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In an ordered list of nine measurements, the median is the fifth measurement. Five is its position, not automatically its value. If that fifth measurement is fourteen centimetres, the median is fourteen centimetres. Reporting five centimetres would confuse where the middle lies with what is found there.
The same distinction matters in frequency tables and cumulative-frequency graphs. A position tells you how far through the ordered observations to look. The median value is the measurement or numerical outcome associated with the middle position, using the usual average of the two central values when the observation count is even.
This guide develops the distinction through original datasets, worked frequency tables and complete practice answers. Use the 2027 G3 Mathematics syllabus for official curriculum context and your school’s conventions for grouped-data estimates. The exact-data method comes first; graphical and grouped extensions are identified separately.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–30
Sections 31–33
For the ordered values 3, 7, 8, 12, 14, 18, 20, 24, 30, there are nine observations. The middle position is (9 + 1)/2 = 5. The value in position five is fourteen, so the median is fourteen. The position formula locates an observation; it does not calculate its measurement value.
For eight ordered values, inspect positions four and five and average their values. In 2, 5, 6, 10, 16, 19, 21, 25, the central values are ten and sixteen. Their mean is thirteen. The median position can be described as halfway between the fourth and fifth observations, but thirteen is the numerical median.
Always count all observations, including repeats and zeros. In a frequency table, the total number of observations is the sum of frequencies, not the number of rows. Use cumulative counts to find which values occupy the central positions, then report the median with the data’s unit where appropriate.
An index tells you an item’s place in an ordered sequence. A measurement tells you the item’s numerical value. In a queue, the third person is identified by position three, but that person’s height is not three centimetres. The distinction is equally important in a numerical dataset.
If the ordered measurements are eleven, thirteen, seventeen, twenty and twenty-eight millimetres, the middle position is three. The value there is seventeen millimetres. The position has no millimetre unit; the median value retains the unit of the measurements. This unit difference is a useful check on the final answer.
Write two labels while learning: median position and median value. The extra words make it harder to substitute the position formula’s output directly into the answer box. Once the distinction is secure, the working can be shortened without losing the underlying two-step process.
The list 18, 4, 12, 9, 20 has five observations, but its third written value is not necessarily the median. Sort it into 4, 9, 12, 18, 20. The third value is twelve, so the median happens to agree with the original third entry in this particular example.
That coincidence is not a method. For 18, 12, 4, 9, 20, the third written entry is four, while the ordered median remains twelve. Changing the recording order does not change the dataset’s median. The statistic depends on numerical order rather than the order in which values were collected or printed.
Keep repeated values when sorting. If two observations both equal nine, they occupy two positions. Sorting rearranges observations; it does not remove duplicates or replace a dataset with its distinct values. Losing a repeat changes the observation count and can shift the central position.
With an odd number n of ordered observations, the central position is (n + 1)/2. For nine observations, four lie before position five and four after it. The extra one in the formula accounts for the central observation itself, leaving equal numbers on either side.
For fifteen observations, the median position is eight, with seven observations on each side. The formula does not depend on whether the values are small, large, positive or negative. It depends only on the count because it locates a rank within an ordered list.
After calculating the position, read the value at that rank. If the eighth observation is forty-two, the median is forty-two. A student who stops at eight has completed the locating step but not the reading step. Preserve the correct position calculation and finish the second part.
With an even count, there is no single observation with equal numbers strictly before and after it. For ten observations, the two central positions are five and six. The usual school numerical median is the mean of the values in those positions.
If the fifth value is eighteen and the sixth is twenty-four, the median is (18 + 24)/2 = 21. The answer need not be an observed value. It is the midpoint of the central pair’s numerical values, not simply the larger of the pair or the positional number 5.5.
The expression (n + 1)/2 gives 5.5 for n = 10, which can serve as a reminder to look between positions five and six. It is not a command to read a nonexistent half observation directly. Translate the positional description into the two actual central values before averaging.
Find the median of 17, 6, 13, 9, 25, 13, 4, 21, 11. There are nine observations, including two separate thirteens. In ascending order they are 4, 6, 9, 11, 13, 13, 17, 21, 25. The middle position is five.
The fifth value is thirteen, so the median is thirteen. The sixth value also happens to be thirteen, but that does not change the odd-count rule. There is one central position for nine observations, and its value determines the median. No average of positions five and six is required.
Check by crossing off one smallest and one largest observation repeatedly. After four pairs are removed, the fifth observation remains. This provides a visible check on the rank. If the data represent measurements, attach their unit to thirteen; if they are abstract scores, report the score meaning clearly.
The table records the ordered dataset from the worked example. Each repeated value retains its own position. Read across from the central position to its value, rather than treating the top-row label as the answer. The two rows describe different kinds of information about the same observations.
| Position in ordered list | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| Observed value | 4 | 6 | 9 | 11 | 13 | 13 | 17 | 21 | 25 |
The table is deliberately simple enough that the position-value distinction is visible. For a larger dataset, writing every position may be unnecessary; a frequency table can compress repeated values. The central question remains the same: which value or pair of values occupies the middle rank or ranks?
Avoid deleting one of the thirteens to make the list look shorter. That would create eight observations rather than nine and change the statistical object being summarised. Repeated measurements are information about frequency, not accidental duplication that can be removed without explanation.
The dataset 2, 2, 2, 7, 9 has five observations but only three distinct values. Its median is the third observation, two. If you keep only the distinct values 2, 7, 9, you obtain a different dataset with median seven. The change comes from discarding frequency information.
This error can be hidden in a frequency table because each distinct value appears only once in the value column. The frequency column tells you how many times it occurs. The total count is the sum of those frequencies, not the number of different value labels.
Before locating a median in a table, calculate the total frequency and write the required central position. That separates the counting job from the value lookup. A correct ordered value column alone does not mean the table can be treated as one observation per row.
Suppose value two has frequency three, value five has frequency four, value eight has frequency two, and value eleven has frequency two. The total count is eleven. The median position is six. Cumulative counts are three, seven, nine and eleven as you move through the ordered values.
Positions one to three contain two. Positions four to seven contain five. Therefore position six contains five, and the median is five. You do not need to expand all eleven observations if the cumulative ranges are clearly understood.
The first cumulative count reaching or exceeding the required position identifies its value row. Here seven reaches past position six, while the previous cumulative count of three does not. This method is a compact rank lookup, not an average of the values in the table’s middle row.
In that table, cumulative frequency seven at value five means seven observations have values at most five. It does not mean the observed value is seven. Five is the data value; seven is the count accumulated through that value. The two quantities belong on different axes in a cumulative-frequency graph.
If the required position is exactly three, it lies in the first row because the first cumulative range includes positions one through three. If it is four, it lies in the next row. Testing a boundary position can prevent an off-by-one error when moving between adjacent rows.
Write a short range such as positions four to seven beside a row if you are unsure. It makes the cumulative total concrete. Once that interpretation is stable, you can use the shorter rule of finding the first cumulative count at least as large as the target position.
Suppose the values three, six and ten have frequencies two, six and two. The total is ten, so inspect positions five and six. Value three occupies positions one and two, value six occupies positions three through eight, and value ten occupies positions nine and ten.
Both central observations therefore equal six. Their average is six, so the median is six. You still considered two positions even though they produced the same value. An even count does not guarantee a median halfway between two different data values.
This is a useful contrast with the habit of averaging the two middle row labels. The three value rows have a middle label of six here, but that coincidence is not the reason the answer is six. The frequency ranges locate the central observations and justify the result.
Now let values three, six and ten have frequencies five, three and two. The total remains ten. Position five is the final three, while position six is the first six. The median is (3 + 6)/2 = 4.5, even though 4.5 does not appear in the value column.
The cumulative frequency after the first row is exactly five, so it reaches the first central position but not the second. You must read the next row for position six. Stopping at the first located row would lose half of the even-count calculation.
This boundary case is an effective check of table understanding. It requires both central ranks and both corresponding values. A method that works only when the pair falls inside one repeated-value block has not yet captured the full even-median rule.
For 0, 0, 4, 7, 9, the total count is five and the median is the third observation, four. The zeros are genuine numerical observations and occupy the first two ordered positions. Removing them without a stated reason would change the dataset and its median.
By contrast, a frequency of zero means no observation occurs at that listed value. A row for value two with frequency zero adds no positions to the ordered data. The zero in the frequency column and a zero in the value column have different roles.
This connects with the series lesson on zero observations and zero frequency. Statistical calculations depend on what each entry represents. Read the column heading before deciding whether a printed zero is an observed measurement or a count saying that no observations of a listed value occurred.
For −8, −3, −1, 2, 6, the middle value is −1. Negative numbers do not get ignored or treated as positive magnitudes. Order them by their position on the number line. Negative eight is smaller than negative three even though eight has the greater magnitude.
For 1.2, 1.05, 1.8, 1.15, 1.4, the ordered list is 1.05, 1.15, 1.2, 1.4, 1.8. The median is 1.2. Comparing decimal places as digit strings can create an ordering error before the central position is even considered.
The position formula is unchanged in both examples because the observation count is five. The values require correct ordering according to their numerical meaning. A median error may therefore come from directed-number or decimal comparison, rather than from the median method itself.
An original dataset records the number of books read by twenty students during a fictional project. The values zero, one, two, three and four have frequencies two, four, five, six and three respectively. Find the median and explain why the number of table rows is irrelevant to the central positions.
The total frequency is twenty. The central positions are ten and eleven. Cumulative counts are two, six, eleven, seventeen and twenty, so both central positions lie in the value-two row, which covers positions seven through eleven. The median is therefore (2 + 2)/2 = 2 books.
There are five value rows, but they summarise twenty observations. Taking the third row because five rows have a middle happens to produce the same answer here, but it is not a justified method. Change the frequencies and that shortcut can fail while the value labels remain unchanged.
Question A: find the median of 14, 3, 9, 6, 20, 11, 8. State the middle position separately. Question B: find the median of 2, 5, 8, 12, 18, 24. State both central values. Question C: explain why the median of nine observations is not necessarily five.
Question D: values one, four and seven have frequencies two, five and two. Find the median. Question E: values one, four and seven have frequencies four, two and two. Find the median, using both central positions. Question F: distinguish a zero observation from a zero-frequency row.
Question G: a learner sorts the distinct values 2, 8, 10 from data 2, 2, 2, 8, 10 and reports median eight. Explain what information was discarded and give the correct median. Your answer should identify the changed observation count rather than merely replace the final number.
For A, the ordered values are 3, 6, 8, 9, 11, 14, 20. Position four is central, and its value is nine. For B, positions three and four contain eight and twelve, so the median is ten. Ten need not be an observed value in an even-sized numerical dataset.
For C, five is only the middle rank. The fifth value can be any number compatible with the ordered data. For D, there are nine observations and position five lies within the value-four block, so the median is four. For E, the total is eight; position four contains one and position five contains four, giving median 2.5.
For F, a zero observation occupies a rank while a zero frequency contributes no observations. For G, the original five observations have median two. Removing repeated twos changes the dataset to three observations and loses the fact that two occurs three times. The new median does not summarise the original data.
On a standard cumulative-frequency graph for measurements, the horizontal axis represents measurement values and the vertical axis represents cumulative counts. A median read-off begins with a central cumulative level, then moves to the graph and down to the measurement axis. The starting count is not the final measurement.
For a grouped distribution of one hundred observations, the usual graphical estimate uses cumulative frequency fifty. If the corresponding horizontal reading is thirty-two minutes, the estimated median is thirty-two minutes, not fifty minutes. Fifty identifies the halfway count in the graph’s vertical scale.
Read both axis labels before tracing a line. A student can follow the graph accurately yet report the wrong axis value if the rank-value distinction is unclear. The same two-step idea from an ordered list remains: locate the centre in terms of observations, then read the associated measured value.
For an exact ordered list with one hundred observations, the usual numerical median averages the fiftieth and fifty-first values. A grouped cumulative-frequency curve commonly estimates the median by reading at half the total cumulative frequency. These are related procedures used with different information and representations.
The grouped graph does not show every exact observation, so the read-off is an estimate based on the grouped display and its interpolation convention. Do not claim that its horizontal reading recovers both exact central observations. Follow the graph-reading method specified in your course or question.
This distinction prevents two valid conventions from being mixed carelessly. Use exact rank lookup for raw or exact discrete-frequency data. Use the stated graphical estimation convention for grouped cumulative displays. The available information determines what kind of answer can honestly be produced.
Suppose grouped data have thirty observations below ten minutes and seventy observations below twenty minutes, out of one hundred in total. The central cumulative level lies in the ten-to-twenty-minute class. That identifies a median class, but not an exact individual central measurement.
A suitable grouped estimate may use interpolation or a cumulative-frequency curve, according to the question. The midpoint fifteen is not automatically the exact median merely because it is the midpoint of the class. The actual observations could be distributed unevenly within the interval.
State whether you are reporting a class, an estimate or an exact median. Those are different objects. A grouped table retains counts in intervals while discarding individual positions within each interval. Statistical precision should reflect that information limit rather than the number of decimal places a calculator can display.
Repeated central values can make several neighbouring positions equal to the numerical median. For 2, 5, 5, 5, 9, the median is five, and three observations share that value. The statistic is still determined even though more than one observation has the median value.
The phrase middle observation refers to position three in this odd dataset. It does not mean only one observation may equal its value. Distinguish the selected rank from the numerical value shared by several ranks. This matters when interpreting claims about how many observations are below or above the median.
With ties, it is not always true that exactly half the observations are strictly below the median and exactly half strictly above. Some can equal it. A more careful description allows at least half to be at or below and at least half at or above, with the usual numerical-median convention.
The median uses central order. The mean redistributes the total equally across the observation count. The mode identifies a most frequent value or values where defined. They answer different summary questions, so one formula should not be substituted for another just because all are called averages informally.
For 1, 2, 2, 3, 12, the median is two, the mean is four and the mode is two. The twelve strongly affects the total used for the mean, while the central ordered value remains two. The matching median and mode are a coincidence of this dataset’s structure.
When a question asks for the median, do not add all values and divide by the count. That is a mean calculation. If you are comparing which summary is useful, discuss the context and distribution rather than declaring one statistic universally better. Each retains different information about the data.
In 2, 4, 7, 9, 11, the median is seven. Replacing eleven with one hundred leaves the ordered central value seven, so the median is unchanged. The largest value changed substantially, but its rank remains above the centre and does not cross it.
This does not mean the median never changes when data change. Replacing seven with eight changes the middle value. Removing or adding observations can change the central positions even if many values stay the same. Inspect both the count and the ordering after any stated alteration.
The series guide on removing one value develops that second case. Here the useful distinction is between changes in numerical magnitude and changes affecting the central ranks. A summary can be insensitive to one type of change without being insensitive to every possible change in the dataset.
If five is added to every observation, the ordered ranks remain in the same order and every central value increases by five. Therefore the median increases by five. For an even count, the average of the two shifted central values also increases by five.
For 2, 4, 8, 10, the median is six. Adding five gives 7, 9, 13, 15, whose median is eleven. The count and central positions remain four observations with positions two and three; the values at those positions have changed. This is another clear separation between rank and measurement.
The transformation needs to apply to every observation as stated. Adding one new observation with value five is a different operation and may change the count and ordering differently. Read whether the task changes each value or adds an extra data point before using a translation shortcut.
If a student reports the median as (n + 1)/2, ask what value occupies that position in the ordered list. The formula may have been used correctly; the solution simply stopped at the locating stage. Preserve that correct work and complete the read-off rather than reteaching the entire topic.
If the central value is wrong despite a correct position, inspect sorting and repeated observations. If the total frequency is wrong, inspect whether the student counted rows instead of adding frequencies. If an even median is wrong, check whether both central positions were located and their values averaged.
Each error has a specific repair. A note such as “position six lies in the value-five block” or “there are twenty observations, not five rows” gives a concrete next action. It is more useful than saying the learner needs to be careful with statistics in general.
Keep the observation count fixed and change the central values. For example, compare two nine-value lists whose fifth entries are four and forty. Their median positions agree, but their median values differ. This directly tests whether the learner has separated the count-based locating step from the value-based answer.
Next keep the distinct value labels fixed and change their frequencies. Ask which rows now contain the central positions. Then include an even total with the pair in one row and another with the pair straddling two rows. These contrasts exercise the cases a one-pattern worksheet may hide.
Finally ask for an explanation with a unit. A position has an ordinal role, while a median waiting time or length retains minutes or centimetres. The unit check can expose a rank reported as a measurement and helps the student read the statistic as a meaningful quantity.
The broader mean, median and mode guide compares the jobs of common averages. The statistics learning route connects ordered data with tables and displays. These destinations help when the difficulty is choosing or interpreting a summary rather than only finding a central position.
Return to the Mathematics learning hub for the series on cumulative frequency, grouped information and changing datasets. Read the exact-data foundation before assuming a grouped graph gives an exact central observation. The distinction between count and value remains useful across all those representations.
Families considering structured support can read the Secondary 1 Mathematics Clementi tutorial page for the approach. A useful checkpoint here is whether a learner can state both the central rank and the numerical value found there in a fresh list or table.
Question H: find the median of −5, 3, −1, 8, 0. Question I: find the median of 1.4, 1.05, 1.2, 1.15. Question J: values two, six and nine have frequencies three, four and five. Find the median and identify both central positions.
Question K: add seven to every value in J. What happens to the median? Question L: a grouped cumulative-frequency graph represents eighty observations. At cumulative frequency forty, the horizontal read-off is twenty-three seconds. State the usual graphical median estimate and identify which number is a count.
For H, the ordered list is −5, −1, 0, 3, 8, giving median zero. For I, the central values are 1.15 and 1.2, giving 1.175. For J, positions six and seven both contain six, so the median is six. For K, it becomes thirteen. For L, the estimate is twenty-three seconds; forty is the cumulative count used to locate it.
The lists 1, 3, 5 and 2, 3, 100 both have median three. Their spreads, totals and largest values differ. Knowing the median does not determine those other features. It summarises central order while leaving many possible datasets compatible with the same reported value.
Likewise, a median of three does not tell you that three occurred most often. In an even dataset, it may not have occurred at all. The values one, two, four and nine have median three, though no observation equals three. The midpoint convention creates a valid numerical centre between the central pair.
When reading a report, ask what extra information is needed for the claim being made. A median can support a statement about central location, but not every statement about variability or individual performance. Correct calculation is the beginning of responsible interpretation, not permission to infer details the statistic does not retain.
If each observation is one student’s weekly reading time, the median summarises students’ reading-time values. If each observation is one day’s total for a whole class, the median summarises daily class totals. Similar units on the axis do not make those datasets interchangeable because the observation unit differs.
Before counting n, identify what one observation represents. A frequency table might count students, journeys, items or days. The median position counts those observations, and the median value describes the measured variable attached to them. This two-part reading prevents a table of group summaries from being mistaken for raw individual data.
For example, the median of five class medians is not automatically the median of all students across the five classes. Group sizes and internal values matter. A summary of summaries can be useful, but its meaning must be stated accurately rather than assumed to reproduce the combined raw-data statistic.
For the nine-measurement opening example, say: “There are nine ordered observations, so the fifth is central; its value is fourteen centimetres, giving a median of fourteen centimetres.” The sentence separates the locating count from the measured answer without becoming long or technical.
For an even example, name both central values before averaging. For a frequency table, name the cumulative range containing each rank. For a grouped graph, call the read-off an estimate and report the measurement-axis value with its unit. These small wording choices preserve the type of evidence used.
You do not need to repeat the whole definition in every final answer. Use enough explanation to show the decisive step, especially when the problem is designed to test a common confusion. Clear working lets you and another reader see whether an incorrect number came from counting, ordering, locating or interpreting.
Suppose the ordered list is 2, 6, x, 14, 20 and its median is ten. With five observations, position three is central, so x = 10. Check that the result respects the stated ordering: ten lies between six and fourteen. The position identified which unknown value the median statement constrained.
For the ordered list 3, 7, x, 19 with median eleven, use the two central values seven and x. Their average satisfies (7 + x)/2 = 11. Multiplying by two gives 7 + x = 22, so x = 15. Fifteen lies between seven and nineteen, so the ordering remains valid. The position 2.5 was a locating description; it was not the unknown value to substitute into the median equation.
If an algebraic answer breaks the given ordering, recheck the model or reject an inconsistent candidate. For example, the first list cannot have median seventeen while x remains between six and fourteen. The requested summary and the order condition must hold together. Knowing a median can determine a particular central unknown in a tightly specified list, even though a median alone cannot reconstruct an unrestricted dataset.
These reverse questions are useful for checking understanding because they turn the usual process around. Instead of locating a known value to report the median, you use the median to constrain the value at a known rank. The distinction between position and measurement remains the same in both directions.
These answers collect the distinctions that often disappear when the median is taught as one formula. The formula identifies a position, while the ordered data provide the value. Return to that separation whenever a frequency table or graph makes the process feel unfamiliar.
Is the median always an observed value?
For an odd number of exact numerical observations, it is the central observed value. For an even number, the usual median averages the two central values and may lie between them without appearing in the dataset. Repeated central values can also give an observed median in an even-sized dataset.
Do repeated values count more than once?
Yes. Each observation occupies a position, even if several have the same value. A frequency table records those repetitions compactly. Count the total observations by adding frequencies, then locate the central rank or ranks. Removing duplicates changes the dataset and can change the median.
Why does the graph start from half the total count?
The usual grouped cumulative-frequency estimate locates the halfway cumulative level, then reads the associated value on the measurement axis. That count is a locating coordinate, not the median measurement itself. Exact discrete data use their actual central observation positions rather than assuming a smooth grouped read-off is exact.
What should I remember first?
Order, count, locate, then read the value. For an even count, locate both central observations and average their values. Keep the data unit on the median answer and keep counts separate from measurements. Those steps carry the same meaning from a short list to a larger frequency table.
Try a final frequency comparison. Values two, five and nine have frequencies one, five and one. There are seven observations, so position four is central and contains five. Change the frequencies to five, one and one while leaving the value labels unchanged. There are still seven observations and the same central position, but position four now contains two. The median changes because the values occupying the ranks changed.
This pair also shows why the middle printed row is not enough. In both tables the middle value label is five, yet only the first table has median five. Write the position ranges for each row to justify the answer. If you can explain both tables without expanding every observation, you understand how frequency compresses an ordered list while retaining the information needed for an exact median. The rank is located from the total count, then translated through the frequencies into a value.