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Solutions — Full Answer Key
MathematicsYear 10 · Statistics
Solutions · Full Answer Key
Pack A answers · Pack B answers · Problem-solving worked solutions
Pack A — Answers
Bronze
1.7
2.7
3.5
4.3
5.10
6.18
7.mode = 5; range = 4
8.(a) IQR = 12 (b) median = 18
9.7
10.
Silver
11.3.8
12.13th value
13.,
14.14
15.min=4, =7, med=12, =17, max=20
16.Boxplot B (smaller IQR)
17.2.9
18.3
19.
20.Skewed (positively/right-skewed)
Gold
21., median = 18,
22.18
23.18
24.Yes — 75 > 70 (upper fence)
25.Same median; A more consistent (smaller IQR)
26.20
27.Third class (covers cf 22 to 45; median position 30 lies here)
28.
29.25%
30.435
Platinum
31.64.5
32.
33. (cf = 120 is 60th percentile for 200)
34.B has higher median and more concentrated middle (smaller IQR: 10 vs 13)
35.70 is an outlier
36.72.5
37.; frequency density (freq/width) gives true comparison
38.B
39.66.0
40. (or similar)
Pack B — Answers
Bronze
1.8
2.10
3.10
4.7
5.17
6.12
7.mode = 7; range = 7
8.(a) IQR = 12 (b) median = 15
9.12 (smallest)
10.
Silver
11.2.8
12.mean of 25th and 26th values
13.,
14.19
15.min=2, =4.5, med=9, =13.5, max=18
16.Boxplot B
17.1.6
18.30
19.
20.Positively (right) skewed
Gold
21., median = 11,
22.23
23.7.75
24.No — 5 > (lower fence)
25.Same median; B more consistent
26.30
27.Third class (cf 25 to 50; median at 40)
28.
29.25%
30.900
Platinum
31.60
32.
33.
34.B has higher median; both have IQR 10
35.50 and 60 are outliers
36.65.6
37.; with unequal widths use density
38.B
39.69.0
40. (or similar)
Problem-solving — Worked Solutions
1Problem 1
Answer
(a) mean 13.75, median 13.5, mode 16, range 8 (b) mean 14.75, median 15, mode 16, range 9 (c) See working
Full working
(a) Sort: 10, 11, 12, 13, 14, 16, 16, 18.
- **Mean** = .
- **Median**: 8 values → mean of 4th and 5th = .
- **Mode** = 16 (appears twice).
- **Range** = .
(b) Replace 11 with 19. New data sorted: 10, 12, 13, 14, 16, 16, 18, 19.
- New sum = . Mean = .
- Median = mean of 4th and 5th = .
- Mode still 16.
- Range = .
(c) The **median** changed by 1.5 (from 13.5 to 15) while the mean changed by 1.0. Here, removing a low value (11) and adding a high value (19) shifted the central order so much that the median moved noticeably. In general, the mean is sensitive to changes in *any* value, while the median only changes when the central values shift.
- **Mean** = .
- **Median**: 8 values → mean of 4th and 5th = .
- **Mode** = 16 (appears twice).
- **Range** = .
(b) Replace 11 with 19. New data sorted: 10, 12, 13, 14, 16, 16, 18, 19.
- New sum = . Mean = .
- Median = mean of 4th and 5th = .
- Mode still 16.
- Range = .
(c) The **median** changed by 1.5 (from 13.5 to 15) while the mean changed by 1.0. Here, removing a low value (11) and adding a high value (19) shifted the central order so much that the median moved noticeably. In general, the mean is sensitive to changes in *any* value, while the median only changes when the central values shift.
2Problem 2
Answer
(a) ~25 (b) , , IQR ~ 15 (c) ~22%
Full working
For : at cf 20; median at cf 40; at cf 60.
(a) Reading off: median at cf 40 → between (cf 28) and (cf 52). Linear interpolation: . **Median ≈ 25**.
(b) at cf 20: between cf 8 (mark 10) and 28 (mark 20). . .
at cf 60: between cf 52 (mark 30) and 72 (mark 40). . .
**IQR ≈ 34 − 16 = 18**.
(c) cf at mark 35: between cf 52 (mark 30) and 72 (mark 40). Linear: . So 62 students scored ≤ 35, meaning students scored more than 35. Percentage: .
(a) Reading off: median at cf 40 → between (cf 28) and (cf 52). Linear interpolation: . **Median ≈ 25**.
(b) at cf 20: between cf 8 (mark 10) and 28 (mark 20). . .
at cf 60: between cf 52 (mark 30) and 72 (mark 40). . .
**IQR ≈ 34 − 16 = 18**.
(c) cf at mark 35: between cf 52 (mark 30) and 72 (mark 40). Linear: . So 62 students scored ≤ 35, meaning students scored more than 35. Percentage: .
3Problem 3
Answer
Team A: more consistent (lower IQR, higher median). Team B: more unpredictable but higher max
Full working
**Comparison.**
- Median: A = 2 > B = 1 → Team A typically scores more per match.
- IQR: A = ; B = . Team A is more consistent.
- Range: A = 5; B = 8. Team B more variable.
**Backing decision.** Depends on goal:
- Want a high-probability return → **Team A** (median 2 vs 1; more likely to score consistently around the median).
- Want a chance of a big score → **Team B** (max 8, longer upper whisker).
For a single match prediction, statistician would back **Team A** (higher median, more consistent middle 50% above zero — Team B has , meaning 25% of B's matches end in 0 goals).
- Median: A = 2 > B = 1 → Team A typically scores more per match.
- IQR: A = ; B = . Team A is more consistent.
- Range: A = 5; B = 8. Team B more variable.
**Backing decision.** Depends on goal:
- Want a high-probability return → **Team A** (median 2 vs 1; more likely to score consistently around the median).
- Want a chance of a big score → **Team B** (max 8, longer upper whisker).
For a single match prediction, statistician would back **Team A** (higher median, more consistent middle 50% above zero — Team B has , meaning 25% of B's matches end in 0 goals).
4Problem 4
Answer
79
Full working
Mean × count = sum: .
Sum of given 9 scores: .
; ; ; ; ; ; ; .
10th score = .
Sum of given 9 scores: .
; ; ; ; ; ; ; .
10th score = .
5Problem 5
Answer
(a) ≈ 11.67 min (b) [10, 15) (c) [10, 15)
Full working
Total frequency: .
(a) Use midpoints: 2.5, 7.5, 12.5, 17.5, 22.5.
Sum:
.
Mean: min.
(b) Highest frequency 22 → **modal class is [10, 15)**.
(c) Median position = 30 (between 30th and 31st). Cumulative: 8, 22, 44, 56, 60. 30 lies between 22 and 44 → **median class is [10, 15)**.
(a) Use midpoints: 2.5, 7.5, 12.5, 17.5, 22.5.
Sum:
.
Mean: min.
(b) Highest frequency 22 → **modal class is [10, 15)**.
(c) Median position = 30 (between 30th and 31st). Cumulative: 8, 22, 44, 56, 60. 30 lies between 22 and 44 → **median class is [10, 15)**.
6Problem 6
Answer
(a) Both 70 (b) A: 0; B: 40 (c) B much more variable
Full working
(a) Set A: all 70, mean = 70. Set B: . Both means are 70.
(b) Set A range: . Set B range: .
(c) Set B has much higher variability (spread). The same mean does **not** mean the same shape — Set A has zero spread (all identical), Set B varies from 50 to 90. On a dot plot Set A is one cluster at 70; Set B is spread out.
(b) Set A range: . Set B range: .
(c) Set B has much higher variability (spread). The same mean does **not** mean the same shape — Set A has zero spread (all identical), Set B varies from 50 to 90. On a dot plot Set A is one cluster at 70; Set B is spread out.
7Problem 7
Answer
(a) 11 (b) Positively skewed (c) Yes — values above 41.5 are outliers
Full working
(a) IQR = .
(b) Whisker lengths: lower = ; upper = . Upper >> lower → **positive (right) skew**. Also: median (18) closer to (14) than to (25), confirming right skew.
(c) Fences: lower = ; upper = . Max = 50 > 41.5 → **outlier at the high end**. The exact value isn't shown by the boxplot, but the maximum 50 lies in outlier territory.
(b) Whisker lengths: lower = ; upper = . Upper >> lower → **positive (right) skew**. Also: median (18) closer to (14) than to (25), confirming right skew.
(c) Fences: lower = ; upper = . Max = 50 > 41.5 → **outlier at the high end**. The exact value isn't shown by the boxplot, but the maximum 50 lies in outlier territory.
8Problem 8
Answer
(a) 71.2 (b) No — weighted average (c) Equal sizes (24 each)
Full working
(a) Sum A: . Sum B: . Combined sum: 2848. Combined mean: .
(b) Simple average . The combined mean (71.2) is **not** the simple average — it is closer to 68 because Class A has more students. The combined mean is a **weighted average**, where each class is weighted by its size.
(c) For combined mean = 72 (the simple average of 68 and 76), the two classes would need to have **equal weights** = equal sizes. So 24 in each, for example. (Or any other equal-size pair.)
(b) Simple average . The combined mean (71.2) is **not** the simple average — it is closer to 68 because Class A has more students. The combined mean is a **weighted average**, where each class is weighted by its size.
(c) For combined mean = 72 (the simple average of 68 and 76), the two classes would need to have **equal weights** = equal sizes. So 24 in each, for example. (Or any other equal-size pair.)
9Problem 9
Answer
(a) Mean = £346k; median = £245k (b) Median fairer (c) £1.3M flat skews the mean upward
Full working
(a) Sum: (thousand). Mean = (thousand) = **£346,000** — consistent with the advertised £350k (rounded).
Median: 10 values → mean of 5th and 6th sorted = (thousand) = **£245,000**.
(b) The **median is fairer** for "typical": 9 out of 10 flats are below £300k. The mean is pulled upward by the single £1.3M outlier.
(c) The outlier (£1.3M flat) drags the mean from ~£240k upwards to £346k — a 44% inflation. The median is immune. The agent has chosen the statistic that suits their advertising.
Median: 10 values → mean of 5th and 6th sorted = (thousand) = **£245,000**.
(b) The **median is fairer** for "typical": 9 out of 10 flats are below £300k. The mean is pulled upward by the single £1.3M outlier.
(c) The outlier (£1.3M flat) drags the mean from ~£240k upwards to £346k — a 44% inflation. The median is immune. The agent has chosen the statistic that suits their advertising.
10Problem 10
Answer
(a) e.g. (b) Multiple — list in working
Full working
Constraints: mean 10 → sum 50. Median 9 → 3rd value (sorted) = 9. Mode 7 → 7 appears more than any other value. With 5 distinct positive ints — wait, **distinct** contradicts mode 7 (which requires repetition).
**Re-read the problem:** "5 distinct positive integers" is incompatible with a mode. So we interpret the question as "5 positive integers (not necessarily distinct) with mode 7" — i.e. 7 must appear at least twice (so it's the unique mode).
Sort: with sum 50, . Mode 7 means 7 must appear ≥ 2 times.
Case 1: . Then , with , , and no other value repeated. Options: (but then 9 appears twice — equals mode 7's count of 2: not unique mode unless we exclude this). ; ; ; . So sets: .
Case 2: One of = 7, and one of = 7? No — , so 7 can't appear above the median.
**Possible sets**: at least 4, as listed above.
**Re-read the problem:** "5 distinct positive integers" is incompatible with a mode. So we interpret the question as "5 positive integers (not necessarily distinct) with mode 7" — i.e. 7 must appear at least twice (so it's the unique mode).
Sort: with sum 50, . Mode 7 means 7 must appear ≥ 2 times.
Case 1: . Then , with , , and no other value repeated. Options: (but then 9 appears twice — equals mode 7's count of 2: not unique mode unless we exclude this). ; ; ; . So sets: .
Case 2: One of = 7, and one of = 7? No — , so 7 can't appear above the median.
**Possible sets**: at least 4, as listed above.
11Problem 11
Answer
(a) mean 65, IQR 12 (b) mean 120, IQR 24 (c) mean 130, IQR 24
Full working
(a) Adding a constant to all values shifts the mean by the same amount but doesn't change the spread:
- New mean = .
- IQR unchanged = 12.
(b) Multiplying all values by 2 scales both the centre and spread by 2:
- New mean = .
- New IQR = .
(c) Add 5, then double. New value = .
- New mean = .
- New IQR: only the contributes to spread (the +10 is constant). New IQR = .
- New mean = .
- IQR unchanged = 12.
(b) Multiplying all values by 2 scales both the centre and spread by 2:
- New mean = .
- New IQR = .
(c) Add 5, then double. New value = .
- New mean = .
- New IQR: only the contributes to spread (the +10 is constant). New IQR = .
12Problem 12
Answer
(a) Sampling variability (b) Robust to high earners (c) Larger sample / stratified sampling
Full working
(a) **Sampling variability** — any random sample is just one possible draw from the population. The sample mean fluctuates around the true mean. With the estimate may be off by several percent of the true value.
(b) **Income distributions are typically right-skewed** (a few very high earners). The mean is pulled upward by these high values, giving a misleadingly high "typical" income. The median is robust to outliers and better reflects the typical household.
(c) **Improvements:**
- Increase sample size (smaller sampling error).
- **Stratified sampling** by neighbourhood, age, or housing type, so each sub-group is represented proportionally.
- Combine mean and median — report both, along with IQR or range, for a fuller picture.
(b) **Income distributions are typically right-skewed** (a few very high earners). The mean is pulled upward by these high values, giving a misleadingly high "typical" income. The median is robust to outliers and better reflects the typical household.
(c) **Improvements:**
- Increase sample size (smaller sampling error).
- **Stratified sampling** by neighbourhood, age, or housing type, so each sub-group is represented proportionally.
- Combine mean and median — report both, along with IQR or range, for a fuller picture.