Sec 2 G2 Maths Bukit Timah | Missing Data Values

Did you know that the average of a group can reveal a missing test score, even when nobody remembers the score itself? Suppose five numbers 4, 6, x, 10 and 12 have a mean of eight. Their total must be 5×8=40. The four known numbers sum to 32, so x is eight. This is a practical way to turn a statistics statement into a simple equation.

But not every missing-data question has one numerical solution. A median can remain unchanged while the largest observation moves dramatically. A mode may stay the same for several possible missing values. A Secondary 2 G2 Mathematics student should learn to recognise which information a mean, median or mode actually controls—and whether a unique answer is justified.

This guide for Secondary 2 G2 Mathematics tuition in Bukit Timah includes total-score equations, ordered positions, tied modes, missing frequencies, data corrections and interpretation. A tutor can use it to diagnose whether a learner is struggling with calculation, data sorting or the deeper idea of what the given summary does and does not determine.

Quick answer: how to use mean, median or mode

Given information	What you can conclude
Mean μ of n data values	Total sum = n×μ.
One missing x and known sum S	x = n×μ−S.
Odd-sized dataset median	The central ordered observation.
Even-sized dataset median	The average of two central ordered observations.
Mode	The most frequently occurring observed value or tied values.
Missing value with too few constraints	More than one answer may be possible; show counterexamples.
A frequency table of exact scores	Mean = sum(value×frequency)/sum(frequency).

Secondary 2 G2 Mathematics: understanding statistics rather than guessing

G2 describes a subject level under Singapore’s Full Subject-Based Banding system. Lower-secondary Mathematics develops data handling and the interpretation of statistical displays and measures of central tendency. The exact order of school assignments differs, so a tutor should begin with the student’s own classwork. Some missing-data puzzles below are extensions of the basic mean, median and mode definitions; their purpose is to teach reasoning, not to claim every school assesses all variants.

Official reference pages: Yishun Secondary School Mathematics content outlines and MOE’s G2 Mathematics syllabus. These identify curriculum context rather than the live assessment scope of a particular school.

Twenty-three foundations for missing-data reasoning

1. Mean is total divided by count

The arithmetic mean of n numerical observations is their sum divided by n. For 4,6,8,10,12, the sum is 40 and the mean is eight. When a missing value appears, the meaning of mean has not changed: all the observations must still add up to n times the mean. Starting with a plain sentence—’the total equals the number of values times their average’—often makes the algebra much easier.

Teaching move: Ask the student to find total and count separately before calculating any missing x.

Watch out: The total number of observations includes the unknown value even though its number is not yet known.

2. Reverse the mean formula to find a missing total

If five scores have a mean of eight, then their combined score is 5×8=40. This does not reveal how that total is distributed among the scores, but it gives a powerful constraint. A learner should write total=mean×count before trying to guess the missing score. The count includes all recorded measurements, not just known entries.

Teaching move: Draw five empty score boxes whose sum must be forty.

Watch out: Multiplying the mean by only four known scores changes the target total.

3. One missing value can be found by subtraction

For the five values 4,6,x,10,12 with mean eight, the total is 40. Known values sum to 32, so x=40−32=8. This is an example of building an equation from the mean, not reverse engineering an answer by trial and error. A correct check restores the full five-value list and calculates its mean again.

Teaching move: Write 4+6+x+10+12=5×8, then solve.

Watch out: Only adding known values and dividing by four would calculate a different mean.

4. A mean need not equal any observed score

The values 3,5,6,10 sum to 24 and have mean six. In this particular list six happens to occur, but 3,5,7,9 also have mean six with no observed six. The mean describes a balance point of the total, not necessarily a member of the dataset. This matters when finding a missing item, because its value is constrained by the total, not by an assumption that it must match the mean.

Teaching move: Compare a set containing the mean with one that does not.

Watch out: A student’s missing-score answer should not automatically be the given mean.

5. A single missing value may be negative unless context forbids it

The algebraic dataset 5,8,x has mean four, so 5+8+x=12 and x=−1. That is mathematically valid numerical data. But if the story describes examination marks on a zero-to-one-hundred scale, a negative result is impossible and the premises would need reconsideration. The context decides permitted values, which is why a final answer needs more than algebra alone.

Teaching move: Explain why a signed temperature dataset may allow a negative observation while a test-score dataset may not.

Watch out: Rejecting every negative x without considering the meaning of the measurements is also incorrect.

6. Median depends on sorted positions

The median is the middle of the ordered dataset, not the middle number as printed on an unsorted page. For 3,9,4,12,6, sort first into 3,4,6,9,12, then median six. If an unknown value is present, where it falls in the ordering matters. An algebraic median equation may require a condition such as 4≤x≤9 to be valid.

Teaching move: Have the student number the positions after sorting.

Watch out: Using the centre item from the original unsorted list produces unreliable answers.

7. Odd-sized datasets have one middle observation

For five ordered values, the third is the median. In the ordered list 2,4,x,9,12, a stated median of six means x=6 if the list is indeed ordered as shown. The 4≤x≤9 condition is satisfied, so the result is valid. This is a different kind of reasoning from the mean: position rather than total determines the value.

Teaching move: Mark positions one through five and circle the third.

Watch out: An unknown at one end of the sorted list cannot automatically be inferred from a given median.

8. Even-sized datasets average two central values

For six ordered values, the median is the mean of positions three and four. In 1,3,5,x,9,11 with x between five and nine, a median of six means (5+x)/2=6, so x=7. The order condition must be checked after solving; otherwise the assumed middle pair could be wrong. Students should show the two central entries explicitly rather than use the mean formula for all six.

Teaching move: Draw six slots and mark the two central positions.

Watch out: Dividing the total of all six observations is the mean calculation, not the median calculation.

9. A median may not determine a missing extreme

Suppose four values are 2,4,8,x and x is at least eight. The sorted middle values are four and eight, so median is six whatever larger x is chosen. The median does not reveal the missing maximum. This is a useful counterexample to the assumption that every one-unknown statistics question has a unique answer. Identify which positions the median actually depends upon.

Teaching move: Test x=9 and x=20 and compare the medians.

Watch out: Reporting x=8 solely because median is six ignores multiple valid larger values.

10. Mode is the most frequent observed value

In 2,4,4,5,10, the value four appears twice, more often than any other, so the mode is four. A question about a missing value may instead constrain which frequencies become largest. The mode is a value from the dataset, not its frequency count. A learner should make a tally before declaring that a missing x must equal the given mode.

Teaching move: Prepare a frequency count of each distinct value.

Watch out: Reporting mode ‘two’ because the highest frequency is two confuses value and frequency.

11. A stated mode often does not determine an unknown uniquely

In 2,4,4,5,7,x, four is already the unique mode for many choices of x not equal to two, five or seven. The unknown could equal four, increasing its frequency, or it could be an entirely new value without changing that unique mode. A statement ‘mode is four’ alone may not identify one value. Students must reason about the actual frequencies and recognise missing information.

Teaching move: Choose x=4 and x=9 and check that both preserve the same mode.

Watch out: Guessing x=4 whenever the mode is four is not universally justified.

12. Sometimes the unique-mode condition does fix x

For values 2,2,3,3,x, the starting two values are tied for highest frequency. To make two the unique mode, x must be two. Then two occurs three times and three twice. If x is a new unrelated value, the tie remains; if x=3, three becomes unique mode instead. Here the unique-mode condition supplies enough information.

Teaching move: Build the frequency tally before and after adding x.

Watch out: A prompt saying merely ‘a mode is two’ may be weaker than saying ‘two is the only mode’.

13. Tied modes are legitimate

A dataset can have two or more values sharing the highest frequency, such as 2,2,3,3,5. It is bimodal with modes two and three. Other sets can have no repeated values, in which case school conventions on ‘no mode’ should be followed. A missing-value puzzle must distinguish the phrase ‘the mode’, ‘one of the modes’ and ‘unique mode’. Those distinctions affect whether a solution is unique.

Teaching move: Construct a tie in a tally chart and then change one observation to break it.

Watch out: Telling a child there must always be exactly one most common number is false.

14. Range depends on the minimum and maximum

The range is maximum minus minimum, not the mean or median. If the smallest value is three and the range is fourteen, the maximum is seventeen. A missing observation known to be the largest can then be found directly. However, if x lies somewhere inside the dataset, the same range may not determine it. Students should locate which observations can be endpoints.

Teaching move: Ask whether x is a minimum, maximum or interior value.

Watch out: Subtracting two middle values to calculate range confuses the measure with an interquartile idea.

15. Mean and median together can cross-check one unknown

If five values are 2,4,x,8,10 with mean six, their total must be thirty and known sum is 24, so x=6. When ordered, 2,4,6,8,10 has median six too. The mean condition finds x; the median condition confirms that x lies where expected. Two given measures can provide separate constraints, but not every pair yields a unique consistent dataset.

Teaching move: Check the value through both sum and sorted middle position.

Watch out: Treating the two constraints as unrelated numbers to average together is not meaningful.

16. Changing one observation changes the mean by the change divided by n

For five observations, replacing one score eight with thirteen increases the total by five and the mean by one. This holds without knowing the other four values. If the original mean was twelve, the corrected mean is thirteen. Recognising this linear response is a useful shortcut and a strong plausibility check when data-recording errors appear.

Teaching move: Write the original sum as n×mean, then add the score correction before dividing.

Watch out: A corrected score should not be treated as an extra sixth observation when it replaces an existing one.

17. Median is usually less sensitive to a changed extreme

For 2,4,6,8,10, replacing the maximum ten with one hundred changes mean from six to twenty-four, but the median stays six because the central ordered position does not change. This contrast helps explain why the mean and median have different purposes. It also warns pupils not to assume a new mean must imply a changed median.

Teaching move: Sort the two lists and compare their centre values and sums.

Watch out: The median is not completely immune to every data change; a value crossing the centre can alter its position.

18. A frequency table represents repeated observations compactly

If score two occurs three times, score five occurs twice and score eight once, the table describes six individual scores. Its total is 3×2+2×5+1×8=24 and mean is 24/6=4. A student can find a missing frequency by writing sum of products and total count expressions. The frequency itself is how many observations, not another score value.

Teaching move: Expand the table into a raw list once to demonstrate what frequencies mean.

Watch out: Dividing the total weighted sum by the number of distinct score categories is incorrect.

19. A missing frequency changes both the weighted sum and count

For scores 2,4,6 with respective frequencies 2,f,1, the weighted sum is 4+4f+6=10+4f and count is f+3. If the mean is 3.5, solve (10+4f)/(f+3)=3.5 to get f=1. Check the four raw observations 2,2,4,6 have mean 3.5. Not every target mean creates a valid non-negative integer f.

Teaching move: Name f as a frequency and state f must be a whole number zero or greater.

Watch out: Treating the denominator as three categories rather than f+3 gives a wrong equation.

20. Some frequency means cannot determine a count

For scores three, five and seven with frequencies two, f and two, the mean is five for any non-negative f, because the three and seven values balance symmetrically around five. The equation simplifies to an identity, not a unique answer. Such cases demonstrate that solving algebra and checking uniqueness are both parts of the question.

Teaching move: Substitute f=0 and f=4 to see the unchanged mean.

Watch out: Assuming a single f-value must exist simply because a letter is shown can create a false conclusion.

21. Combined means require weighting by sample sizes

Suppose group A contains four students averaging six points, total 24, and group B contains six students averaging eight points, total 48. Together the ten students total 72, so combined mean is 7.2. The simple average of six and eight is seven, which is incorrect because the groups are different sizes. A missing-value question with combined groups uses each group’s actual count.

Teaching move: Write total points above total number of students before averaging.

Watch out: Averaging two reported means equally assumes equal sample sizes when none are given.

22. Grouped class means are estimates rather than exact originals

When individual values are placed into intervals, the exact observations inside each class may no longer be known. A class midpoint times frequency supports an estimated mean, but it does not reconstruct a unique original raw dataset. This is distinct from a frequency table listing exact scores. The presence of classes such as 0≤x<5 should prompt students to consider midpoint approximation.

Teaching move: Contrast a frequency table of exact values with a grouped interval table.

Watch out: Claiming an estimated grouped mean determines every original score is unjustified.

23. A missing result must respect the question’s context

Test scores may need to be whole numbers within a range such as zero to twenty; measured lengths or times can be decimal; temperatures can be negative. Algebra can produce a value that violates those constraints. A student should check whether the computed unknown is plausible in the named context and explain an inconsistent premise rather than rounding to a convenient but false answer.

Teaching move: State permitted values for x before or after solving according to the question.

Watch out: A fractional number of people or a negative examination mark should not be accepted without support.

24. A complete statistical explanation distinguishes what is known from what is inferred

A mean fixes a total, a median constrains central order and a mode constrains frequencies. Sometimes those facts uniquely identify an unknown; sometimes they allow several answers. The most important habit is to state why an answer is unique or why there is insufficient information. That moves the student beyond a formula into genuine statistical reasoning.

Teaching move: Ask for one alternative dataset satisfying the same known median or mode.

Watch out: A teacher’s example with one solution does not prove every missing-data question behaves the same way.

Twenty-five worked examples of unknown data and statistics

Try to name the summary measure before calculating: mean controls the total, median controls the ordered middle, and mode controls frequency. Every worked example includes a check, and sometimes the right conclusion is that the information is not sufficient.

Worked example 1: Find a missing mean score

Question: Five scores are 4,6,x,10,12 with mean eight. Find x.

Method: Required total=5×8=40. Known sum=4+6+10+12=32. Hence x=8.

Check: Full set sums 40, mean 8.

Common mistake: Include x in the count even before its value is known.

Independent variation: Change the mean to nine and find x.

Worked example 2: A four-score mean equation

Question: Scores are 3,5,x,10 with mean six. Find x.

Method: Total=4×6=24; known values sum 18; x=6.

Check: 3+5+6+10=24.

Common mistake: Do not divide known sum 18 by only three to finish.

Independent variation: Compute the median of the finished list.

Worked example 3: Another five-score missing value

Question: Scores 3,4,5,5,6,x have mean five. Find x.

Method: Six values require total 30. Known sum=23, so x=7.

Check: 3+4+5+5+6+7=30.

Common mistake: A mean five does not mean missing x must be five.

Independent variation: Find median and mode afterwards.

Worked example 4: A negative algebraic observation

Question: Three numerical data values are 5,8,x with mean four. Find x.

Method: Total required=3×4=12. Known sum 13, so x=−1.

Check: 5+8−1=12, mean four.

Common mistake: The answer may be valid for signed data but impossible as a nonnegative score.

Independent variation: Discuss which real-world contexts allow negative observations.

Worked example 5: Missing observation from total

Question: Five values have mean ten; four are 7,8,9,11. Find fifth.

Method: Total 5×10=50. Known sum35, so missing=15.

Check: 7+8+9+11+15=50.

Common mistake: No reason requires missing value to lie between existing values.

Independent variation: Find range and median of completed set.

Worked example 6: Odd median with an ordered unknown

Question: Five ordered values are 2,4,x,9,12. Median is six. Find x.

Method: The third item is median, so x=6. It meets 4≤6≤9.

Check: Ordered list median is indeed six.

Common mistake: Do not apply the mean-sum formula to a median problem.

Independent variation: Change stated median to seven.

Worked example 7: Even median with unknown fourth item

Question: Six ordered values are 1,3,5,x,9,11. Median is six. Find x.

Method: (5+x)/2=6, so 5+x=12 and x=7.

Check: The sorted list 1,3,5,7,9,11 has median (5+7)/2=6.

Common mistake: The two middle observations are third and fourth, not first and last.

Independent variation: Try median 6.5.

Worked example 8: Median does not fix the maximum

Question: Four values are 2,4,8,x, with x≥8. Their median is six. Find x.

Method: There is no unique value. The middle pair remains 4 and 8 for every x≥8, so median=(4+8)/2=6.

Check: x=9 and x=20 both satisfy the same median.

Common mistake: Do not invent x=8 because the question mentions six.

Independent variation: Give two distinct valid x-values and compare their means.

Worked example 9: Mean and median cross-check

Question: Five values 2,4,x,8,10 have mean six. Find x, then median.

Method: Total 30; known sum 24; x=6. The list is ordered 2,4,6,8,10, so median six.

Check: Both reported measures agree.

Common mistake: The median alone might constrain central placement, but the mean equation determines x here.

Independent variation: Change the value ten to twelve and recalculate.

Worked example 10: A single clear mode

Question: Data are 2,4,4,5,10. Find mode.

Method: Four appears twice; all others once, so mode=4.

Check: Frequency table has highest count two at value four.

Common mistake: Mode is four, not two.

Independent variation: Add another five and describe the new modes.

Worked example 11: Unique mode fixes a tie

Question: Data are 2,2,3,3,x. What x makes two the unique mode?

Method: Choose x=2, giving three twos and two threes.

Check: Any new different x keeps the tie; x=3 makes three the unique mode.

Common mistake: Use ‘unique’ or ‘only’ carefully.

Independent variation: What x makes three the only mode?

Worked example 12: Mode alone is insufficient

Question: Data are 2,4,4,5,7,x and the only mode is four. Must x equal four?

Method: No. x=4 makes four occur three times; x=9 also leaves four with highest frequency two. Both valid.

Check: Neither example creates a tie at count two.

Common mistake: Do not force the missing observation to equal a stated mode.

Independent variation: Explain why x=5 would create two modes.

Worked example 13: Bimodal dataset

Question: Data are 2,2,3,3,5. Identify modes.

Method: Two and three each appear twice, more than five. Thus the data are bimodal with modes 2 and 3.

Check: Both top frequencies are equal.

Common mistake: Do not pick the lower number as the sole mode.

Independent variation: Add one three and state new unique mode.

Worked example 14: Range and missing maximum

Question: The smallest observed value is three, largest x, and range fourteen. Find x.

Method: Range=x−3=14, so x=17.

Check: 17−3=14.

Common mistake: This solution assumes x is the maximum.

Independent variation: If the missing value is interior rather than maximum, is x determined?

Worked example 15: Range alone not enough for an interior value

Question: Data are 2,5,x,12 with 2≤x≤12 and range ten. Find x.

Method: There is no unique x. Minimum two and maximum twelve already fix the range, regardless of an interior x.

Check: x=7 and x=9 both work.

Common mistake: The range does not locate middle observations.

Independent variation: Compare the mean for two possible x-values.

Worked example 16: Corrected-score mean change

Question: Five scores average twelve. One score eight was recorded but should have been thirteen. Find corrected mean.

Method: Original total=5×12=60. Replacing eight with thirteen adds five, so corrected total 65 and mean 13.

Check: Mean rises by 5/5=1.

Common mistake: Replacement is not adding a sixth score.

Independent variation: What if corrected value were eighteen?

Worked example 17: An outlier changes mean but not median

Question: Data are 2,4,6,8,10. Replace ten with one hundred. Compare means and medians.

Method: Original sum30, mean6; corrected sum120, mean24. Median remains six in both sorted lists.

Check: Central position unchanged; total rises ninety over five observations.

Common mistake: Do not assume every summary measure responds in the same way.

Independent variation: What happens to range?

Worked example 18: Frequency table mean

Question: Scores 2,5,8 occur with frequencies 3,2,1. Find mean.

Method: Weighted total=2×3+5×2+8×1=6+10+8=24; count=3+2+1=6. Mean=24/6=4.

Check: Expanded data 2,2,2,5,5,8 also have mean four.

Common mistake: Do not divide by three distinct score values.

Independent variation: Compute mode and median from expanded list.

Worked example 19: Missing frequency from given mean

Question: Scores 2,4,6 have frequencies 2,f,1 and mean 3.5. Find f.

Method: Weighted sum=10+4f; total=f+3. Solve (10+4f)/(f+3)=3.5. Then 10+4f=3.5f+10.5, giving f=1.

Check: Data 2,2,4,6 sum14 over four, mean3.5.

Common mistake: Frequency must be a non-negative whole number.

Independent variation: Try a stated mean of 4 and investigate consistency.

Worked example 20: Mean cannot fix symmetric frequency

Question: Scores 3,5,7 occur with frequencies 2,f,2. Given mean five, find f.

Method: Weighted sum=6+5f+14=20+5f, count=4+f. Dividing gives five for every non-negative f. There is no unique f.

Check: f=0 yields 3,3,7,7 mean five; f=4 also has mean five.

Common mistake: A missing letter does not guarantee one unique numerical solution.

Independent variation: Explain why the two endpoint values balance around five.

Worked example 21: Combine unequal group means

Question: Group A has four observations of mean six; Group B has six observations of mean eight. Find combined mean.

Method: Totals are 24 and 48, making 72 across ten observations. Combined mean=72/10=7.2.

Check: Seven-point-two lies between six and eight, nearer eight because B is larger.

Common mistake: Do not take simple mean (6+8)/2=7 when group sizes differ.

Independent variation: Find combined mean if group B has four observations instead.

Worked example 22: Find a missing data value in an even list

Question: Data 4,6,8,x have mean seven. Find x and median.

Method: Required total=4×7=28, known sum18, x=10. Sorted list 4,6,8,10 has median (6+8)/2=7.

Check: Both mean and median equal seven.

Common mistake: Median formula needs the sorted two central values.

Independent variation: Change mean to eight and calculate new x.

Worked example 23: A dataset with no repeated values

Question: Values 2,4,6,8,10 are all distinct. What is mode?

Method: No value has a higher frequency than others. Under the usual school description, there is no mode.

Check: Each value appears once.

Common mistake: Do not report the middle six as mode; it is the median.

Independent variation: Change the last ten into six and find mode.

Worked example 24: Impossible test score context

Question: Five examination scores on a 0–20 scale are 17,18,19,20,x and mean twelve. Find x.

Method: Total must be 60; known sum74, so x=−14. This is outside the allowed score range. Therefore the stated premise is inconsistent for such scores.

Check: No nonnegative missing mark could reduce total below 74.

Common mistake: Do not quietly accept a negative test score just because algebra gave it.

Independent variation: Find the minimum possible mean if x=0.

Worked example 25: Grouped class mean versus exact data

Question: A class interval 0≤x<10 has frequency five. Does its midpoint five prove all five original scores were five?

Method: No. The midpoint is used for an estimated grouped mean. The original scores could vary within the interval, so exact raw values are not recoverable from class counts alone.

Check: Five different distributions can share the same interval frequency.

Common mistake: Do not claim a class midpoint is the exact value of every observation.

Independent variation: Give two possible raw datasets with the same class frequency.

Eleven diagnostic errors with precise repairs

1. Unknown not included in count

How the mistake appears: The learner computes total from n known values rather than n+1 complete observations.

Repair: Name the full dataset size first and multiply mean by that count.

Changed question: Write an equation for a fresh five-score question with one x.

2. The mean is mistaken for the missing number

How the mistake appears: The student automatically reports x=8 because the five-value mean is eight.

Repair: Calculate required total, subtract known sum and check the full list.

Changed question: Construct a dataset whose missing value is not equal to its mean.

3. Median found before sorting

How the mistake appears: A learner chooses the middle printed entry of an unsorted list.

Repair: Sort, number positions and then select one central observation or the average of two.

Changed question: Reorder the same list and show that its median stays unchanged.

4. Two central observations ignored for even n

How the mistake appears: The pupil selects the third of six numbers without averaging positions three and four.

Repair: Write the two central slots and use (a+b)/2.

Changed question: Solve a six-value dataset with one unknown fourth value.

5. The median is treated as a sum rule

How the mistake appears: A child multiplies median by n to find total, as though it were the mean.

Repair: Explain that the median is a positional measure and may remain unchanged as the extremes change.

Changed question: Find two datasets with the same median but different totals.

6. Stated mode assumed to fix x

How the mistake appears: A problem says mode four, so the child assigns x=4 without checking frequencies.

Repair: Make a tally and test at least two possible values of x.

Changed question: Explain when the phrase ‘unique mode’ supplies enough information and when it does not.

7. Mode value and mode frequency confused

How the mistake appears: A learner reports the highest count rather than the score that has that count.

Repair: Keep separate columns named value and frequency.

Changed question: Ask for both modal value and frequency in a new table.

8. Frequency table total incorrectly counted

How the mistake appears: Student divides a weighted score sum by the number of distinct score categories.

Repair: Add all frequencies to find n and calculate sum of score×frequency.

Changed question: Expand a short frequency table to its raw list.

9. Changing a score treated as adding a person

How the mistake appears: Correcting a recorded eight to thirteen is counted as a new sixth score.

Repair: Replace the observation, adjust the sum by plus five and keep sample size fixed.

Changed question: Compare before-and-after means with an independently calculated total.

10. Integer and physical score restrictions ignored

How the mistake appears: An algebraic x=−14 is reported as a test score despite the stated 0–20 range.

Repair: Check whether the result is permitted; if not, identify inconsistent premises.

Changed question: Construct a valid numerical dataset with a signed negative temperature instead.

11. Grouped estimates treated as exact observations

How the mistake appears: Student claims every score in 0≤x<10 is five because midpoint is five.

Repair: Explain midpoint approximation and information lost by grouping.

Changed question: Give two different raw five-score lists within the interval.

Six small-group teaching investigations

Investigation 1: A student who knows division but cannot reverse an average

The pupil can find the mean of 4,6,8,10,12 yet freezes when one value is x. Ask them to imagine five empty boxes that must total forty for a mean of eight. Fill four boxes with the known entries and find the remainder. Only then introduce 4+6+x+10+12=40. The algebra becomes a short description of the total, not a new rule to memorise.

What counts as progress: The learner can independently use n×mean to calculate total and then subtract a known sum in a different question.

Investigation 2: A pupil who confuses median and mean

Give 2,4,6,8,10 and then replace ten with one hundred. The mean changes from six to twenty-four while median stays six. Have the learner sort both sets and calculate each summary separately. The large outlier shows that mean depends on the sum whereas median depends on positions. Ask which measure better reflects the typical central score under the chosen interpretation.

What counts as progress: The child states why an altered extreme observation can change the mean without moving the central ordered value.

Investigation 3: A child who assumes every missing-value puzzle has one answer

Show the ordered list 2,4,8,x with x≥8 and median six. Ask for x=9 and x=20; both satisfy. Discuss what additional information, such as the mean or range, would pin down x. This develops evidence-based reasoning and prevents filling in an arbitrary value because a blank space appears on the worksheet.

What counts as progress: The student explicitly states ‘not enough information’ and supplies at least two valid counterexamples where appropriate.

Investigation 4: Three pupils compare mean median and mode

Give a dataset 2,2,4,6 and let three students calculate a different summary. Mean is 3.5, median is three, and mode is two. (The ordered middle pair 2 and 4 averages to three.) Ask each to explain what information their measure uses. Then change the largest six to sixteen and rotate roles. The exercises show that related statistics are not interchangeable and sometimes respond differently to an outlier.

What counts as progress: Each pupil can compute and interpret all three measures alone after group discussion rather than relying on the assigned role.

Investigation 5: A missing-frequency equation with a domain check

Present exact values 2,4,6 with frequencies 2,f,1 and mean 3.5. Let the learner write the weighted total 10+4f and count f+3, solve f=1 and reconstruct 2,2,4,6. Then change target mean to four; no finite f satisfies it, as the equation becomes contradictory. Explain why valid frequencies must be non-negative integers and why some stated averages cannot occur.

What counts as progress: A correct learner checks both the algebraic result and whether the reconstructed physical data table produces the mean.

Investigation 6: A parent checking a real school worksheet

Ask the child first what the problem gives: a mean, a median, a mode, a frequency table, or several constraints. Then ask whether that information controls a sum, a middle position or repetition counts. The child should form an equation or tally only after deciding the meaning. If the task is ambiguous, encourage a valid second example rather than forcing a guess. This makes the mathematics conversational without surrendering precision.

What counts as progress: The child identifies the relevant summary measure and explains whether the unknown is unique.

Fifty-three quick statistics questions with answers

Use these as short retrieval practice. A wrong answer should prompt a question about the first missing fact: total count, sorted order, frequency tally or contextual restriction. Retest with different values after a delay, rather than copying the model answer on the same page.