📚 Interdisciplinary Statistics Questions for Year 10 Edexcel | 跨学科综合题型训练（Year 10 Edexcel 统计）
In the Edexcel Year 10 Statistics course, applying statistical techniques to real-world scenarios across different subjects is essential for mastering the curriculum. This article provides a series of interdisciplinary questions that blend statistics with science, geography, business, and more, helping you practise data handling, probability, and interpretation skills. Each section presents a contextual problem, outlines the relevant statistical tools, and walks you through the solution step by step.
在Edexcel十年级统计课程中，将统计技术应用到不同学科的实际情境中，对于掌握课程内容至关重要。本文提供了一系列跨学科问题，将统计与科学、地理、商业等学科相结合，帮助你练习数据处理、概率分析和解读技能。每一节都给出一个情境问题，列出相关的统计工具，并逐步引导你完成解答。
1. Science Experiments: Handling Measurement Errors | 科学实验：处理测量误差
In a physics experiment, students repeated the measurement of the period of a pendulum five times: 2.01 s, 1.98 s, 2.03 s, 1.97 s, 2.02 s. They need to calculate the mean, range, and estimate the uncertainty. Why is repeating measurements important for reducing random error?
在一个物理实验中，学生们重复测量了单摆的周期五次：2.01 s, 1.98 s, 2.03 s, 1.97 s, 2.02 s。他们需要计算平均值、极差并估计不确定度。为什么重复测量对于减少随机误差很重要？
First, compute the mean: (2.01 + 1.98 + 2.03 + 1.97 + 2.02) ÷ 5 = 10.01 ÷ 5 = 2.002 s. The range is 2.03 − 1.97 = 0.06 s. A simple estimate of uncertainty is half the range: 0.03 s. The reported result can be 2.00 s ± 0.03 s (rounded to two decimal places). Repeating measurements allows random errors to average out, making the mean a more reliable estimate of the true value.
首先计算平均值：(2.01 + 1.98 + 2.03 + 1.97 + 2.02) ÷ 5 = 10.01 ÷ 5 = 2.002 秒。极差为 2.03 − 1.97 = 0.06 秒。不确定度的一种简单估计方法是极差的一半：0.03 秒。报告结果可表示为 2.00 s ± 0.03 s（四舍五入到两位小数）。重复测量可以使随机误差相互抵消，使平均值成为真实值更可靠的估计。
Extension: If the same experiment is done with a larger sample of 20 measurements, the standard deviation can be used to express precision. The formula for sample standard deviation s = √[Σ(x − x̄)² ÷ (n − 1)] provides a more formal measure of spread.
拓展：如果对同样的实验进行20次测量，可以用标准差来表示精密度。样本标准差公式 s = √[Σ(x − x̄)² ÷ (n − 1)] 给出了更正式的离散程度度量。
2. Geography: Climate Data and Box Plots | 地理：气候数据与箱线图
A geography class collected monthly rainfall data (in mm) for two cities over a year. City A: 56, 48, 52, 60, 65, 70, 75, 78, 72, 63, 58, 50. City B: 80, 82, 78, 85, 90, 95, 100, 98, 92, 88, 85, 84. Construct box plots and compare the distribution of rainfall, mentioning median, interquartile range (IQR), and overall range.
地理课上收集了两个城市一年的月降雨量数据（单位：毫米）。城市A：56, 48, 52, 60, 65, 70, 75, 78, 72, 63, 58, 50。城市B：80, 82, 78, 85, 90, 95, 100, 98, 92, 88, 85, 84。绘制箱线图并比较降雨量分布，提及中位数、四分位距（IQR）和全距。
Sort each dataset. City A: 48, 50, 52, 56, 58, 60, 63, 65, 70, 72, 75, 78. Median = (60+63)÷2 = 61.5 mm. Q1 = (52+56)÷2 = 54 mm. Q3 = (70+72)÷2 = 71 mm. IQR = 17 mm. Range = 78 − 48 = 30 mm. City B: 78, 80, 82, 84, 85, 85, 88, 90, 92, 95, 98, 100. Median = (85+88)÷2 = 86.5 mm. Q1 = (82+84)÷2 = 83 mm. Q3 = (92+95)÷2 = 93.5 mm. IQR = 10.5 mm. Range = 100 − 78 = 22 mm. The box plot for City B is shifted higher and is more compact (smaller spread), while City A shows greater variability but lower typical rainfall.
对每个数据集排序。城市A：48, 50, 52, 56, 58, 60, 63, 65, 70, 72, 75, 78。中位数 = (60+63)÷2 = 61.5 mm。下四分位数Q1 = (52+56)÷2 = 54 mm。上四分位数Q3 = (70+72)÷2 = 71 mm。IQR = 17 mm。全距 = 78 − 48 = 30 mm。城市B：78, 80, 82, 84, 85, 85, 88, 90, 92, 95, 98, 100。中位数 = (85+88)÷2 = 86.5 mm。Q1 = (82+84)÷2 = 83 mm。Q3 = (92+95)÷2 = 93.5 mm。IQR = 10.5 mm。全距 = 100 − 78 = 22 mm。城市B的箱线图整体位置更高且更紧凑（散度较小），而城市A虽然变异性更大，但典型降雨量较低。
3. Business: Cumulative Frequency and Profit Analysis | 商业：累积频率与利润分析
A small business records the daily profit (in £) for 30 days. The results are grouped: £0−50 (4 days), £50−100 (8 days), £100−150 (10 days), £150−200 (5 days), £200−250 (3 days). Draw a cumulative frequency graph and estimate the median profit and the interquartile range. What percentage of days had a profit above £180?
一个小型企业记录了30天的每日利润（单位：英镑）。结果分组如下：£0−50（4天），£50−100（8天），£100−150（10天），£150−200（5天），£200−250（3天）。绘制累积频率图，并估算利润中位数和四分位距。利润超过£180的天数百分比是多少？
Cumulative frequencies: up to £50: 4; up to £100: 12; up to £150: 22; up to £200: 27; up to £250: 30. Plot the upper class boundaries (50, 100, 150, 200, 250) against cumulative frequency. From the graph, the median (at 15) is approximately £115. Q1 (at 7.5) ≈ £72, Q3 (at 22.5) ≈ £162, so IQR ≈ £90. To find the percentage above £180: at £180, cumulative frequency ≈ 25 (interpolating), so 30 − 25 = 5 days are above £180, which is (5/30)×100 ≈ 16.7%.
累积频数：低于£50：4；低于£100：12；低于£150：22；低于£200：27；低于£250：30。以组上限（50, 100, 150, 200, 250）为横轴、累积频数为纵轴描点作图。从图上读出中位数（对应15）约 £115。Q1（对应7.5）≈ £72，Q3（对应22.5）≈ £162，因此IQR ≈ £90。利润超过£180的天数：£180处累积频数约为25（插值），所以超过£180的天数为30 − 25 = 5天，占比(5/30)×100 ≈ 16.7%。
4. Sports: Comparing Athlete Performance Using Mean and Standard Deviation | 体育：用均值和标准差比较运动员表现
Two basketball players, A and B, scored the following points in 8 matches: Player A: 18, 22, 20, 19, 24, 17, 21, 23. Player B: 25, 12, 30, 15, 28, 10, 22, 18. Calculate the mean and standard deviation for each player. Who is the more consistent scorer?
两名篮球运动员A和B在8场比赛中的得分如下：球员A：18, 22, 20, 19, 24, 17, 21, 23。球员B：25, 12, 30, 15, 28, 10, 22, 18。计算每位球员的均值和标准差，谁的表现更稳定？
Player A: mean = (18+22+20+19+24+17+21+23) ÷ 8 = 164 ÷ 8 = 20.5. Deviations: −2.5, 1.5, −0.5, −1.5, 3.5, −3.5, 0.5, 2.5. Squared: 6.25, 2.25, 0.25, 2.25, 12.25, 12.25, 0.25, 6.25. Sum = 42. Standard deviation s = √(42 ÷ 7) ≈ 2.45. Player B: mean = (25+12+30+15+28+10+22+18) ÷ 8 = 160 ÷ 8 = 20. Squared deviations from 20: 25, 64, 100, 25, 64, 100, 4, 4; sum = 386. s = √(386 ÷ 7) ≈ 7.43. Player A has a much smaller standard deviation, so their scoring is more consistent, while Player B shows higher variability with both very high and low scores.
球员A：均值 = (18+22+20+19+24+17+21+23) ÷ 8 = 164 ÷ 8 = 20.5。偏差：−2.5, 1.5, −0.5, −1.5, 3.5, −3.5, 0.5, 2.5。平方和：6.25+2.25+0.25+2.25+12.25+12.25+0.25+6.25 = 42。标准差 s = √(42 ÷ 7) ≈ 2.45。球员B：均值 = (25+12+30+15+28+10+22+18) ÷ 8 = 20。与20的偏差平方：25, 64, 100, 25, 64, 100, 4, 4；和 = 386。s = √(386 ÷ 7) ≈ 7.43。球员A的标准差远小于B，因此A的得分更稳定，而B的波动性更大，有非常高和非常低的得分。
5. Population Studies: Time Series of Birth Rates | 人口研究：出生率时间序列
Year 10 statistics students analyse the crude birth rate (per 1000 people) for a region from 2010 to 2019: 12.5, 12.3, 12.0, 11.8, 11.5, 11.4, 11.2, 11.0, 10.9, 10.7. Plot a time series graph, calculate the 3-point moving averages, and comment on the trend. What might cause this trend?
十年级学生分析某地区2010至2019年的粗出生率（每千人）：12.5, 12.3, 12.0, 11.8, 11.5, 11.4, 11.2, 11.0, 10.9, 10.7。绘制时间序列图，计算三点移动平均数并评论趋势。可能是什么原因导致这种趋势？
Moving averages: (12.5+12.3+12.0)/3 = 12.27; (12.3+12.0+11.8)/3 = 12.03; (12.0+11.8+11.5)/3 = 11.77; (11.8+11.5+11.4)/3 = 11.57; (11.5+11.4+11.2)/3 = 11.37; (11.4+11.2+11.0)/3 = 11.20; (11.2+11.0+10.9)/3 = 11.03; (11.0+10.9+10.7)/3 = 10.87. The moving averages show a clear downward trend, smoothing out minor fluctuations. Possible reasons could be increased access to education, urbanisation, or changes in family planning policies.
移动平均：(12.5+12.3+12.0)/3 = 12.27; (12.3+12.0+11.8)/3 = 12.03; (12.0+11.8+11.5)/3 = 11.77; (11.8+11.5+11.4)/3 = 11.57; (11.5+11.4+11.2)/3 = 11.37; (11.4+11.2+11.0)/3 = 11.20; (11.2+11.0+10.9)/3 = 11.03; (11.0+10.9+10.7)/3 = 10.87。移动平均数显示出明显的下降趋势，并平滑了微小波动。可能的原因包括教育普及、城市化或计划生育政策的变化。
6. Medicine: Probability and Relative Risk | 医学：概率与相对风险
A medical study investigates a new vaccine. In the vaccinated group of 2000 people, 30 developed the disease. In the control group of 2000 unvaccinated people, 120 developed the disease. Calculate the risk in each group, the relative risk, and explain what it means.
一项医学研究调查了一种新疫苗。在2000名接种者中，有30人发病；在2000名未接种的对照组中，有120人发病。计算各组患病风险、相对风险，并解释其含义。
Risk in vaccinated group = 30/2000 = 0.015 (or 1.5%). Risk in control group = 120/2000 = 0.06 (or 6%). Relative risk (RR) = risk in vaccinated ÷ risk in control = 0.015/0.06 = 0.25. A relative risk of 0.25 means the vaccinated group has only a quarter of the risk of the unvaccinated group, suggesting the vaccine is effective. This can also be expressed as a 75% reduction in risk.
接种组风险 = 30/2000 = 0.015（或1.5%）。对照组风险 = 120/2000 = 0.06（或6%）。相对风险（RR）= 接种组风险 ÷ 对照组风险 = 0.015/0.06 = 0.25。相对风险为0.25意味着接种组的患病风险仅为未接种组的四分之一，表明疫苗有效。这也可以表示为风险降低了75%。
Additionally, the absolute risk reduction (ARR) is 0.06 − 0.015 = 0.045 (4.5 percentage points). The number needed to vaccinate (NNV) to prevent one case is 1/ARR ≈ 22.2, meaning about 22 people need to be vaccinated.
此外，绝对风险降低（ARR）为 0.06 − 0.015 = 0.045（4.5个百分点）。预防一例所需接种人数（NNV）为 1/ARR ≈ 22.2，意味着大约需要为22人接种疫苗。
7. Environmental Science: Stratified Sampling for Pollution Estimates | 环境科学：分层抽样估计污染
An environmental survey aims to estimate the average concentration of a pollutant in a lake. The lake is divided into three strata: near the inlet (area A), central (B), and near the outlet (C). The areas and sample results are: A – 20% of lake, sample mean = 45 mg/L, sample size = 10; B – 50%, mean = 32 mg/L, size = 25; C – 30%, mean = 38 mg/L, size = 15. Calculate the overall stratified mean.
一项环境调查旨在估计湖中污染物的平均浓度。该湖被分为三层：进水口附近（A区）、中心区（B区）和出水口附近（C区）。各区域的面积比例和抽样结果为：A区占湖面20%，样本均值 = 45 mg/L，样本量10；B区占50%，均值 = 32 mg/L，样本量25；C区占30%，均值 = 38 mg/L，样本量15。计算总体分层均值。
The stratified mean is a weighted average using the strata proportions: μ = 0.20×45 + 0.50×32 + 0.30×38 = 9 + 16 + 11.4 = 36.4 mg/L. This method gives a more representative estimate than a simple random sample because it ensures each part of the lake is proportionally represented. When strata differ systematically, stratified sampling reduces bias.
分层均值是利用各层比例计算的加权平均值：μ = 0.20×45 + 0.50×32 + 0.30×38 = 9 + 16 + 11.4 = 36.4 mg/L。这种方法比简单随机抽样更具代表性，因为它确保了湖的每一部分都按比例得到体现。当各层存在系统性差异时，分层抽样可以减少偏差。
8. Sociology: Questionnaire Design and Bias | 社会学：问卷设计与偏差
A student designs a questionnaire to investigate how much time peers spend on social media. A question asks: “Do you agree that spending too many hours on social media is harmful and should be limited?” Identify the type of bias in this question and suggest an improved, neutral wording. Also discuss sampling bias if the survey is only given to friends.
一名学生设计了一份问卷，调查同龄人花在社交媒体上的时间。其中一个问题是：”你是否同意花太多时间在社交媒体上是有害的，并且应该被限制？”请指出这个问题中的偏差类型，并提出改进的中性措辞。如果该调查只发给朋友，讨论抽样偏差。
The question is leading because it assumes the respondent views social media as harmful and suggests a desired answer. A neutral revision could be: “On average, how many hours per day do you spend on social media?” followed by a scale. Additionally, only surveying friends introduces selection bias; the sample is not representative of the whole year group, as friends may share similar habits. To avoid this, a random sample or stratified sample from the entire year group should be used.
该问题具有引导性，因为它先入为主地认为社交媒体有害并暗示了期望的回答。一个中性的修改可以是：“你平均每天在社交媒体上花多少小时？”并辅以量表。此外，只调查朋友会引入选择偏差；样本不能代表整个年级组，因为朋友之间可能有相似的习惯。为避免这种情况，应从整个年级组中随机抽样或分层抽样。
9. Engineering: Scatter Graphs and Correlation for Material Strength | 工程：散点图与材料强度相关性
An engineering class tests the force needed to break plastic beams of different thicknesses. Data: thickness (mm): 2, 3, 4, 5, 6, 7, 8; breaking force (N): 40, 65, 85, 110, 130, 160, 185. Plot a scatter graph, describe the correlation, and draw a line of best fit. Estimate the breaking force for a thickness of 5.5 mm. Comment on the reliability of extrapolating for a thickness of 12 mm.
工程课上测试了不同厚度的塑料梁所需的断裂力。数据：厚度（mm）：2, 3, 4, 5, 6, 7, 8；断裂力（N）：40, 65, 85, 110, 130, 160, 185。绘制散点图，描述相关性，画出最佳拟合线。估算厚度为5.5 mm时的断裂力。评论外推至12 mm厚度的可靠性。
The points show a strong positive linear correlation. A line of best fit can be drawn passing roughly through (2,40) and (8,185). Gradient ≈ (185−40)/(8−2) = 145/6 ≈ 24.2 N/mm. Equation: F ≈ 24.2t − 8.4 (approx). For t=5.5 mm, F ≈ 24.2×5.5 − 8.4 ≈ 124.7 N. Extrapolation to 12 mm is unreliable because we have no data beyond 8 mm; the relationship might not remain linear, and the beam could break differently at greater thicknesses.
这些点显示出强正线性相关。最佳拟合线可大致通过(2,40)和(8,185)。斜率 ≈ (185−40)/(8−2) = 145/6 ≈ 24.2 N/mm。方程：F ≈ 24.2t − 8.4（近似）。当厚度 t=5.5 mm 时，F ≈ 24.2×5.5 − 8.4 ≈ 124.7 N。外推到12 mm是不可靠的，因为我们没有超过8 mm的数据；这种关系可能不再保持线性，梁在更大厚度时可能会以不同方式断裂。
10. Integrated Project: Multi-disciplinary Data Report | 综合项目：多学科数据报告
Your task is to design a mini-research project combining geography and economics: investigate the relationship between the distance from the city centre and the price of a cup of coffee in cafes. Outline the data collection plan using systematic sampling, present hypothetical data in a two-way table, calculate the mean price for three distance zones, and draw comparative bar charts. What conclusion can you draw about the ‘distance-price’ relationship?
你的任务是设计一个结合地理和经济学的迷你研究项目：调查咖啡馆离市中心的距离与咖啡价格之间的关系。概述使用系统抽样的数据收集计划，用双向表呈现假设数据，计算三个距离区的平均价格，并绘制比较条形图。关于“距离-价格”关系，你能得出什么结论？
Plan: list all cafes along a main road radiating from the centre, select every 3rd cafe. Record distance (km) and price of a standard latte. Hypothetical data for three zones: 0−1 km (inner): £3.20, £3.50, £3.40, £3.30 (mean £3.35); 1−3 km (middle): £3.00, £2.90, £3.10, £2.80 (mean £2.95); 3−5 km (outer): £2.50, £2.70, £2.60, £2.80 (mean £2.65). The comparative bar chart of means shows a clear decreasing trend as distance increases. This suggests a negative correlation: cafes closer to the centre tend to charge more, likely due to higher rent and demand.
计划：列出从市中心辐射出的主路上的所有咖啡馆，选择每隔两家的咖啡馆（系统抽样）。记录距离（公里）和标准拿铁的价格。三个区域的假设数据：0−1 km（内区）：£3.20, £3.50, £3.40, £3.30（均值 £3.35）；1−3 km（中区）：£3.00, £2.90, £3.10, £2.80（均值 £2.95）；3−5 km（外区）：£2.50, £2.70, £2.60, £2.80（均值 £2.65）。均值的比较条形图显示，随着距离增加，价格有明显的下降趋势。这表明存在负相关：离市中心较近的咖啡馆往往收费更高，这可能是因为更高的租金和需求。
From this integrated project, you practise sampling methods, measures of central tendency, data presentation, and drawing conclusions from real-world patterns.
通过这个综合项目，你练习了抽样方法、集中趋势的度量、数据呈现以及从实际模式中得出结论。
Published by TutorHao | Statistics Revision Series | aleveler.com
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