The same average does not settle a decision
Two groups in a fictional school garden each have an average observed height gain of 10 cm. That sounds like a tie. But suppose the school asks for a trial that produces reliable minimum gains rather than a high occasional result. The average alone cannot tell us how many pots clear the minimum. We must return to individual observations, choose the specified threshold, count correctly and explain what the comparison does and does not establish.
This volume follows Chemistry Volume 0528 on ammonia and ammonium salts. Knowing a substance’s formula does not tell us how a plant experiment turns out, so we now work with separate imaginary garden observations. The aim is a particular examination skill: **make a conditional decision from a raw distribution when mean, spread and threshold-based pass rates lead to different impressions**. The general methods of standard deviation and pooled means already have owners in Volume 0351 and Volume 0395; this article applies them to a new decision rule, without pretending a dataset proves an actual fertiliser’s effectiveness.
Quick facts that prevent an avoidable wrong answer
• Read the threshold before calculating a success rate. • “At least 9 cm” includes exactly 9; “more than 9 cm” excludes it. • Divide passing observations by the number actually assessed. • A mean reports total divided by count, not the proportion clearing a target. • Use median, range, quartiles and standard deviation to describe distribution rather than as substitutes for the stated criterion. • Equal means can hide unequal low-end outcomes. • Avoid averaging two group means without sample-size weights. • Do not infer a causal benefit from a small, uncontrolled trial.
The process is practical. Copy the data in order, verify the sample size, write a greater-than or greater-than-or-equal-to condition, count values meeting it, calculate the fraction and then explain whether the decision criterion is satisfied. You will often need fewer calculations than a student who starts finding every statistic without reading the question.
Official 2027 G3 Mathematics context
SEAB lists G3 Mathematics for the 2027 Singapore–Cambridge Secondary Education Certificate as K310; 4052 is the earlier GCE O-Level Mathematics reference. The 2027 K310 syllabus includes descriptive statistics, use and interpretation of the mean and standard deviation, cumulative-frequency and quartile ideas, graphical summaries and comparisons in real-world contexts. K310 has two written papers of 90 marks each, each lasting two hours 15 minutes and contributing 50% of the final grade. Students must use appropriate working, representations and accuracy according to the question.
This is original practice constructed to teach reasoning, not an official SEC question and not a prediction that a school-garden scenario will appear. The appropriate code and the applicable syllabus year matter: consult the SEAB 2027 G3 index rather than assuming a dated O-Level resource is identical in every administrative detail. An examination may supply its own method or rounding instruction, which should take priority over a general habit.
The entire fictional raw dataset
A class records a height gain, in centimetres, for each of eight labelled pots in Group A and eight pots in Group B after the same observation interval. A’s sorted gains are 8, 9, 9, 10, 10, 11, 11 and 12. B’s sorted gains are 6, 8, 9, 10, 10, 11, 12 and 14. The figures are invented for Mathematics practice. They are not scientific measurements from a real fertiliser experiment or a recommendation to apply a chemical product.
Both groups have eight values. We have not been told which material, if any, was used on each group, whether samples were randomly assigned, whether water, soil and sunlight were controlled or how repeat observations were collected. The calculations can be exact for the listed numbers while the biological explanation remains uncertain. That distinction is essential when communicating a decision based on data.
Check the two means before interpreting them
Add A: 8 + 9 + 9 + 10 + 10 + 11 + 11 + 12 = 80. The mean is 80 ÷ 8 = 10 cm. Add B: 6 + 8 + 9 + 10 + 10 + 11 + 12 + 14 = 80, also giving 80 ÷ 8 = 10 cm. This equality is genuine, but it describes only one feature of each distribution.
The statement “both groups have mean gain 10 cm” is correct. The statement “every plant gained about the same amount” does not follow. Group B includes the lowest individual result, 6, and the highest, 14. A has a narrower range. Before making a decision, find out whether the goal is total average gain, avoiding low results, obtaining high maximum values or meeting a minimum share of successful pots.
The median also ties, but that is not the whole story
Each list has eight observations, so the median is the mean of the fourth and fifth ordered values. In A those values are 10 and 10, so the median is 10. In B they are also 10 and 10, so its median is 10. The two most familiar measures of centre therefore agree: mean 10 and median 10 for both groups.
Now imagine a brochure announcing “the same results”. That phrase is too strong. The same mean and median do not require the same minima, maxima, quartiles, pass counts or variability. A stronger interpretation is that the centres match for these listed values. The learner must then describe the feature relevant to the question instead of flattening a distribution into a single average.
Compute ranges as a first view of spread
The range of A is maximum minus minimum: 12 − 8 = 4 cm. The range of B is 14 − 6 = 8 cm. B’s range is twice A’s. This is a legitimate numerical comparison of overall extremes in the observed sets. It does not say that individual B observations are always worse or that B’s larger maximum makes the group universally better.
A range uses only the two ends. It is sensitive to unusual extreme observations and ignores how the middle six are arranged. That is why the median and quartiles matter if a question asks about a typical or central interval. Yet if the decision concerns whether any pot falls below 7 cm, that very minimum may be the critical piece of information.
Read quartiles using a stated convention
For these eight sorted values, use the median-of-halves method, a clear convention for this worked example. A’s lower half is 8, 9, 9, 10, so Q1 is (9 + 9) ÷ 2 = 9. Its upper half is 10, 11, 11, 12, so Q3 is 11. Therefore A’s interquartile range is 11 − 9 = 2 cm.
B’s lower half is 6, 8, 9, 10, giving Q1 = (8 + 9) ÷ 2 = 8.5. Its upper half is 10, 11, 12, 14, giving Q3 = (11 + 12) ÷ 2 = 11.5. B’s interquartile range is 3 cm. The central half of B’s observations is more spread out by this measure. If an examination gives a specific quartile convention or calculator procedure, follow it consistently; do not mix methods across the two groups.
A box plot communicates five numbers, not every result
A basic box-and-whisker diagram built from these lists would show A’s minimum 8, Q1 9, median 10, Q3 11 and maximum 12. B would show minimum 6, Q1 8.5, median 10, Q3 11.5 and maximum 14. A reader could immediately compare centres, central spread and overall extent on a shared scale.
But a conventional box plot does not show each original observation and does not automatically display the mean. Nor does it establish cause, quality or a population-wide percentage. If the question asks how many observed gains are at least 9 cm, the raw list is easier: count values directly. Do not pretend the box boundaries always reveal exact counts at an arbitrary threshold inside a quartile interval.
Standard deviation supports the pattern without replacing the criterion
For the full listed set A, the mean is 10. The squared deviations from 10 add to 12: 4 + 1 + 1 + 0 + 0 + 1 + 1 + 4. Dividing by eight gives variance 1.5 and a standard deviation of √1.5 ≈ 1.22 cm. For B, the squared deviations are 16 + 4 + 1 + 0 + 0 + 1 + 4 + 16 = 42. Dividing by eight gives variance 5.25 and standard deviation √5.25 ≈ 2.29 cm.
These calculations treat each displayed list as the complete set being described with a divisor of n, a convention consistent with describing the eight values. Both statistics support the observation that B is more variable. They do not answer how many values pass a particular target without examining the distribution. Use standard deviation when spread matters, but do not replace a count requested by the stem with a spread measure just because it seems more advanced.
State the threshold in mathematical language
The school may define a satisfactory observed gain as **at least 9 cm**, which is x ≥ 9. The boundary value 9 counts as a success. In A, only the value 8 fails, so seven of eight values pass. In B, the values 6 and 8 fail, so six of eight pass. A’s observed pass proportion is 7 ÷ 8 = 0.875, or 87.5%; B’s is 6 ÷ 8 = 0.75, or 75%.
That result differs even though the means match. If the committee had written “more than 9 cm”, the correct condition would be x > 9: A would have five passes, not seven, and B would also have five, not six. The inequality symbol is doing real work. An examiner does not reward an answer to the wrong boundary condition because the subsequent arithmetic is neat.
Change the threshold and watch the ranking change
At a minimum of 8 cm, all eight A values pass, while seven of eight B values pass. At 9 cm, A has seven passes and B six. At 10 cm, both have five. At 11 cm, both have three. At 12 cm, A has one and B has two. At 13 cm, A has none and B has one. Thus the ranking of pass counts depends on the threshold. A leads at lower standards, the groups tie in the middle, and B has more very high observations at 12 or 13.
This is the core skill of the volume. Rather than asking which group is “best” with no criterion, specify what counts as success and find the corresponding proportion. A larger standard deviation explains some of the pattern, but the actual count gives the evidence for the chosen standard.
A proportion is not a probability guarantee
The observed 7/8 and 6/8 values are proportions in these invented samples. They are not probabilities that every future plant will respond in the same way. Eight pots are a small dataset, and we have no information about random sampling or repeat trials. A statement like “87.5% of all future plants will succeed” would turn one observed rate into a universal prediction unsupported by the given information.
Mathematics exam tasks often simplify uncertainty so that students can calculate correct fractions. A learner should still distinguish the exact statistic of the table from an inference about an unobserved population. In a public report, “seven of eight observed A pots met the 9 cm criterion” is stronger and more responsible than “A guarantees 87.5% success”.
Use a two-condition decision rule carefully
Suppose the teacher approves a classroom display if the group has a mean gain of at least 10 cm **and** at least 87.5% of observed pots gain at least 9 cm. Group A has mean 10 and seven of eight at or above 9, so it meets both conditions. Group B has mean 10 but only six of eight at or above 9, so it does not. Both parts matter. Repeating the means and ignoring the second condition misses the reason the decision changed.
If the rule instead says at least 75% must pass, B passes exactly because 6 ÷ 8 = 75%. “At least” includes the boundary. If the rule says “more than 75%”, B fails. Always translate the words into inequalities before starting your calculation. The same logic applies to examinations, budgets, safety thresholds and ordinary planning tasks.
Do not average percentages from unequal groups
The original A and B each contain eight observations, making direct comparison of counts straightforward. But a new exercise might show eight observations at one site and twelve at another. Suppose a separate, invented Site C has 8 results with mean gain 10 cm, while Site D has 12 results with mean 11 cm. The combined total is 8 × 10 + 12 × 11 = 80 + 132 = 212 cm across 20 pots. The pooled mean is 212 ÷ 20 = 10.6 cm, not the unweighted average 10.5 cm.
This is a distinct additional dataset, not a hidden revision of A or B. The general method belongs to the pooled-means owner. Here it matters because a trial report could misleadingly average site summaries without telling readers about differing group sizes. A correct overall statement must attach each mean to its original number of observations.
A missing pot is not automatically a zero
Suppose the observation sheet contains an empty cell for one pot. The cell might mean the pot was removed, not measured, damaged or recorded elsewhere. Replacing missing data with zero without an instruction can lower the mean and alter the pass rate. Excluding the pot without explanation can also make a weak trial look better. We need to know the measurement rule.
In examination practice, follow the stated convention. If a question clearly specifies that a missing pot recorded zero gain, include zero. If it says only seven usable observations remain, calculate with seven when requested. If it supplies no reason for a blank and asks you to critique a report, identify missingness as an uncertainty. Do not make up facts to force the data into a preferred conclusion.
An extreme observation changes the mean more than the median
Imagine a **new** ninth B observation of 30 cm, added only as a mathematical thought experiment. B’s original total is 80 cm, so the new total is 110 cm for nine values. The mean becomes 110 ÷ 9 ≈ 12.22 cm. The median, however, remains the fifth ordered value, 10 cm. One extreme point can therefore pull the mean upward while the middle value barely changes.
No real plant has been measured at 30 cm in this lesson. The exercise shows why the mean and median tell different stories. If a poster cites only the raised mean, ask whether a single extreme result is controlling the headline. If it cites only the unchanged median, ask whether the extreme observation might still matter to a question about maximum performance or measurement errors.
Do not remove inconvenient observations silently
Removing the low B value of 6 leaves seven values with total 74 cm, so the new mean is 74 ÷ 7 ≈ 10.57 cm. That increase does not prove the treatment improved; it is an arithmetic consequence of dropping a lower-than-average observation. Unless the question states a sound reason for exclusion, trimming a result after seeing it risks biasing the comparison.
By contrast, a measurement could genuinely be invalid if its scale was recorded incorrectly or a pot was never part of the assigned trial. Such an exclusion needs a consistent rule applied to both groups. The general lesson is that data cleaning is not the same as hiding data. A learner should show what changed and why before calculating a revised summary.
Check denominator and rounding in a frequency question
For eight pots, one passing observation represents 1/8 = 12.5 percentage points. Moving from six passes to seven is a rise from 75% to 87.5%, an increase of 12.5 percentage points. It is not a 12.5% relative increase; the relative increase is (87.5 − 75) ÷ 75 × 100% ≈ 16.7%. The question’s wording determines which comparison is requested.
A response like “A beats B by 12.5%” is ambiguous and can be misleading. Write “12.5 percentage points” for the difference between two percentages. If the examiner asks for relative percentage improvement, show the original base in the denominator. This is the same mathematical discipline students first meet in primary school, now applied to a more complex dataset.
The graph’s axis can change a reader’s impression
A bar chart of 87.5% and 75% may look dramatically different if its vertical axis begins at 70%. That does not make the underlying numbers wrong, but it amplifies the apparent visual separation. A line graph can be inappropriate if the categories are merely two unrelated groups rather than stages on a continuous ordered scale. A careful reader checks both the labels and what the graphic makes easy to infer.
A well-chosen display could show a simple table of pass counts plus the threshold definition. If the goal is variability, paired box plots on the same horizontal scale are useful. If the goal is exact successes at 9 cm, a short frequency table is clearer. Representations are tools for the question, not decorations for a claim already decided.
Why two datasets do not establish causation
The two group labels tell us nothing about assignment or controls. Without information about starting heights, light, water, soil, pot size and treatment conditions, it is not responsible to say an ammonium compound caused the observed differences. Even a well-designed small experiment can support only conclusions justified by its data and procedures. Numerical precision cannot substitute for experimental design.
The previous Chemistry lesson explains ammonium species. It does not prove any particular crop response. Keeping this separation respects both subjects. Mathematics can calculate the means and pass proportions exactly; Science asks whether a proposed cause has been isolated; English should communicate the resulting distinction accurately. That distinction is worth more than the superficial confidence of a large percentage.
How to write the final comparison in four sentences
First describe the common centre: both samples have mean and median 10 cm. Second name the spread: A ranges from 8 to 12 cm while B ranges from 6 to 14 cm. Third state the decision rule: at least 9 cm counts as success, with seven of eight A observations and six of eight B observations passing. Fourth restrict the conclusion: A has the higher observed pass proportion at that threshold in this fictional sample, but the evidence does not establish a universal biological advantage.
A learner can compress these ideas for a short examination answer, but the logic should survive compression. Do not jump from a mean to a recommendation, or from an observed fraction to a general causal promise. The best short answer states the criterion, number and limitation in the smallest clear space.
Revision that leads to transferable accuracy
Cover the solution and reconstruct the two sorted lists. Calculate both sums, medians and ranges. Next, without calculating a mean again, find the pass counts for thresholds 8, 9, 10, 11, 12 and 13. Then change “at least” to “more than” in one question and see which numbers change. Finish by explaining why the school cannot conclude a fertiliser is better from these tables.
If a particular step fails, use a focused owner rather than restarting the whole topic: Volume 0351 for standard deviation, Volume 0359 for quartiles and box plots, Volume 0395 for unequal group sizes. The PSLE average-and-group-size lesson gives a useful early bridge. Our present owner remains the threshold-driven decision based on a raw distribution.
Ten threshold-decision cases
Each worked case uses the same fictional eight-pot groups and changes the words defining success. Solve before reading. A change in the rule can legitimately change the eligible group even when no observation changes.
Decision case 1: At least 8 cm
A has eight values at or above 8 and B has seven. The proportions are 100% and 87.5%. A leads under this particular threshold, but the statement does not say which group has a higher maximum or what would happen under a tougher criterion.
Decision case 2: At least 9 cm
A passes seven of eight, 87.5%; B passes six of eight, 75%. The threshold includes observations equal to 9. The 12.5-percentage-point difference is exact for these data, but it is not a population probability or evidence of a specific chemical cause.
Decision case 3: More than 9 cm
The boundary observations of exactly 9 no longer pass. A has five values above 9 and B also has five, giving 62.5% for each. The ranking changes simply because the success definition changed. This is why a mathematical inequality must be translated from the words before counting.
Decision case 4: At least 10 cm
Both A and B have five passing values out of eight, so both show 62.5%. Their means are also 10 cm, but that does not make the full lists identical. The match at this threshold is a statement about five observations, not a general equality of performance.
Decision case 5: At least 11 cm
A has 11, 11 and 12; B has 11, 12 and 14. Each passes three of eight, or 37.5%. The groups tie under this rule although B has the larger maximum. The decision rule determines the conclusion, not a headline about extremes.
Decision case 6: At least 12 cm
A has only one passing value, 12, while B has two, 12 and 14. Pass proportions are 12.5% and 25%. B now leads. It would be wrong to keep recommending A solely because A led at the lower 9 cm threshold.
Decision case 7: More than 12 cm
A has no values above 12; B has one value, 14. The rates are 0% and 12.5%. This narrow observed comparison is valid, but does not show that B is generally preferable when the school is concerned about low results or variability.
Decision case 8: At least 75% of pots must clear 9 cm
A clears the 9 cm target in seven of eight cases and B in six. Both meet an at-least-75% acceptance rule, because B reaches the boundary exactly. If the decision sought the higher passing proportion rather than merely eligibility, A would rank above B.
Decision case 9: More than 75% of pots must clear 9 cm
A’s 87.5% is strictly greater than 75%, while B’s 75% equals but does not exceed it. Only A passes. This one-word change in the condition changes the eligibility result even though not one observation changed.
Decision case 10: Mean at least 10 and pass rate at least 87.5% at 9 cm
Both means equal 10, satisfying the first condition. A’s pass rate is 87.5% and satisfies the second; B’s 75% fails. Therefore only A is eligible under this two-part rule. The conclusion follows from the explicitly stated conjunction, not a general biological claim.
Practice questions and worked reasoning
These questions are designed to rehearse one precise step at a time. Check the unit, boundary, denominator and scope of every result. The calculations refer to invented classroom data unless a separate example is explicitly introduced.
Practice 1: How many observations are in each original group?
Eight for A and eight for B. The number assessed is the denominator in the sample pass proportion. Do not count the group labels as observations or include a later hypothetical ninth value unless the question explicitly changes the dataset.
Practice 2: What is the total of A’s gains?
The sum is 8 + 9 + 9 + 10 + 10 + 11 + 11 + 12 = 80 cm. An efficient check is to pair opposite ends: 8 + 12, 9 + 11, 9 + 11 and 10 + 10 each make 20.
Practice 3: What is the total of B’s gains?
The sum is 6 + 8 + 9 + 10 + 10 + 11 + 12 + 14 = 80 cm. The same total as A, combined with the same number of values, ensures their means are equal even though the distributions differ.
Practice 4: What is the mean of each original group?
Both are 80 ÷ 8 = 10 cm. The mean is a measure of centre. It cannot tell you directly how many observations pass a stated threshold, what the minimum is or whether the treatment caused the recorded differences.
Practice 5: Why are the medians equal?
The fourth and fifth ordered values in each eight-value list are 10 and 10. Their mean is 10 cm. Median equality does not require the same range or quartiles, so it is insufficient for declaring the groups’ results identical.
Practice 6: How is the range of A found?
Subtract the smallest listed value, 8, from the largest, 12. The result is 4 cm. A range compares extremes, not the full pattern of observations between them. It is useful when extreme outcomes matter.
Practice 7: How is the range of B found?
Subtract 6 from 14 to obtain 8 cm. It is twice A’s 4 cm range, showing B has a wider observed extent. The result alone does not identify which group meets a school-defined minimum success fraction.
Practice 8: Which group has a higher maximum?
B reaches 14 cm, while A reaches 12 cm. B’s maximum is two centimetres higher. A higher maximum can matter for a goal focused on top individual results but says little about the proportion clearing a lower safety or reliability threshold.
Practice 9: Which group has a higher minimum?
A’s lowest is 8 cm, while B’s lowest is 6 cm. A therefore has the higher observed minimum by 2 cm. This may matter to a decision concerned with avoiding low gains, but it does not make every A observation greater than every B observation.
Practice 10: What is Q1 for A using the median-of-halves method?
The lower four values are 8, 9, 9 and 10. The median of those is (9 + 9) ÷ 2 = 9 cm. State the method if the context needs it, and use the same convention consistently when comparing B.
Practice 11: What is Q3 for A?
The upper four values are 10, 11, 11 and 12. Their median is 11 cm. The interquartile range is therefore 11 − 9 = 2 cm. This describes the central spread by the stated convention.
Practice 12: What is Q1 for B?
The lower half is 6, 8, 9 and 10. Its median is (8 + 9) ÷ 2 = 8.5 cm. A different quartile convention could be instructed in a different setting; the task’s method must be respected.
Practice 13: What is Q3 for B?
The upper half is 10, 11, 12 and 14. Its median is (11 + 12) ÷ 2 = 11.5 cm. That makes B’s interquartile range 11.5 − 8.5 = 3 cm.
Practice 14: Which group has the larger IQR?
B has IQR 3 cm against A’s 2 cm, under the same stated calculation method. This describes greater central spread in B’s listed values, but does not directly give the exact percentage of observations at least 9 cm.
Practice 15: What is the standard deviation of A in this example?
Squared deviations from the mean 10 sum to 12. Divide by eight and take the square root: √(12/8) = √1.5 ≈ 1.22 cm. The calculation describes the listed observations using a divisor of eight.
Practice 16: What is the standard deviation of B?
Squared deviations from 10 sum to 42. The standard deviation is √(42/8) = √5.25 ≈ 2.29 cm. B is more dispersed by this statistic, consistent with its larger range and interquartile range.
Practice 17: Why is a larger standard deviation not necessarily worse?
A larger spread means observations are more dispersed around the mean. Whether that is desirable depends on the decision. It may be undesirable if consistent minimum performance matters, but a goal rewarding exceptionally high results could view part of the same variation differently.
Practice 18: What is the pass count for A at 9 cm?
Seven values are at least 9; only the value 8 is below the requirement. Write the condition as x ≥ 9 to include the two observations equal to 9. The pass fraction is 7/8.
Practice 19: What is the pass count for B at 9 cm?
Six values are at least 9: 9, 10, 10, 11, 12 and 14. The 6 and 8 fail. The observed pass fraction is 6/8 = 3/4 = 75%.
Practice 20: What is 7/8 as a percentage?
Divide 7 by 8 to get 0.875, then multiply by 100 to obtain 87.5%. State that this is the sample pass proportion for a defined threshold. It is not a guaranteed chance for a future plant.
Practice 21: What is the difference between 87.5% and 75%?
It is 12.5 percentage points. If the question instead asks for relative percentage increase from B’s rate to A’s, the base is 75%, giving 12.5/75 × 100% ≈ 16.7%. Do not confuse the two comparisons.
Practice 22: What does ‘at least’ mean?
It means greater than or equal to. A value exactly at the boundary passes. In an examination question, changing an inclusive inequality into a strict one can change the count, the percentage and the final recommendation.
Practice 23: What does ‘more than’ mean?
It means strictly greater than. A value equal to the threshold does not pass. For x > 9, A’s two values equal to 9 are excluded, and both A and B end up with five passes.
Practice 24: Why does a target of 10 cm produce a tie?
Both lists contain five values equal to or greater than 10. That is a 5/8, or 62.5%, pass rate for each group. The tie says nothing about different minima, maxima or tails of the distributions.
Practice 25: How does the ranking change at 12 cm?
A has one observation of at least 12, whereas B has two. B’s 25% observed pass proportion exceeds A’s 12.5%. Ranking depends on the defined criterion, so a lower-threshold decision cannot simply be copied.
Practice 26: Why is the sample pass proportion not a future probability?
It is calculated from eight invented observations. Without a sampling model and appropriate evidence, it does not justify a claim about all future plants. Use wording such as “seven of eight observed pots” when reporting the actual scope.
Practice 27: Does equal average prove two treatments have equal effects?
No. The figures are a descriptive comparison, not an experiment establishing cause. Starting conditions, assignment, other environmental variables and sample uncertainty matter to causal interpretation. The mean can be equal even when observed distributions differ.
Practice 28: What is a fair summary of A?
A’s eight values have mean 10, median 10, range 4 and seven results meeting the at-least-9 criterion. This describes observed performance under the stated mathematical rule. It should not be relabelled as proof of superior fertiliser chemistry.
Practice 29: What is a fair summary of B?
B’s eight values have mean 10, median 10, range 8 and six results meeting the at-least-9 criterion. It has a higher maximum but a lower minimum. Which property matters depends on the decision rule.
Practice 30: Why is a box plot insufficient to recover every threshold count?
A box plot summarises selected order statistics. It hides many individual values, so an arbitrary boundary within a quartile segment may not have a uniquely recoverable count. When raw data are available, use them for exact pass frequencies.
Practice 31: Why should two box plots use the same scale?
A common numerical axis makes differences in medians and spread visually comparable. Different axis widths can exaggerate or hide variation. A sound graphical comparison keeps units and scale consistent unless the difference is clearly disclosed.
Practice 32: How is a pooled mean found for unequal group sizes?
Multiply each group mean by its size, add those totals and divide by the combined count. For an independent example with eight at mean 10 and twelve at mean 11, the pooled mean is (80 + 132)/20 = 10.6 cm.
Practice 33: What is wrong with averaging 10 and 11 directly in that unequal-size example?
The simple result 10.5 gives the groups equal weight despite one having eight observations and the other twelve. It is not the mean of all twenty values. A group mean must be weighted by the group’s actual count.
Practice 34: Why does a blank result need explanation?
A blank might mean missing measurement, exclusion, damaged equipment or another condition. It cannot automatically become zero without a stated rule. Both omission and forced zero-filling can distort the mean and pass proportion.
Practice 35: What happens to B’s mean if a ninth observation of 30 is added?
The original total 80 becomes 110 for nine observations, so the mean is 110/9 ≈ 12.22 cm. The original eight-value dataset has not changed; the hypothetical added value illustrates sensitivity of the mean to an extreme observation.
Practice 36: What happens to B’s median in that new nine-value set?
After adding 30, the ordered values still place 10 in the fifth position. The median remains 10 cm. This contrast with the increased mean shows why both centre measures can be informative when an extreme value appears.
Practice 37: What happens if B’s lowest value 6 is deleted?
Seven values remain with total 74, so the new mean is 74/7 ≈ 10.57 cm. The change is arithmetic, not proof that a treatment improved. Removing a value requires a defensible, consistently applied reason.
Practice 38: When should a learner use a frequency table?
Use it when the question asks how many observations fall in categories or pass cut-offs. Frequencies retain the counts needed for proportions. A mean or box plot can describe the distribution, but may not answer the exact passing-count question.
Practice 39: When is a cumulative-frequency method useful?
It helps count the observations at or below a boundary and can support estimates of medians, quartiles and percentiles for suitable datasets. The cumulative direction and any inclusive boundary must be checked. In the eight-value raw lists here, direct counting is simpler.
Practice 40: Why is an apparent ‘winner’ sometimes undefined?
If no objective or threshold is supplied, different measures support different preferences. B has the higher maximum, A the higher minimum, and both the same mean. The honest answer specifies the criterion before selecting an option.
Practice 41: How does a two-part acceptance rule work?
Both compulsory conditions must hold. If mean must be at least 10 and the 9 cm pass rate at least 87.5%, A meets both but B misses the pass-rate condition. One successful calculation cannot substitute for checking the other.
Practice 42: Can we treat 87.5% as an exact fraction?
Yes, for eight observed values it equals 7/8 exactly. But numerical exactness for the sample is not the same as certainty about a larger population. Keep the calculation’s exactness and the inference’s limitation separate.
Practice 43: What is the fastest final answer to the garden question?
“Both groups have mean gain 10 cm. At the 9 cm minimum, A passes 7/8 observations, or 87.5%, while B passes 6/8, or 75%. A has the higher observed pass share at that threshold; causation is not established.”
Practice 44: What should be practised without notes?
Recreate the two sorted datasets, calculate one centre and one spread measure, find passes under three changed inequalities and explain why conclusions depend on the rule. Finally state what cannot be inferred from a small fictional sample.
Practice 45: What makes this volume different from a general standard-deviation lesson?
The central skill is not computing the standard deviation; that has an earlier owner. It is deciding which group meets a changing threshold from raw observations despite equal means, then reporting the narrow result without making a biological claim. Spread measures support that decision rather than defining it.
Related study and primary sources
Use Volume 0528 for the Chemistry distinction between ammonia and ammonium; the dataset here does not test chemical efficacy. For the underlying statistical techniques, visit Volume 0351 on standard deviation, Volume 0359 on quartiles and box plots, Volume 0395 on pooling unequal groups and the Secondary 4 Mathematics grouped-data guide. The PSLE average-and-group-size guide and PSLE caution about pooled samples offer simpler bridges. For subject study use Science and English. Confirm the examination code and topics with the SEAB 2027 G3 index and K310 syllabus. This is independently written teaching material, not an official paper or a claim about real plants.
