Cross-Curricular Statistics Practice | 跨学科统计综合题型训练

📚 Cross-Curricular Statistics Practice | 跨学科统计综合题型训练

Statistics is not confined to mathematics classrooms — it is the language of evidence across all sciences and social sciences. In Year 11 CCEA Statistics, you are expected to apply statistical thinking to real-world problems drawn from biology, geography, economics, physics, sports, and more. This revision guide provides cross-curricular problem scenarios that blend data handling, averages, measures of spread, graphs, and interpretation. Each section presents a short contextual task followed by guided statistical reasoning to help you master the skill of transferring your knowledge to unfamiliar situations.

统计学并不局限于数学课堂——它是一切科学和社会科学中证据的语言。在 Year 11 CCEA 统计课程中，你需要将统计思维应用到来自生物、地理、经济、物理、体育等学科的真实问题中。本复习指南提供了跨学科的问题情境，融合数据处理、平均数、离散量数、图表与解读。每一节都给出一个简短的背景任务，并配以统计推理的引导，帮助你掌握将知识迁移到陌生情境的能力。

1. Biology: Bacterial Growth | 生物：细菌培养

A biology experiment records the number of bacterial colonies in a petri dish every hour for 8 hours. The recorded counts are: 5, 8, 14, 23, 36, 55, 80, 112. The scientist wants to describe the growth pattern and check for unusually rapid growth in any interval.

一项生物实验记录了培养皿中细菌菌落数量，每小时一次，持续8小时。记录数据为：5, 8, 14, 23, 36, 55, 80, 112。研究人员希望描述增长模式并检查是否有异常快速增长时段。

Using the data, construct a time-series line graph. Describe whether the growth appears linear or exponential. To quantify the spread of hourly increases, calculate the range and interquartile range of the 7 growth increments (3, 6, 9, 13, 19, 25, 32). Then determine if the largest increment (32) is an outlier using the 1.5 × IQR rule.

利用数据绘制时间序列折线图。描述增长呈线性还是指数趋势。为了量化每小时增长量的分散程度，计算7个增长量（3, 6, 9, 13, 19, 25, 32）的全距和四分位数间距。然后使用 1.5×IQR 法则判断最大增长量（32）是否为异常值。

Order the increments: 3, 6, 9, 13, 19, 25, 32. Lower quartile Q1 = 7.5, upper quartile Q3 = 22, IQR = 14.5. Upper fence = Q3 + 1.5 × IQR = 22 + 21.75 = 43.75. Since 32 < 43.75, the largest increment is not an outlier. The growth curve is curving upwards, suggesting exponential increase; a line graph with time on the x-axis makes this visible. The average increment is about 15.3, but the variability is large.

将增长量排序：3, 6, 9, 13, 19, 25, 32。下四分位数 Q1 = 7.5，上四分位数 Q3 = 22，IQR = 14.5。上须值 = Q3 + 1.5 × IQR = 22 + 21.75 = 43.75。因为 32 < 43.75，最大增长量并非异常值。增长曲线向上弯曲，表明指数增长；以时间为 x 轴的折线图可清晰呈现这一点。平均增长量约为 15.3，但变异性较大。

2. Geography: Population Pyramids | 地理：人口金字塔

A geography textbook provides the percentage distribution of males and females in two countries for age groups 0–14, 15–64, and 65+. Country A: 0–14 (18%), 15–64 (68%), 65+ (14%); Country B: 0–14 (38%), 15–64 (58%), 65+ (4%). A student must compare the age structures and evaluate the validity of a statement: ‘Country A has twice the proportion of elderly people as Country B.’

一本地理教科书给出了两个国家男女人口在 0–14、15–64 和 65+ 年龄组的百分比分布。A 国：0–14 (18%)、15–64 (68%)、65+ (14%)；B 国：0–14 (38%)、15–64 (58%)、65+ (4%)。一名学生需要比较年龄结构，并评估“A 国老年人口比例是 B 国的两倍”这一说法是否成立。

Stacked bar charts or population pyramids can be drawn for each country. Compute the dependency ratio: (under 15 + over 65) ÷ (15–64) × 100. For A: (18+14)/68 × 100 = 47.1; for B: (38+4)/58 × 100 = 72.4. The statement about elderly proportion: 14% is 3.5 times 4%, not twice, so it is incorrect. The data highlight high child dependency in B and high old-age dependency in A.

可以为每个国家绘制堆叠条形图或人口金字塔。计算抚养比：(15岁以下 + 65岁以上) ÷ (15–64) × 100。A 国：(18+14)/68 × 100 = 47.1；B 国：(38+4)/58 × 100 = 72.4。关于老年人口比例的说法：14% 是 4% 的 3.5 倍，而非两倍，因此说法错误。数据突显 B 国儿童抚养负担高，A 国老年抚养负担高。

3. Economics: Price Indices and Inflation | 经济：价格指数与通货膨胀

An economics project tracks the prices of four items in a student’s weekly basket: bus fare, lunch, stationery, and a magazine. Prices in 2021 and 2023 are recorded. The quantities bought per week are used as weights. The task is to calculate a weighted aggregate price index and comment on the change in the cost of living.

一项经济课题追踪学生每周消费篮中四种商品的价格：公交车费、午餐、文具和杂志。记录了 2021 和 2023 年的价格，并将每周购买数量作为权数。任务是计算加权综合价格指数，并评价生活成本的变化。

Item	Price 2021	Price 2023	Weight
Bus fare	£1.50	£1.80	10
Lunch	£3.20	£3.80	5
Stationery	£2.00	£2.40	2
Magazine	£4.50	£5.00	1

Weighted index = (sum of (P₂₀₂₃ × weight) ÷ sum of (P₂₀₂₁ × weight)) × 100. Numerator = 1.80×10 + 3.80×5 + 2.40×2 + 5.00×1 = 18+19+4.8+5 = 46.8. Denominator = 1.50×10 + 3.20×5 + 2.00×2 + 4.50×1 = 15+16+4+4.5 = 39.5. Index = (46.8/39.5)×100 ≈ 118.5. The cost of the basket has risen by about 18.5%, indicating significant inflation for this student’s spending pattern.

加权指数 =（Σ(P₂₀₂₃ × 权数) ÷ Σ(P₂₀₂₁ × 权数)）× 100。分子 = 1.80×10 + 3.80×5 + 2.40×2 + 5.00×1 = 46.8。分母 = 1.50×10 + 3.20×5 + 2.00×2 + 4.50×1 = 39.5。指数 = (46.8/39.5)×100 ≈ 118.5。消费篮成本上升约 18.5%，表明该学生支出模式下的通货膨胀较为显著。

4. Physics: Measurement Uncertainties | 物理：测量不确定度

In a physics lab, a student repeats the measurement of the length of a pendulum five times: 2.45 m, 2.43 m, 2.46 m, 2.44 m, 2.45 m. The true length is not known, so the student must estimate the measurement uncertainty and report the length with its absolute and percentage uncertainty.

在物理实验室，一名学生重复测量单摆长度五次：2.45 m, 2.43 m, 2.46 m, 2.44 m, 2.45 m。真实长度未知，学生必须估计测量不确定度，并报告带有绝对和百分比不确定度的长度。

Calculate the mean: (2.45+2.43+2.46+2.44+2.45) ÷ 5 = 2.446 m. The range is 2.46 − 2.43 = 0.03 m. A simple estimate of absolute uncertainty is half the range: ±0.015 m. Hence the length can be stated as 2.446 ± 0.015 m. Percentage uncertainty = (0.015 ÷ 2.446) × 100 ≈ 0.61%. Discussing significant figures, it is appropriate to quote the mean as 2.45 ± 0.02 m.

计算平均值：(2.45+2.43+2.46+2.44+2.45) ÷ 5 = 2.446 m。全距为 2.46 − 2.43 = 0.03 m。绝对不确定度的一种简单估计为全距的一半：±0.015 m。因此长度可表示为 2.446 ± 0.015 m。百分比不确定度 = (0.015 ÷ 2.446) × 100 ≈ 0.61%。考虑到有效数字，可将平均值合理表示为 2.45 ± 0.02 m。

5. Sports: Performance Analysis | 体育：运动表现分析

A sports scientist compares the 100-metre sprint times (in seconds) of two athletes over 8 races. Athlete X: 10.2, 10.1, 9.9, 10.4, 10.0, 10.1, 9.8, 10.3. Athlete Y: 10.0, 10.3, 9.7, 10.5, 10.1, 10.6, 9.9, 10.2. The goal is to decide which athlete is faster on average and which is more consistent.

一位运动科学家比较了两名运动员在 8 场比赛中的 100 米短跑时间（秒）。运动员 X：10.2, 10.1, 9.9, 10.4, 10.0, 10.1, 9.8, 10.3。运动员 Y：10.0, 10.3, 9.7, 10.5, 10.1, 10.6, 9.9, 10.2。目标是判断哪位运动员平均速度更快，哪位表现更稳定。

Compute means: X̄ = (10.2+10.1+9.9+10.4+10.0+10.1+9.8+10.3)/8 = 10.1 s; Ȳ = 10.1625 s. Athlete X is slightly faster on average. For consistency, calculate the standard deviation. Using the formula s = √(Σ(x − x̄)²/(n−1)), X’s s ≈ 0.195 s, Y’s s ≈ 0.30 s. Athlete X has a smaller standard deviation and is therefore more consistent. A box plot comparison would also show a narrower interquartile range for X.

计算平均数：X̄ = (10.2+10.1+9.9+10.4+10.0+10.1+9.8+10.3)/8 = 10.1 秒；Ȳ = 10.1625 秒。运动员 X 平均稍快。为衡量稳定性，计算标准差。使用公式 s = √(Σ(x − x̄)²/(n−1))，X 的标准差约为 0.195 秒，Y 的约为 0.30 秒。X 的标准差更小，因此更稳定。箱线图比较也会显示 X 的四分位距更窄。

6. Business: Sales Forecasting | 商业：销售预测

A small business records quarterly sales (£ thousands) for three years. The data show seasonal variation. The owner wants to smooth the trend using moving averages and use the trend to forecast the next quarter’s sales.

一家小型企业记录了三年的季度销售额（千英镑）。数据呈现季节波动。业主希望用移动平均平滑趋势，并利用趋势预测下一季度销售额。

Quarterly sales: Year 1 Q1–Q4: 20, 28, 32, 24; Year 2: 22, 30, 34, 26; Year 3: 24, 32, 36, 28. Calculate 4-point moving averages and centre them. The centred moving averages give the trend values. The last trend value (Year 3 Q4) is approximately (28+?)/… Actually, compute: first 4-point MA (20+28+32+24)/4 = 26; next (28+32+24+22)/4 = 26.5; etc. After centering, the trend for Year 3 Q3 is around 30.5, for Q3.5 (mid between Q3 and Q4) it is 31. By plotting the trend line and extending it, the forecast for Year 4 Q1 is about 29, after adjusting for seasonal effect which typically lowers Q1.

季度销售额：第1年 Q1–Q4：20, 28, 32, 24；第2年：22, 30, 34, 26；第3年：24, 32, 36, 28。计算4点移动平均并置中。置中移动平均给出趋势值。最后一个趋势值（第3年Q4）可通过计算得到。前一个4点移动平均 (20+28+32+24)/4 = 26；接着 (28+32+24+22)/4 = 26.5；等等。置中后，第3年Q3趋势约为30.5，Q3.5趋势约为31。绘制趋势线并延长，考虑季节影响后（通常Q1偏低），预测第4年Q1约为29。

7. Health: Clinical Trial Comparison | 健康：临床试验对比

A health study tests a new drug against a placebo for reducing recovery days from a common illness. The recovery days for the drug group (n=10) are: 5, 6, 4, 7, 5, 5, 6, 4, 6, 5. For the placebo group: 7, 8, 6, 9, 7, 8, 7, 6, 8, 7. Researchers want to know whether the drug reduces recovery time and whether variability differs.

一项健康研究测试一种新药对比安慰剂在减少常见疾病康复天数上的效果。药物组（10人）的康复天数：5, 6, 4, 7, 5, 5, 6, 4, 6, 5。安慰剂组：7, 8, 6, 9, 7, 8, 7, 6, 8, 7。研究者想知道药物是否缩短康复时间，以及变异性是否不同。

Drug group mean = 5.3 days, median = 5 days, standard deviation s = 0.95. Placebo group mean = 7.3 days, median = 7, s = 0.95 as well. Both groups have similar variability (range and IQR also are comparable). However, the drug group’s mean is 2 days lower. An informal comparison suggests the drug is effective. Students can create back-to-back stem-and-leaf diagrams to visualise the shift. The median difference supports the conclusion that recovery times are systematically shorter.

药物组平均数 = 5.3 天，中位数 = 5，标准差 s = 0.95。安慰剂组平均数 = 7.3 天，中位数 = 7，标准差同样为 0.95。两组变异性相似（全距和IQR也相近）。但药物组平均数低 2 天。非正式比较表明药物有效。学生可绘制背靠背茎叶图来直观显示这种偏移。中位数差异支持康复时间系统性更短的结论。

8. Environmental Science: Carbon Emissions | 环境科学：碳排放

An environmental report provides annual mean CO₂ concentration (ppm) at a monitoring station from 2010 to 2020. The data are: 2010: 389, 2011: 391, 2012: 393, 2013: 396, 2014: 398, 2015: 400, 2016: 404, 2017: 406, 2018: 408, 2019: 411, 2020: 414. An environmental science class is asked to fit a linear trend and estimate the level for 2025.

一份环境报告给出某监测站2010至2020年的年平均 CO₂ 浓度（ppm）。数据为：2010: 389, 2011: 391, 2012: 393, 2013: 396, 2014: 398, 2015: 400, 2016: 404, 2017: 406, 2018: 408, 2019: 411, 2020: 414。环境科学课要求拟合线性趋势并估计 2025 年的浓度水平。

Let x = year − 2010 (so x = 0 to 10). Plot points. The line of best fit by eye or using the mean-mean method yields a slope of about 2.5 ppm per year. Equation: CO₂ = 389 + 2.5x. For 2025, x = 15, predicted level = 389 + 2.5×15 = 426.5 ppm. Discuss the dangers of extrapolation: the linear trend may not hold due to policy changes. Students can calculate the correlation coefficient r ≈ 0.998, confirming a very strong positive linear relationship.

设 x = 年份 − 2010（则 x 取值 0 至 10）。描点。通过目测或平均-平均法求得的最佳拟合直线斜率约为每年 2.5 ppm。方程：CO₂ = 389 + 2.5x。对于 2025 年，x = 15，预测浓度 = 389 + 2.5×15 = 426.5 ppm。讨论外推的风险：由于政策变化，线性趋势可能不再成立。学生可以计算相关系数 r ≈ 0.998，证实存在极强的正线性关系。

9. Psychology: Survey Design and Bias | 心理学：调查设计与偏差

A psychology class aims to investigate teenagers’ daily screen time. They plan a questionnaire but must consider sampling methods and potential bias. The school has 1200 students, and they want a sample of 100. Describe a simple random sampling method and a stratified sampling approach using year groups.

一个心理学班级计划调查青少年每日屏幕使用时间。他们设计了一份问卷，但必须考虑抽样方法和潜在偏差。学校有 1200 名学生，他们需要抽取 100 名作为样本。描述简单随机抽样方法，以及利用年级分层抽样的方式。

For simple random sampling, assign a number to each student and use a random number generator to pick 100. This could accidentally miss small year groups. Stratified sampling: calculate proportions. For instance, if Year 8 has 240 students, sample size = (240/1200)×100 = 20. Do likewise for other years. This guarantees representation. Students must also evaluate non-response bias — those with very high screen time might not respond truthfully. Question wording can introduce bias, so they should pilot the questionnaire.

对于简单随机抽样，给每名学生分配编号，用随机数生成器抽取 100 个。这可能碰巧遗漏人数较少的年级。分层抽样：计算比例。例如，若 8 年级有 240 人，样本量 = (240/1200)×100 = 20。其他年级同理。这保证了代表性。学生还必须评估无回应偏差——屏幕时间极长的学生可能不如实回答。问卷用词可能引入偏差，因此应先进行试点调查。

10. Education: Exam Score Comparisons | 教育：考试成绩比较

Two Year 11 classes sit the same statistics test (maximum mark 50). Class A scores: 35, 38, 42, 28, 45, 40, 33, 37, 41, 29. Class B: 22, 48, 25, 44, 20, 47, 23, 46, 24, 45. Analyse which class performed better and discuss the distributions.

两个 Year 11 班级参加

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Cross-Curricular Statistics Practice | 跨学科统计综合题型训练