In this guide
Two school-fair queues each have an estimated mean wait of ten minutes. “Then they are equally quick,” says a pupil. Another asks how many people waited at least twelve minutes. That question opens a richer statistics lesson.
Try it in Class 10: allow 45 minutes after teaching grouped mean and the median formula. Pairs analyse two original fictional datasets, compare an estimate with a distribution, and write a claim that the data can support. Provide the table, paper and a calculator if appropriate to your classroom practice.
Read the intervals before calculating
The following invented records describe 20 waits at each of two queues. Each interval includes its lower endpoint and excludes its upper endpoint: a wait of exactly 4 minutes belongs to 4≤t<8, not the first group. All times are in minutes. No pupil names, observation of real children or collection of personal information is required.
| Wait t(minutes) | Midpoint | Queue A frequency | Queue B frequency |
|---|---|---|---|
| 0≤t<4 | 2 | 4 | 2 |
| 4≤t<8 | 6 | 4 | 6 |
| 8≤t<12 | 10 | 4 | 6 |
| 12≤t<16 | 14 | 4 | 2 |
| 16≤t<20 | 18 | 4 | 4 |
| Total | — | 20 | 20 |
Before giving out a formula, ask pupils to describe one visible difference. Queue A spreads its observations evenly across the five groups. Queue B has more waits in the middle two groups. Both have four waits in the final group. These statements use the frequencies directly and can be checked without estimating anyone’s exact wait.
Build the calculation, rather than hiding it
Ask pairs to add two columns for frequency×midpoint. One partner computes A while the other computes B; they then exchange the work. Each partner must explain why a group such as 8≤t<12 uses 10 as its representative midpoint. The midpoint stands in for unknown values; it is not a claim that everyone in that group waited exactly 10 minutes.
| Midpoint | A:frequency×midpoint | B:frequency×midpoint |
|---|---|---|
| 2 | 8 | 4 |
| 6 | 24 | 36 |
| 10 | 40 | 60 |
| 14 | 56 | 28 |
| 18 | 72 | 72 |
| Sum | 200 | 200 |
The grouped mean estimate is 200÷20=10 minutes for each queue. Invite learners to explain both the common estimate and its limitation in one sentence. A useful version is: “Using the class midpoints, both means are 10 minutes, although the exact individual waiting times were not recorded in this table.”
Ask the median a different question
Add cumulative frequencies. For A they are 4,8,12,16,20; for B they are 2,8,14,16,20. With N=20, the N/2 position is 10, which lies in 8≤t<12 for both datasets. In the formula median=l+[(N/2−cf)/f]×h, l=8,h=4 and cf=8. The class frequency differs.
Teacher check: two medians and a careful conclusion
Queue A:8+[(10−8)/4]×4=10 minutes. Queue B:8+[(10−8)/6]×4=8+4/3=9⅓minutes, about 9.33 minutes. The result estimates a median from grouped observations; it is not the recovered exact middle waiting time.
A has 8 of 20 waits, or 40%, at least 12 minutes. B has 6 of 20, or 30%, at least 12 minutes. Thus B has fewer observations in those longer-wait groups in this fictional sample. Both have 4 of 20, or 20%, at least 16 minutes. Those two thresholds answer different questions.
A defensible conclusion is: “The grouped mean estimates tie at 10 minutes. B has a lower grouped median estimate and a smaller share of waits at least 12 minutes in this sample. That supports comparing B favourably on those measures, but does not prove every person would wait less in B.”
Make the missing information visible
Challenge pairs to choose possible exact times for the first group without changing its frequency. For A, the four exact waits could be(1,1,1,1)minutes or, in a different possible dataset,(3,3,3,3)minutes. Each alternative has four observations in 0≤t<4, yet their totals are different. This shows why a grouped mean is an estimate. Changing a midpoint calculation cannot recreate information that grouping removed.
Then ask: “What else would help us decide how to organise next year’s queue?” Learners might request several days of observations, arrival rates, the number of service counters or variation during busy periods. These are proposals for useful evidence. They are not facts already present in the table. Avoid turning a small fictional sample into a general claim about a real school service.
Use 45 minutes well
- 0–5: interpret intervals and make a first comparison.
- 5–15: calculate the grouped means with visible working.
- 15–25: find cumulative frequencies and estimate medians.
- 25–35: compare the 12-minute and 16-minute thresholds.
- 35–42: test how grouping hides exact times.
- 42–45: write a claim, evidence and one limitation.
A pupil who needs arithmetic support can receive the midpoint-product table with two entries missing. Keep the interpretation questions unchanged. A learner can explain a cumulative total orally or highlight the median class. For a stretch, keep every frequency fixed and move the possible exact waits towards the lower ends of their intervals. Ask whether the actual mean must still be 10 minutes, and why the midpoint estimate stays unchanged. Learners should explain the information lost through grouping, without inventing exact observations.
The exit task is: “A friend says the queues are identical because their means match. Write a two-sentence reply using one calculation and one frequency comparison.” Look for the difference between an average, the pattern of observations and the strength of a claim.
This activity supports the statistics focus in CBSE’s Class X Mathematics curriculum for 2026–27. The mathematical-thinking guide develops the habit of reading data with context. The task-and-rubric example offers a transferable way to assess reasoning rather than a final number alone.
AI assistance was used to finalize this contribution. The views expressed are personal views of the author; official statements are identified through linked official sources.
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