Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 11 · Chapter companion part
Explain an equal-share average, calculate it by several checked methods and distinguish exact recorded means from grouped estimates.
Class 8 English Version: part C of the Chapter 11 companion. Eight teaching sections and 30 original written tasks. Selected companion coverage, not complete textbook coverage; use all four parts. Written interpretation, chart drawing and complete working need guardian or teacher review.
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The arithmetic mean is the total of numerical observations divided by how many observations there are. It is the amount each would have if the total were shared equally. Our ten invented session lengths sum to 180 minutes, so their mean is 18 minutes per session. This describes that recorded total and count; it does not say every session lasted 18 minutes or that 18 is best. A mean needs numerical quantities with a meaningful unit, not arbitrary identity labels.
Write x̄=(x₁+…+xₙ)/n=Σxᵢ/n. The bar labels the mean, n is the observation count, xᵢ is record i and Σ means add the listed terms. Sum all records, including repeats, before dividing by n. For 2,4,6,8 blocks, total 20 shared over four bags is 5 blocks per bag. Five may be a virtual equal share; the actual bags still contained 2,4,6,8. The mean lies between the smallest and largest record when the list is nonempty.
An assumed mean a is a convenient reference, not a claim about the answer. Let dᵢ=xᵢ−a. Then x̄=a+Σdᵢ/n because Σxᵢ=na+Σdᵢ. For our ten sessions, choose a=20 minutes. Their signed deviations sum−20 minutes; the correction is −2 minutes per session and the mean is 20−2=18 minutes. Choosing a=16 instead gives correction+2 and the same 18. Negative differences are essential.
Printed pages 177–178 call the running sum of xᵢ−a cumulative frequency. That label is wrong: the column adds signed differences, so it can decrease or be negative and has the measurement unit. True cumulative frequency adds nonnegative record counts and ends at n. For our a=20 example, the running deviations begin−8,−14,−20 and finish−20 minutes. They do not count a negative number of sessions.
For distinct recorded values xᵢ with frequencies fᵢ, total count is n=Σfᵢ and total quantity is Σfᵢxᵢ. Mean=Σfᵢxᵢ/Σfᵢ. For values 4,8,12 with counts 1,2,1, the expanded list is 4,8,8,12. Its total 32 and count 4 give mean 8. Averaging table rows equally ignores how many observations each row represents. Count all records, not merely the number of different values.
When only interval counts are known, replace each interval by its midpoint m=(lower limit+upper limit)/2 as a stated approximation. Our inclusive classes 10–14,15–19,20–24,25–29 have midpoints 12,17,22,27 and frequencies 3,4,2,1. Their midpoint total is 175 and estimated mean 17.5 minutes. The actual ten records have exact mean 18 minutes. Grouping hides the positions within each interval; another raw list could have the same counts and a different mean.
For these integer width-five classes, every record is at most 2 minutes from its midpoint, so the true mean is within 2 minutes of the midpoint estimate if all records lie in their stated classes. This bound is not a probability or a promise of closeness within a smaller amount. Printed page 180 gives 89.5×4 as 348; it should be358. The corrected products give the displayed total 6190 and midpoint estimate 61.9. The estimate does not recover an unknown exact mean.
If n records have mean m, their total is nm. For five records with mean 14, total 70; if four known values total 52, the missing value is 18. For groups of sizes 4 and 6 with means 6 and 11, totals 24 and 66 combine to 90 over 10 records, giving mean 9. An unweighted average of 6 and 11 would wrongly give equal influence to unequal group sizes. Recheck units: when the source wage table says 100 taka, an invented mean estimate 62.5 table units would mean 6250 BDT. Distinguish exact recorded calculations from grouped estimates.
Our original ten records have exact mean 18 minutes. Grouping them into four width-five classes and replacing every record by its midpoint gives 17.5 minutes, an estimate 0.5 minute lower here. Neither result certifies a learner’s ability or a universal study target.
The source’s cumulative-frequency heading on its shortcut table should say running sum of signed deviations.
On printed page 180,89.5×4=358, not 348. The corrected product agrees with the displayed 6190 total, which gives a grouped estimate rather than an exact raw mean.
The ten invented reading-session lengths in minutes.
| Value x (minutes) | Frequency f | Product fx (minutes) |
|---|---|---|
| 12 | 1 | 12 |
| 14 | 2 | 28 |
| 16 | 1 | 16 |
| 18 | 3 | 54 |
| 20 | 1 | 20 |
| 22 | 1 | 22 |
| 28 | 1 | 28 |
| Total | 10 | 180 |
The 18-minute row contributes three sessions and 54 minutes. Total 180/count 10 gives mean 18 minutes per session. Seven distinct rows do not mean seven records.
Sorted fictional durations. These differences are in minutes; the last column does not count observations.
| Record | x (minutes) | d=x−20 (minutes) | Running signed sum (minutes) |
|---|---|---|---|
| 1 | 12 | −8 | −8 |
| 2 | 14 | −6 | −14 |
| 3 | 14 | −6 | −20 |
| 4 | 16 | −4 | −24 |
| 5 | 18 | −2 | −26 |
| 6 | 18 | −2 | −28 |
| 7 | 18 | −2 | −30 |
| 8 | 20 | 0 | −30 |
| 9 | 22 | 2 | −28 |
| 10 | 28 | 8 | −20 |
The final−20 means the actual total is 20 minutes below ten copies of 20. Dividing by ten gives correction−2 minutes per session. The book’s cumulative-frequency heading on a corresponding signed table is wrong.
These counts come from the same ten fictional sessions, but an interval alone does not give each raw value.
| Inclusive class (minutes) | Midpoint m (minutes) | Frequency f | Product fm (minutes) |
|---|---|---|---|
| 10–14 | 12 | 3 | 36 |
| 15–19 | 17 | 4 | 68 |
| 20–24 | 22 | 2 | 44 |
| 25–29 | 27 | 1 | 27 |
| Total | Not added as values | 10 | 175 |
The midpoint estimate 175/10=17.5 minutes is 0.5 minute below the exact raw mean 18 here. The discrepancy and its direction are not universal; choose and state the class origin and width.
Printed Example 5 and Exercise 6 use mathematical classroom data, not verified current financial evidence.
| Source item | Correction or interpretation | Consequence |
|---|---|---|
| Printed 180 product 89.5×4 | 358, not 348 | Other products sum 5832;5832+358=6190 |
| Printed 180 total 6190 over 100 | 61.9 midpoint units per observation | An interval-based estimate, not a recovered raw mean |
| Exercise 6 unit:100 taka | One table unit means 100 BDT | An invented estimate 62.5 table units means 6250 BDT |
Check arithmetic and units separately. None of these classroom amounts is a claim about present wages, account terms or a payment obligation.
Ten fictional sessions have total 180 minutes.
x̄=Σxᵢ/n x̄=180/10=18 minutes per session
18 minutes describes these ten records; it does not mean every session lasted 18 or prescribe an ideal duration.
Use a=20 minutes for the ten-session list.
dᵢ=xᵢ−a Σxᵢ=na+Σdᵢ x̄=a+Σdᵢ/n=20+(−20)/10=18 minutes
Choosing a=16 instead gives a+20/10=18. The answer is independent of the chosen reference.
Use distinct recorded values xᵢ; for grouped estimates replace them by declared midpoints mᵢ.
n=Σfᵢ Exact-value mean=Σfᵢxᵢ/Σfᵢ Grouped midpoint estimate=Σfᵢmᵢ/Σfᵢ
Our exact-value mean is 180/10=18 minutes, while the interval midpoint estimate is 175/10=17.5 minutes.
Four records have mean 6 units; six have mean 11 units.
Combined total=4×6+6×11=90 units Combined count=4+6=10 Combined mean=90/10=9 units per record
The simple average(6+11)/2=8.5 wrongly gives equal influence to groups of different sizes.
Four fictional bag counts. Move through the redistributions without changing the total.
No autoplay. Manual controls, static alternatives and reduced-motion handling remain available. This is a mathematical sharing model, not a change to stored observations.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original diagram using fictional classroom records, not measured personal information or current statistics. Read the quantities, units and conventions beside the figure; colour alone does not carry the meaning.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original diagram using fictional classroom records, not measured personal information or current statistics. Read the quantities, units and conventions beside the figure; colour alone does not carry the meaning.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original diagram using fictional classroom records, not measured personal information or current statistics. Read the quantities, units and conventions beside the figure; colour alone does not carry the meaning.
Show your working on paper. Check your answer, then choose Next.
18
Total 180 divided by 10 gives 18 minutes per session: the equal share, not every actual duration.
180
Mean×count=18×10=180 minutes.
18
Total 5×14=70; known sum 52, leaving 18.
18
20+(−20)/10=18 minutes; the negative correction lowers the reference.
18
16+20/10=18 minutes; a new reference must give the same mean.
8
Weighted total 4+16+12=32 and count 4 give 8 per observation.
4
1+2+1=4 records, although there are only three distinct table values.
32
1×4+2×8+1×12=32 units in total.
17.5
(12×3+17×4+22×2+27×1)/10=175/10=17.5 minutes; it is a midpoint estimate.
0.5
18−17.5=0.5 minutes; the estimate is half a minute low for these particular records.
27
(25+29)/2=27, a representative value used in a stated estimate.
6250
62.5×100=6250 BDT. This remains a grouped estimate, not a verified real wage.
14
Old total 40 plus 30 gives 70 over five records; the mean becomes 14.
9
(4×6+6×11)/(4+6)=90/10=9 units per record.
2
Each record differs from its class midpoint by at most 2, so the average difference has magnitude at most 2. This is a bound, not a probability.
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Their total 180 minutes divided by 10 is 18 minutes per session. It is the equal-share summary of these records; individual durations vary and it does not measure ability or prescribe an ideal study time.
Name the total, denominator, unit and interpretation.
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x̄ is the mean, xᵢ is an individual bag count, n=4 is the number of bags and Σ adds all four counts. Total 20/4 gives 5 blocks per bag under equal sharing, while the actual bags keep their listed counts.
Translate the formula into words and preserve the unit.
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12+14+14+16+18+18+18+20+22+28=180. Divide by 10 to get 18 minutes. It lies between 12 and 28, and 18×10 restores the total 180.
A plausible size check supports the arithmetic but is not a substitute for summing.
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Each record xᵢ=a+dᵢ. Summing n records givesΣxᵢ=na+Σdᵢ. Divide by n>0 to obtain x̄=a+Σdᵢ/n. Signed differences retain information below the reference.
Define a, dᵢ, n and show why the reference is added back.
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Deviations are−8,−6,−6,−4,−2,−2,−2,0,2,8 minutes. Their sum is −20, meaning the total lies 20 minutes below ten copies of 20. The correction−20/10=−2 gives mean 18 minutes.
Label the running column as a signed sum, never cumulative frequency.
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Its entries add xᵢ−a, so they are running signed deviations with the value unit. A cumulative frequency instead adds observation counts, cannot decrease when counts are nonnegative and ends at n. The source’s final 115 is a deviation total, not 115 people out of 20.
Describe both columns and how their meanings differ.
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Relative to 16 the deviations total 20, giving 16+20/10=18. Relative to 20 they total−20, giving 20−20/10=18. Changing the reference changes the correction but leaves the original total and count unchanged.
Compare both signed corrections with units.
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Values 12,14,16,18,20,22,28 have frequencies 1,2,1,3,1,1,1. Products are 12,28,16,54,20,22,28, totalling 180. Frequencies total 10, so the exact recorded mean is 18 minutes.
Do not divide by seven distinct rows instead of ten observations.
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Values 2 and 8 with counts 3 and 1 expand to 2,2,2,8. The mean is 14/4=3.5, not(2+8)/2=5. The more frequent value has more records and must have greater weight.
Use a fully stated list and explain the weighting.
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Midpoints 12,17,22,27 give products 36,68,44,27, total 175. Estimated mean 175/10=17.5 minutes. The original records have mean 18, so the estimate is 0.5 minute low in this example; another list with the same class counts need not have that difference.
State approximation, hidden information and unit.
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Each integer in 10–14 differs from midpoint 12 by at most 2, and similarly for every equal class. Averaging those signed replacement differences cannot exceed 2 in magnitude. This is a worst-case bound under the stated grouping, not a chance statement or a guarantee that every estimate is nearly exact.
Do not apply the same 2-unit bound to wider or continuous bins without checking them.
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For midpoint 89.5 with frequency 4, the product is 358, not 348. With the other displayed products,358 yields total 6190 and estimated mean 61.9 over 100 records. The exact raw mean is still unavailable from intervals alone.
Show the corrected product and preserve the estimate label.
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The total must be5×14=70. Known values sum 52, so the missing value is 18. Adding 18 gives 70 and dividing by 5 checks 14.
Recover a total before subtracting the known values.
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Four records at mean 6 contribute 24 units and six at mean 11 contribute 66. The combined mean is 90/10=9. Averaging 6 and 11 directly gives 8.5 and wrongly makes four records count as much as six.
State both totals, total count and resulting unit.
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For an empty list n=0, division by n is undefined, so there is no arithmetic mean to report. For a nonempty list of 7s, total 7n/n=7; the equal-share value agrees with every record.
Distinguish no data from a genuine zero-valued mean.
Compare three routes to one mean and report the grouping loss.
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This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 11, printed pages 171–187. It does not replace the complete chapter or its practical teaching.
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