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Two datasets can have exactly the same grouped frequency table and still have different exact means. The table records how many observations fall in each interval, not where every observation lies inside that interval. Grouping keeps some information and discards some detail.
For G3 Mathematics, a mean calculated using class midpoints is therefore generally an estimate of the original mean. The arithmetic can be exact for the midpoint model while the result remains an estimate of the unseen raw data. More calculator digits cannot recover information that the table does not contain.
If your child obtains the expected midpoint answer but calls it the exact original mean, the calculation may be correct while the interpretation needs a small repair. Ask what values were assumed for observations inside each interval. That question reveals why the answer should be described as an estimate.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–27
Suppose a table records two observations in 0 ≤ x < 10 and two in 10 ≤ x < 20. Dataset A could be 1, 2, 11, 12. Dataset B could be 8, 9, 18, 19. Both fit the table, but their exact means are 6.5 and 13.5.
The midpoint calculation replaces the two observations in the first interval with 5 and 5, and the two in the second with 15 and 15. It gives an estimated mean of 10 for either dataset. The same table necessarily produces the same midpoint estimate.
There is no contradiction. The exact means use information that differs between the raw datasets. The midpoint estimate uses only the shared intervals and frequencies. It cannot distinguish where the observations were located within those intervals.
All datasets here are invented teaching examples. Unless otherwise stated, observations are real numbers and intervals include their lower boundary but exclude their upper boundary. That convention makes the possible values and the endpoint checks explicit.
The SEAB G3 Mathematics syllabus for 2027 includes calculating means for grouped data and interpreting statistical representations. The present guide focuses on what a grouped calculation can and cannot tell us about the original observations.
The syllabus also specifies histograms with equal class intervals. No unequal-width histogram technique is required here. Our examples use interval tables and ordinary mean calculations so that the information lost through grouping stays easy to inspect.
This is a reasoning issue as well as a calculation issue. A learner needs to know not just which columns to multiply, but what the resulting number represents. A midpoint is a representative value chosen for an interval, not necessarily an observation that actually occurred.
Subject levels should not become labels for a child’s mathematical identity. A specific question about grouped means can be clarified through a few small datasets. Once the student can explain the difference between retained counts and missing within-class values, the method becomes much easier to use honestly.
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Raw data list the individual recorded values. If the observations are 1, 2, 11 and 12, their exact total is 26 and their exact mean is 26/4 = 6.5. Nothing inside that calculation requires a midpoint assumption.
A grouped table might instead report only that two values lie below 10 and two lie from 10 up to 20, using the stated intervals. It tells us the count in each interval but not the exact total of the values inside it.
The table is still useful. It summarises the broad distribution compactly, makes interval counts easy to compare and can support an approximate measure of centre. Its usefulness does not depend on preserving every feature of the raw data.
The limitation appears when a reader asks a question requiring discarded detail. The exact raw-data total is such a question. To calculate it uniquely, we need the individual values or equivalent information about their sums. Frequencies and interval endpoints alone generally do not supply that information.
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The interval 0 ≤ x < 10 includes zero and values below ten, but not ten itself. Its midpoint is (0 + 10)/2 = 5. That midpoint lies halfway between the two boundaries and is a convenient representative for the class.
The interval 10 ≤ x < 20 includes ten but excludes twenty. Its midpoint is 15. Assigning a value of ten to the second class avoids double counting at the shared boundary. Every allowed observation belongs to exactly one of these adjacent classes.
Do not confuse the midpoint with the class width. The first width is ten, while its midpoint is five. For the second class, the width remains ten while the midpoint is fifteen. These quantities answer different questions.
Before calculating, ask the learner to place a few sample values into the classes: zero, nine, ten and nineteen. A boundary-reading mistake can affect the frequencies before any mean calculation begins. Clear interval interpretation is the first part of the statistical model.
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Use a separate column for the representative value and another for its frequency-weighted contribution. This keeps the assumption visible rather than disguising each midpoint as an exact original observation.
| Class interval | Frequency f | Midpoint m | Contribution fm |
|---|---|---|---|
| 0 ≤ x < 10 | 2 | 5 | 10 |
| 10 ≤ x < 20 | 2 | 15 | 30 |
| Total | 4 | Not a total of midpoints | 40 |
The estimated total is 40 and the number of observations is four, so the estimated mean is 40/4 = 10. The frequency column provides the denominator. The sum of the distinct midpoints is not the total number of observations.
It is helpful to say what the model does: it replaces every observation in a class by that class’s midpoint. The weighted calculation then finds the exact mean of this replacement dataset. The replacement dataset is a model of the original data, not a recovered copy of them.
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Return to Dataset A, 1, 2, 11, 12. The first two values lie in the first class, and the last two in the second. Their sum is 26, giving an exact mean of 6.5. Both class counts match the table.
Dataset B, 8, 9, 18, 19, has exactly the same class counts. Its sum is 54 and its exact mean is 13.5. No interval boundary has been crossed, so the grouped representation has not changed.
This pair is decisive evidence that the exact mean is not uniquely determined by the table. We do not merely suspect that information is missing. We can exhibit two allowed originals with different answers to the exact-mean question.
The same reasoning appears in the guide to using two valid examples to show insufficient information. Both examples must obey every condition. Here they do, including the class counts and the boundary convention, while their exact means differ.
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The midpoint procedure uses only the class boundaries and frequencies. If those inputs are identical, the products and their totals are identical. Therefore the estimate is ten for both Dataset A and Dataset B.
This does not show that the procedure has made an arithmetic mistake. It shows that the procedure deliberately uses a compressed representation. Two originals that become indistinguishable after compression cannot be separated by a calculation that sees only the compressed information.
A student might try to adjust the midpoint because the exact mean of Dataset A is known to be lower. That would use extra raw-data information not present in the table. Once the original data are available, simply calculate the exact mean from them.
When only the table is given, apply the requested midpoint method consistently and describe the result as an estimate. Do not choose different representatives merely to obtain a preferred answer unless the question explicitly supplies a different model or additional within-class information.
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An interval can contain observations on either side of its midpoint without containing the midpoint itself. In Dataset A, neither five nor fifteen appears. Nevertheless, those are the two representatives used by the grouped-mean calculation.
For integer-valued observations, a midpoint can even be a non-integer. The class 1 ≤ x < 4 has midpoint 2.5, although an integer count in that class can be only one, two or three. The representative is not a claim that half a counted item occurred.
This is a familiar feature of averages more generally. A mean can be non-integer even when every observation is an integer. What matters is whether the quantity is being described as an actual observation or as a summary value.
Label the midpoint column clearly. Writing “representative” or “midpoint” reduces the risk that the student treats it as recovered data. Good notation supports interpretation: it helps the reader see which numbers were recorded, which were counted, and which were introduced for the estimate.
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Now let the first class contain three observations and the second contain one. The midpoint products are 3 × 5 = 15 and 1 × 15 = 15. The estimated mean is (15 + 15)/4 = 7.5.
The ordinary mean of the two midpoints is still ten, but it gives the classes equal weight. That ignores the fact that three quarters of the observations lie in the first class. The observations, not the number of class labels, determine the denominator.
For raw values 1, 2, 3 and 11, the exact mean is 17/4 = 4.25. For 7, 8, 9 and 19, it is 43/4 = 10.75. Both datasets fit the new frequencies, and both share the midpoint estimate 7.5.
This example combines two separate safeguards. Use frequencies correctly to calculate the estimate, and then interpret that correctly calculated estimate as a model-based summary. A student can get either part wrong independently, so check the weighting and the meaning separately.
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The method uses a representative at the centre of each interval. It does not demonstrate that observations are evenly spaced, uniformly distributed, or symmetrically arranged within the class. Those features are not recoverable from the grouped counts alone.
For example, a class could contain ten observations all near its lower boundary. Another dataset could place the same ten observations near its upper boundary. Both would give the same midpoint contribution despite very different exact within-class totals.
When teaching, avoid saying “the values are spread evenly” as though it were a known fact. A more accurate explanation is “we use the midpoint to represent the values because their individual positions are not supplied.” That states the approximation without inventing unseen data.
Additional information could justify another representative. If the actual mean within each class were known, multiplying that mean by the class frequency would recover the exact class total. But the original interval table alone does not normally provide those within-class means. Keep assumptions and given information distinct.
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Consider the raw values 4, 6, 14 and 16. Their exact total is 40 and exact mean is ten. They fit the original two-and-two grouped table, whose midpoint estimate is also ten.
No observation equals its class midpoint, yet the deviations cancel. In the first class, four is one below five and six is one above. The second class has the same balanced pattern around fifteen. Their totals equal the midpoint replacement totals.
Therefore “estimated” does not mean “guaranteed to be numerically different from the exact answer.” It describes how the number was obtained and what the available information justifies. An estimate can coincide with the truth without the table proving that it does.
This distinction matters in answer writing. Do not infer exactness because a teacher later reveals matching raw data. The table’s evidential limit remains the same. The midpoint method did not know whether the within-class deviations cancelled, because those deviations were not supplied to it.
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For each raw observation, compare its actual value with the midpoint used to replace it. The difference actual minus midpoint is that observation’s deviation from the replacement value. Adding these deviations gives the difference between the actual total and the estimated total.
For Dataset A, the deviations are −4, −3, −4 and −3. Their total is −14. Dividing by four gives −3.5, which is exactly the difference between the exact mean 6.5 and the estimated mean ten.
For Dataset B, the deviations are 3, 4, 3 and 4. Their total is 14, so the exact mean exceeds the estimate by 3.5. For the balanced dataset 4, 6, 14, 16, the total deviation is zero.
This is an explanation, not a correction method when raw values are absent. Without those values, the deviations are unknown. The calculation shows what information would be needed to turn the estimated mean into the exact one, and why the frequency table alone cannot usually supply it.
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Suppose a third class 20 ≤ x < 30 has frequency zero. Its midpoint is 25, but its weighted contribution is 0 × 25 = 0. It adds neither an observation to the count nor a value to the estimated total.
Do not include 25 as though a new observation had appeared. The row says that no observations fall in that interval. The existence of a class label is different from the existence of recorded values in that class.
The guide to zero observations and zero frequencies develops this distinction for exact-value tables. In grouped data, the midpoint introduces one more role, but frequency still determines how many observations are represented.
A zero-frequency interval can be useful for displaying the range of possible categories or an empty part of a distribution. Its presence should not change the mean simply because another midpoint has been printed. Check that the denominator is the sum of frequencies, not the number of rows.
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For the original two-and-two table, the smallest possible total is 0 + 0 + 10 + 10 = 20. Lower endpoints are included, so this total is attainable. The exact mean can therefore be as low as five.
Every first-class observation is below ten, and every second-class observation is below twenty. The total is strictly less than 10 + 10 + 20 + 20 = 60. Thus the exact mean is less than fifteen, not less than or equal to fifteen.
The table therefore permits exact means in the interval from five inclusive to fifteen exclusive. The midpoint estimate ten lies at the centre of these boundary-derived limits. Dataset A and Dataset B lie within the permitted range.
This bound does not recover a unique mean. It states what the available information guarantees. Be precise about included and excluded endpoints: replacing a strict upper boundary with an attainable maximum would change the conclusion. The data type can also tighten the bounds, as the next section shows.
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If the original observations are known to be integers, the first class contains values from zero through nine, and the second from ten through nineteen. The largest possible total is 9 + 9 + 19 + 19 = 56, giving a maximum mean of fourteen.
The lower bound remains five, attained by 0, 0, 10 and 10. The continuous-data upper limit of fifteen is no longer the sharpest statement because integers cannot approach the excluded endpoints through values such as 9.99 and 19.99.
The standard interval midpoint calculation still uses the stated class boundaries unless the task defines another representative. It gives ten for the original table. Do not silently switch to a different midpoint convention simply because you have noticed that the observations are integers.
This section’s purpose is to show that extra domain information changes what can be guaranteed. Read whether the data represent measurements, counts, rounded measurements or another quantity. A bound is only as accurate as the interpretation of the possible values within each interval.
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A table has classes 0 ≤ x < 10, 10 ≤ x < 20 and 20 ≤ x < 30, with frequencies one, two and one. The midpoints are five, fifteen and twenty-five. Their weighted products are five, thirty and twenty-five.
The estimated total is sixty and the frequency total is four, so the estimated mean is fifteen. A possible raw dataset is 1, 11, 12 and 21. Its exact total is forty-five and exact mean is 11.25.
Another possible raw dataset is 9, 18, 19 and 29. Its exact total is seventy-five and exact mean is 18.75. Both reproduce the same three class frequencies and therefore the same midpoint estimate, despite their different exact means.
For a continuous-data bound, the smallest total is 0 + 10 + 10 + 20 = 40, giving ten. The total is strictly below 10 + 20 + 20 + 30 = 80, giving an upper mean limit of twenty. Both proposed exact means fit these limits, as they should.
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A table records three observations in 0 ≤ x < 4 and one observation in 4 ≤ x < 8. For A, calculate the midpoint estimate of the mean. For B, calculate the exact mean of the compatible raw list 0, 1, 2, 4.
For C, calculate the exact mean of another compatible raw list 1, 2, 3, 7. Explain why B and C do not contradict A. For D, state a lower bound and a strict upper bound for the exact mean when observations are real numbers.
For E, suppose the observations are integers. Find the largest possible exact mean. For F, give a compatible raw dataset whose exact mean equals the midpoint estimate. For G, explain whether the table proves that either midpoint occurred as an actual observation.
Finally, H asks what would happen if a zero-frequency class 8 ≤ x < 12 were added. Would the estimate change? Show the contribution of that row and explain why the number of printed classes is not the number of observations.
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For A, the midpoints are two and six. The weighted total is 3 × 2 + 1 × 6 = 12, and the count is four. The estimated mean is three. For B, the exact total is seven and the exact mean is 1.75.
For C, the exact total is thirteen and the exact mean is 3.25. Both raw lists fit the same frequencies. A uses representative midpoints, while B and C use the actual values, so their different results are compatible with the grouped summary.
For D, the minimum total is four and the total is strictly below twenty. The exact mean is therefore at least one and less than five. For E, the largest integer total is 3 + 3 + 3 + 7 = 16, giving a maximum mean of four.
For F, 2, 2, 2, 6 works and has mean three. For G, no actual midpoint occurrence is guaranteed. For H, the added row contributes zero times ten and adds zero observations, so the estimate remains three. The frequency total, not the row count, stays four.
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If a wide class is divided into narrower classes and the new frequencies are supplied, the table can reveal more about where observations lie. For example, splitting 0 ≤ x < 10 into two classes can distinguish values below five from those at least five.
That additional information may improve a midpoint estimate, but it does not guarantee that every particular refined estimate is closer to the exact mean. Errors in the coarser grouping might happen to cancel especially well. Refinement reduces some uncertainty without promising monotonic improvement in every numerical case.
Do not redraw a table into narrower intervals and invent the missing frequencies. The original wide-class count does not tell us how its observations should be divided between the new classes. New information is required to populate that refinement honestly.
The grouping-sensitivity guide gives a wider discussion of how grouping affects the appearance of distributions. Here the focused lesson is that more detailed class information can constrain possible originals, while a calculation cannot manufacture details absent from the data.
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More observations do not automatically make the midpoint estimate exact. Imagine that every observation in each class lies near its lower boundary. Repeating that pattern many times preserves the systematic difference between actual values and their midpoint replacements.
Duplicating Dataset A produces 1, 2, 11, 12, 1, 2, 11, 12. The exact mean remains 6.5. The frequencies double, but the estimated mean remains ten because both the estimated total and the count double together.
The estimation gap therefore remains 3.5. A larger count has not repaired the missing within-class detail. The issue is the arrangement of values relative to the midpoints, not merely how many rows of raw data might once have existed.
This is not a claim that larger samples are useless in statistics. It is a narrower statement about this deterministic grouping operation. Keep the conclusion within its scope: repeating the same pattern does not remove a midpoint approximation error already built into that pattern.
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If two groups use exactly the same class boundaries and conventions, their frequencies can be added class by class. The midpoint estimate from the combined table then uses those combined frequencies and the common midpoints.
Suppose Group A has frequencies two and two in the original classes, while Group B has frequencies one and three. Their estimated totals are forty and fifty, with four observations each. The combined estimate is ninety divided by eight, or 11.25.
This is also the mean of the two group estimates because their counts are equal. If the group counts differ, weight the estimates by those counts rather than averaging them equally. The same total-and-count principle applies to exact means.
If the class boundaries differ, do not add frequencies by row position as though the intervals matched. A count for 0 ≤ x < 10 cannot be divided reliably into narrower classes without more information. First establish which summaries are genuinely compatible, then decide what calculation the combined information supports.
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SECTION 22 OF 27
21. Do not confuse an exact-value frequency table with an interval table
A table listing the exact values five and fifteen with frequencies two and two describes the raw multiset 5, 5, 15, 15. Its mean is exactly ten within that data model. The value labels are actual possible observations, not interval representatives.
A table listing intervals 0 ≤ x < 10 and 10 ≤ x < 20 with the same frequencies describes many possible raw multisets. Its midpoint estimate is ten, but its exact original mean is not uniquely determined.
The calculations may look identical because both multiply a value-like column by a frequency. Their meanings differ because the first column has a different role. Read the heading and units before deciding whether the result is exact or estimated.
This distinction is particularly useful when a student learns the formula mechanically. A memorised instruction to multiply and divide can produce the expected number while leaving the interpretation unstable. Ask the learner to expand the exact-value table into a list and explain why an interval table cannot be expanded uniquely.
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If a midpoint calculation gives 7.5, writing 7.500000 does not make it a more exact description of the unseen raw-data mean. The trailing digits describe the numerical calculation, not the strength of the underlying information.
Likewise, rounding a midpoint estimate to a requested number of significant figures is a reporting step. It is distinct from the approximation introduced when actual values were replaced by midpoints. There can be two different sources of difference from the original mean.
The guide to rounding items and totals develops the reporting-order issue. In a grouped-mean question, retain sensible calculation precision and use the requested final accuracy, while still describing the midpoint result as an estimate.
A clear answer might say, “Using the class midpoints, the estimated mean is 7.5 units.” That wording names the method and the status of the result. It does not apologise for a legitimate approximation or overstate what the grouped table proves.
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If a student gets the midpoint products wrong, repair the arithmetic or weighting. If they use the sum of class frequencies incorrectly, repair the denominator. If they calculate correctly but call the result exact, repair the interpretation rather than repeating every multiplication.
If the student claims the raw data were uniformly spread, ask where that information appears. If they put an upper-boundary observation in two classes, repair the interval convention. If they treat a zero-frequency midpoint as an observation, repair the distinction between a class and its occupancy.
One short question often reveals the missing relationship: “Could another raw list fit this table?” If the learner can construct two lists and compare their means, the information-loss idea has become concrete. If not, use two classes with only one or two observations each.
Keep feedback specific and optimistic. Grouped data introduce a new representational layer, so a student may need help seeing which numbers were retained and which were substituted. The difficulty is a learnable reading decision, not evidence that statistics is beyond the learner.
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After a worked example, change the frequencies rather than merely changing every number at once. Ask the learner to predict whether the estimated mean should move toward the lower or higher class before calculating. This checks the meaning of frequency weights.
Next, keep the table fixed and ask for two compatible raw lists with different means. Then ask for a compatible list with exactly the midpoint mean. These tasks separate knowing the procedure from understanding why the result is generally an estimate.
The Clementi Secondary 1 Mathematics small-group guide provides the wider teaching context, while the statistics examination guide connects representation reading to written answers.
For a confident learner, ask for boundary-derived limits on the exact mean and require the endpoint convention to be explained. This extends the same idea without introducing an unrelated technique: the grouped table does not determine one exact answer, but it still constrains what answers are possible.
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Two midpoint estimates can be compared as estimates, but their ordering does not always prove the ordering of the exact original means. Consider two groups using the common classes 0 ≤ x < 10 and 10 ≤ x < 20. Group A has frequencies one and three; Group B has frequencies three and one.
The midpoint estimate for A is (1 × 5 + 3 × 15)/4 = 12.5. For B it is (3 × 5 + 1 × 15)/4 = 7.5. Based on the midpoint model, A has the higher estimated mean.
Now choose compatible raw values for A as 0, 10, 10, 10. Their exact mean is 7.5. Choose compatible raw values for B as 9, 9, 9, 19. Their exact mean is 11.5. Every observation lies in its stated class, yet the exact ordering is the reverse of the midpoint-estimate ordering.
This does not prevent a student from making the comparison requested in an examination question. It clarifies the wording: “A has the higher estimated mean using class midpoints” is supported. “A’s exact original mean must be higher” is a stronger claim not guaranteed by these tables.
Extra information can sometimes make a definite ordering possible. For example, if one group’s smallest possible mean exceeded another group’s largest possible mean, their exact ordering would be settled. In the present example, the permitted ranges overlap, leaving room for the reversal demonstrated above.
The practical habit is to match confidence to information. Do the midpoint calculation accurately, use it for the requested summary, and avoid presenting the unseen raw data as though their within-class positions were known. A useful estimate and an honest limitation can coexist in the same clear answer.
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How should unequal-sized groups be combined?
Suppose one grouped table represents four observations with midpoint estimate ten, while another represents eight observations with midpoint estimate fifteen. Their estimated totals are forty and one hundred twenty. Combining those totals gives one hundred sixty across twelve observations, so the combined estimate is 40/3.
The ordinary mean of ten and fifteen is 12.5, which gives the groups equal weight despite their different counts. The correct combined estimate is closer to fifteen because the second group represents twice as many observations. This is the same total-and-count relationship used for ordinary means, now applied to estimated totals.
The calculation assumes compatible representations and the same midpoint model as the original estimates. It does not turn the combined estimate into an exact raw-data mean. Two questions must stay separate: have the available summaries been weighted correctly, and what information did those summaries preserve? Correct weighting repairs the first issue. It cannot recover the missing within-class values responsible for the second. State the combined result as an estimate when that is what its inputs support.
What if the last class has no finite upper boundary?
An open-ended class such as x ≥ 20 has no ordinary midpoint determined by two finite boundaries. Do not invent an upper boundary simply to complete the usual table. The count tells us how many observations lie at or above twenty, but it does not supply a representative value in the same way as a bounded interval.
The question might provide an additional assumption or ask for a different kind of conclusion. If not, explain why the ordinary midpoint calculation cannot be completed from the stated information alone. A neat numerical answer is not preferable to an honest account of a missing input. This limitation follows from the method’s definition, not from a difficulty with multiplication.
Can an exact total remove the uncertainty about the mean?
Yes. If the exact raw-data total and the total number of observations are supplied independently, divide those quantities to obtain the exact mean. The individual raw list need not be known if its exact sum is known. Grouping alone loses within-class sums, but additional information can restore what the particular calculation needs. Check the source and meaning of that total: an estimated midpoint total should not be relabelled as an exact raw-data total.
Why is an estimate still useful?
No. It is a standard way to summarise grouped data when the individual values are unavailable. Apply the method requested and state what the result represents. An approximation can be useful without being an exact recovery of the raw data.
Can the estimate equal the exact mean?
Yes. Within-class deviations can cancel, as in 4, 6, 14, 16 for the original table. The grouped table alone does not guarantee that cancellation, so agreement in a revealed dataset does not make the method universally exact.
Does the table tell us the exact median too?
Not generally. It can locate observations within classes without giving their precise positions. The exact information available depends on the question and the representation. Do not transfer a midpoint assumption to another statistic without justification.
What is the most useful habit to remember?
Name what is known, name what is represented by a midpoint, and distinguish the model’s mean from the original mean. Use the Mathematics explanation directory to continue with another focused interpretation guide.
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