, change one condition. Replace an extreme value and ask which summaries change. Add a constant and predict the new spread before computing. Convert a raw list into intervals and identify which answers have become estimates. These comparisons teach what each measure preserves.

The proposed routine is one worked repair followed by a fresh numerical task and a written interpretation. Do not infer permanent mastery from one correct result. The useful evidence is whether the learner can select the appropriate convention, calculate it and explain its limits in later independent work.

25. Continue through the BTT learning routes

Return to the BTT Mathematics Hub or the BTT Mathematical Lab. For the conceptual library, use Statistics and Data; for prior foundations, use Averages, Spread and Data Interpretation.

Within Batch 05, Simultaneous Equations, Elimination and Modelling distinguishes exact conditions from measured inputs; Quadratic Equations, Factorisation and Roots supports the algebra of squares; and Vectors, Magnitude, Direction and Geometric Reasoning gives another setting where the magnitude of a difference matters.

Sources and scope

Reference points are the R documentation on sample-quantile algorithms and OpenStax’s sections on box plots and measures of spread. The data sets, derivations and practice questions on this page are original.

Use the MOE secondary curriculum doorway and the learner’s school instructions to select appropriate sections and conventions. This is a worked teaching guide, not an official syllabus or a claim that every statistical extension belongs to every secondary subject level.