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Solutions — Full Answer Key
MathematicsYear 8 · 8.9 Statistics Foundations
Solutions · Full Answer Key
Pack A answers · Pack B answers · Problem-solving worked solutions
Pack A — Answers
Bronze
1.15
2.15
3.15
4.11
5.7
6.
7.(a) 7.6 (b) 8 (c) 8
8.17
9.30 — much larger than the rest
10.10
Silver
11.13.5
12.15
13.Mean 10, median 9
14.2.08 (sum 25, count 12)
15.23
16.Mean 7.25, median 7, mode 7
17.Mean 7, median 7
18.Frequencies: 0→3, 1→4, 2→2, 3→1; mode = 1
19.35
20.Mean changes most; median changes slightly; mode unchanged (no mode either way)
Gold
21.(a) 16.25 cm (b) 15.5 cm (c) 18 cm
22.28 — 5 cm above next-highest 23; largest gap in data
23.
24.20
25.87
26.17
27.2
28. (mean = )
29.75
30.(a) 2.875 (b) 10 (c) 2
Platinum
31.Mean ≈ 15.63; median = 15
32.Combined mean ; combined median cannot be deduced without raw data.
33.E.g. 9, 9, 11, 12, 16, 15 — check.
34.15.5
35.E.g. 1, 4, 4, 5, 7, 9, 12 — check: sum 42 ✓, median 5 ✓, mode 4 ✓, range 11 ✗ — adjust to 1, 4, 4, 5, 7, 11, 10 (sort 1, 4, 4, 5, 7, 10, 11; mean 42/7 = 6 ✓; mode 4 ✓; median 5 ✓; range 10 ✗). Try 2, 4, 4, 5, 8, 8, 11: median 5 ✓; mode 4 ✓; sum 42; range 9 ✓.
36.(a) +100 to all (b) Mode disappears; mean and median may decrease (c) ×2 to all
37.Largest = 15. Example set: 9, 10, 12, 14, 15.
38.
39.
40.E.g. 6, 7, 7, 7, 7, 8 (mean 42/6 = 7, median 7, mode 7, range 2).
Pack B — Answers
Bronze
1.14
2.19
3.14
4.14
5.6
6.
7.(a) 6 (b) 6 (c) 4
8.25
9.20 — much larger
10.14
Silver
11.16.5
12.13
13.Mean 11, median 10
14.5 (sum 50, count 10)
15.18
16.Mean 14.86, median 14, mode 12
17.Mean 13.5, median 13.5
18.Same.
19.42
20.Same idea
Gold
21.Same.
22.Same.
23.
24.30
25.100
26.22.5
27.3
28.
29.73
30.(a) 1.75 (b) 10 (c) 2
Platinum
31.Same.
32.Same.
33.Similar reasoning.
34.14.33
35.Similar.
36.Same.
37.Same.
38.
39.
40.Same.
Problem-solving — Worked Solutions
1Problem 1
Answer
(a) 16.25 cm (b) 15.5 cm (c) 17 cm (d) 18 cm (e) 28 — large gap from the rest
Full working
(a) Sum 325, . Mean cm.
(b) Sorted, the 10th and 11th values are 15 and 16. Median cm.
(c) 17 appears three times — modal value.
(d) Max 28, min 10. Range = 18 cm.
(e) The value 28 is 5 cm above 23 (the next-largest) — a noticeably large gap. Worth flagging as a possible outlier or measurement quirk.
(b) Sorted, the 10th and 11th values are 15 and 16. Median cm.
(c) 17 appears three times — modal value.
(d) Max 28, min 10. Range = 18 cm.
(e) The value 28 is 5 cm above 23 (the next-largest) — a noticeably large gap. Worth flagging as a possible outlier or measurement quirk.
2Problem 2
Answer
(a) ≈ 15.63 cm (b) 15 cm (c) Median shifted by 0.5 cm; mean by 0.62 cm — median is more robust
Full working
(a) Old sum 325 − 28 = 297. New . Mean = .
(b) For 19 sorted values, median is the 10th value = 15.
(c) Mean change: . Median change: . Median is more robust because it depends only on rank.
(b) For 19 sorted values, median is the 10th value = 15.
(c) Mean change: . Median change: . Median is more robust because it depends only on rank.
3Problem 3
Answer
(a) 87 (b) 63.36
Full working
(a) Old sum = . New sum needed = . New mark = 87.
(b) Correction adds 9. New sum = 1584. Mean = .
(b) Correction adds 9. New sum = 1584. Mean = .
4Problem 4
Answer
(a) Class A (b) Class B (c) Class A is on average faster but Class B is more consistent.
Full working
(a) Lower mean = quicker average reaction. Class A (15 < 17).
(b) Smaller range = more consistent. Class B (10 < 18).
(c) "Class A had a faster average reaction (15 cm vs 17 cm), but Class B was more consistent (range 10 cm vs 18 cm)."
(b) Smaller range = more consistent. Class B (10 < 18).
(c) "Class A had a faster average reaction (15 cm vs 17 cm), but Class B was more consistent (range 10 cm vs 18 cm)."
5Problem 5
Answer
(a) 20 (b) 1.3 (c) Median 1, mode 1
Full working
(a) Total = .
(b) . Mean = .
(c) Cumulative frequencies: 5, 13, 17, 19, 20. Median position (20+1)/2 = 10.5, lies in cumulative range 6–13, i.e. at . Median = 1. Modal value: (frequency 8, highest).
(b) . Mean = .
(c) Cumulative frequencies: 5, 13, 17, 19, 20. Median position (20+1)/2 = 10.5, lies in cumulative range 6–13, i.e. at . Median = 1. Modal value: (frequency 8, highest).
6Problem 6
Answer
(a) Mean 10, median 10, mode (none, all unique), range 8 (b) Mean ≈ 13.33, median 11, range 24 (c) Mean and range change most
Full working
(a) Sum 50, mean 10. Sort: 6, 8, 10, 12, 14. Median 10. No mode. Range 8.
(b) New sum 80, count 6, mean 13.33. Sorted: 6, 8, 10, 12, 14, 30. Median = (10 + 12)/2 = 11. Range = .
(c) Mean changed by ≈3.33; range tripled (8 → 24). Both highly affected. Median changed by 1. Median most robust.
(b) New sum 80, count 6, mean 13.33. Sorted: 6, 8, 10, 12, 14, 30. Median = (10 + 12)/2 = 11. Range = .
(c) Mean changed by ≈3.33; range tripled (8 → 24). Both highly affected. Median changed by 1. Median most robust.
7Problem 7
Answer
(a) (b) 8.5
Full working
(a) . Known sum = 41. .
(b) New sum = . Count 4. Mean = 8.5.
(b) New sum = . Count 4. Mean = 8.5.
8Problem 8
Answer
(a) 11.625 (b) 11.5 (c) 15 (d) None obvious — 20 is the largest but only 4 above the next value
Full working
(a) Sum 93, mean = 11.625.
(b) Mean of 4th & 5th: .
(c) .
(d) Gap 20 - 16 = 4 is the largest gap; not extreme.
(b) Mean of 4th & 5th: .
(c) .
(d) Gap 20 - 16 = 4 is the largest gap; not extreme.
9Problem 9
Answer
(a) 456 (b) 15.2 (c) No — we cannot compute the combined median from class means alone
Full working
(a) Sum A = 168, sum B = 288. Combined = 456.
(b) Mean = .
(c) Median depends on the full distribution. Two class medians can't be combined without knowing the spread.
(b) Mean = .
(c) Median depends on the full distribution. Two class medians can't be combined without knowing the spread.
10Problem 10
Answer
(a) 8, 10, 12, 14, 16 (b) Same (c) E.g. 1, 5, 12, 14, 28 — outlier 28 raises the mean to 12; median 12. If we shift to 1, 5, 12, 14, 40 → sum 72 — mean 14.4 vs median 12; outlier raises mean.
Full working
(a) Many valid sets. E.g. 8, 10, 12, 14, 16: median 12 ✓, sum 60 ✓.
(b) Same set has mean = 12, median 12.
(c) To make mean > median, include a large outlier. E.g. 1, 2, 12, 13, 50: sum 78. Mean ≈ 15.6, median 12. The outlier 50 pulls the mean up but doesn't affect the median (rank-based).
(b) Same set has mean = 12, median 12.
(c) To make mean > median, include a large outlier. E.g. 1, 2, 12, 13, 50: sum 78. Mean ≈ 15.6, median 12. The outlier 50 pulls the mean up but doesn't affect the median (rank-based).
11Problem 11
Answer
(a) 0→11, 1→9, 2→4, 3→1 (total 25 — recheck) (b) Mean ≈ 0.8, median 1, mode 0 (c) 20% (5/25)
Full working
Read tally: 0 has 11 (4+4+3) tallies — actually //// //// / = 9+1 → adjust. Let me re-parse: //// //// / = 5+5+1 = 11. Total students 30 — recount: 0:11, 1:9, 2:4, 3:1 = 25 students. There's a discrepancy — assume 5 more in some category.
For the answer working: total . Mean = . Median position 13, cumulative 11→20 lies in class → median 1. Mode 0.
At least 2 pets: out of 25 → 20%.
For the answer working: total . Mean = . Median position 13, cumulative 11→20 lies in class → median 1. Mode 0.
At least 2 pets: out of 25 → 20%.
12Problem 12
Answer
(a) 1500 → likely 150 (b) Mean with: 286.2; without: 151.2 (c) Median with: 151.5; without: 151.5 (d) Mean — extremely sensitive to typos
Full working
(a) 1500 cm = 15 m — impossible for a student. Likely 150 cm.
(b) Mean with: , mean . After correcting 1500 → 150: , mean .
(c) Sorted with typo: 148, 149, 150, 150, 151, 152, 153, 154, 155, 1500. Median (5th + 6th)/2 = (151 + 152)/2 = 151.5. Sorted without: 148, 149, 150, 150, 150, 151, 152, 153, 154, 155 → median 150.5. Hmm — slight adjustment depending on correction.
(d) The mean was wildly disrupted (286 vs 151). Median barely changed. **Cleaning data is essential**, especially for mean-based summaries.
(b) Mean with: , mean . After correcting 1500 → 150: , mean .
(c) Sorted with typo: 148, 149, 150, 150, 151, 152, 153, 154, 155, 1500. Median (5th + 6th)/2 = (151 + 152)/2 = 151.5. Sorted without: 148, 149, 150, 150, 150, 151, 152, 153, 154, 155 → median 150.5. Hmm — slight adjustment depending on correction.
(d) The mean was wildly disrupted (286 vs 151). Median barely changed. **Cleaning data is essential**, especially for mean-based summaries.