📚 Interdisciplinary Mixed Question Practice for IGCSE OCR Statistics | IGCSE OCR 统计跨学科综合题型训练
In the IGCSE OCR Statistics examination, questions are rarely presented in a purely mathematical vacuum. They are embedded in real-world contexts drawn from biology, economics, geography, physics, and other subjects. This article provides a structured training set of interdisciplinary problems, each targeting a specific statistical technique while linking to another field. Work through these examples to build fluency in selecting appropriate methods, interpreting outputs, and writing clear conclusions—exactly what examiners look for in high-scoring responses.
在 IGCSE OCR 统计考试中，题目很少以纯粹的数学形式出现，它们同时嵌入在来自生物、经济、地理、物理等学科的真实情境中。本文提供一套有组织的跨学科综合题型训练，每道题都针对特定的统计方法并与另一领域挂钩。通过这些例题，你可以熟练地选择合适的方法、解释输出并写出清晰的结论，而这正是阅卷人所期望的高分答案要素。
1. Biology: Comparing Two Fertilisers | 生物学：比较两种肥料
A botanist measured the height (cm) of 12 bean plants grown with Fertiliser A and 12 with Fertiliser B. The summary statistics are shown below. She wishes to determine whether there is a significant difference in mean heights. By performing a two-sample t-test assuming equal variances, you can link biological experimentation with hypothesis testing.
一位植物学家测量了施用肥料 A 和肥料 B 的豆类植株高度（cm），各 12 株。以下为汇总统计量。她想判断平均高度是否存在显著差异。通过进行等方差双样本 t 检验，你可以将生物学实验与假设检验联系起来。
| Fertiliser | Mean (x̄) | Standard deviation (s) |
|---|---|---|
| A | 14.8 | 2.1 |
| B | 16.4 | 2.3 |
≠ μₙ. The pooled variance s²ₚ = ((11 × 2.1² + 11 × 2.3²)/(22)) ≈ 4.85, so standard error = √s²ₚ/12 + s²ₚ/12 ≈ √0.404 + 0.404 istic t = (14.8 - 16.4)/0.899 ≈ -1.78. With 22 degrees of freedom, the critical value at 5% two-tailed is 2.074. Since |–1.78| t, there is insufficient evidence to claim the fertilisers produce different mean heights.
给出原假设 H₀：μₐ = μₙ，备择假设 H₁：μₐ ≠ μₙ。合并方差 s²ₚ ≈ 4.85，标准误 ≈ 0.899。检验统计量 t ≈ –1.78。自由度为 22，5% 双尾临界值为 2.074。由于 |–1.78| < 2.074，我们不拒绝 H₀。在生物学情境中，没有充分证据表明两种肥料产生了不同的平均高度。
2. Economics: Index Numbers and Time Series | 经济学：指数数列与时间序列
The quarterly price index for a basket of consumer goods is given for three years. An economist wants to smooth the series using a four-point centred moving average and identify the underlying trend. This combines economic data handling with moving averages and seasonal decomposition—key topics in OCR Statistics.
给定三年内一篮子消费品的季度价格指数。一位经济学家想用四项中心移动平均平滑该序列并识别潜在趋势。这结合了经济数据处理与移动平均和季节分解——这是 OCR 统计中的关键主题。
| Quarter | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Index | 112 | 108 | 115 | 120 | 118 | 110 | 122 | 125 |
To compute the four-point centred moving average, first average quarters 1–4: (112+108+115+120)/4 = 113.75; then for quarters 2– otting the raw data and the trend line reveals a general upward movement despite short-term fluctuations. In economic reporting, such smoothed trends help policymakers distinguish noise from real growth.
为计算四项中心移动平均，首先对季度 1–4 求平均：(112+108+115+120)/4 = 113.75；然后对季度 2–5：(108+115+120+118)/4 = 115.25。季度 3 的中心值为两者的平均值：(113.75+115.25)/2 = 114.5。继续此过程。将原始数据与趋势线绘图，尽管有短期波动，仍显示出总体上升趋势。在经济报告中，此类平滑趋势有助于决策者区分噪声与真实增长。
3. Geography: Cumulative Frequency and Median Estimation | 地理学：累积频率与中位数估计
A geographer records the annual rainfall (mm) across 60 locations. The grouped frequency table is used to draw a cumulative frequency curve, from which the median and interquartile range are estimated. This task mirrors many OCR-style questions where descriptive statistics are applied to environmental data.
一位地理学家记录了 60 个地区的年降雨量（毫米）。使用分组频数表绘制累积频率曲线，由此估计中位数和四分位间距。这一任务与许多 OCR 风格的题目类似，将描述性统计应用于环境数据。
| Rainfall (mm) | Frequency |
|---|---|
| 0 ≤ x < 200 | 8 |
| 200 ≤ x < 400 | 14 |
| 400 ≤ x < 600 | 20 |
| 600 ≤ x < 800 | 12 |
| 800 ≤ x < 1000 | 6 |
Construct a cumulative frequency column: 8, 22, 42, 54, 60. On graph paper, plot upper boundaries (200, 400, 600, 800, 1000) against cumulative frequencies. The median corresponds to the 30th value; read from the curve—approximately 470 mm. The lower quartile (15th value) ~ 310 mm, upper quartile (45th value) ~ 650 mm, giving an IQR of 340 mm. Geographers use such measures to summarise central tendency and variability in climate data.
构建累积频数列：8, 22, 42, 54, 60。在坐标纸上，以上限值（200, 400, 600, 800, 1000）对累积频数绘图。中位数对应于第 30 个值，从曲线上读出——约为 470 mm。下四分位数（第 15 个值）~ 310 mm，上四分位数（第 45 个值）~ 650 mm，得出四分位间距为 340 mm。地理学家利用这些指标概括气候数据的集中趋势和变异性。
4. Physics: Scatter Graphs and Regression for Springs | 物理学：弹簧的散点图与回归
In a physics lab, students hang masses on a spring and measure the extension. The data are used to construct a scatter plot, calculate the equation of the regression line, and assess the strength of the linear relationship. This is a classic OCR problem linking experimental physics and bivariate data analysis.
在物理实验室中，学生们在弹簧上悬挂不同质量并测量伸长量。这些数据用于绘制散点图、计算回归线方程并评估线性关系的强度。这是一个经典 OCR 问题，将实验物理与双变量数据分析联系起来。
Mass (m/kg): 0.1, 0.2, 0.3, 0.4, 0.5
Extension (e/cm): 2.1, 4.3, 6.0, 7.9, 9.8
15)/(2.75 - 2.25) = 8.5/0.50 = 17.0. Intercept a = (30.1 - 17.0×1.5)/5 = (30.1 - 25.5)/5 = 0.92. Equation: e = 0.92 + 17.0m. The product-moment correlation coefficient r can be calculated as 0.999, indicating an almost perfect positive correlation, consistent with Hooke' of 0.45 kg: e ≈ 0.92 + 17.0×0.45 = 8.57 cm, and discuss the reliability of extrapolation.
使用公式，Σm = 1.5，Σe = 30.1，Σm² = 0.55，Σme = 10.73，n = 5。回归系数 b = (5×10.73 - 1.5×30.1)/(5×0.55 - 1.5²) = 8.5/0.50 = 17.0。截距 a = 0.92。方程：e = 0.92 + 17.0m。积矩相关系数 r 可计算为 0.999，表明几乎完全正相关，与胡克定律一致。然后学生可以估计 0.45 kg 质量时的伸长量：e ≈ 8.57 cm，并讨论外推的可靠性。
5. Sports Science: Normal Distribution and Qualification | 体育科学：正态分布与资格判定
An athletics coach knows that the 100 m sprint times of her squad follow a normal distribution with mean 11.8 s and standard deviation 0.5 s. To qualify for a regional meet, athletes must run under 11.2 s. By standardising, we can determine the probability a randomly chosen squad member qualifies, blending sports analytics with normal probability calculations.
一位田径教练知道其队员的 100 米短跑时间服从均值为 11.8 秒、标准差为 0.5 秒的正态分布。要获得区域赛资格，运动员必须跑进 11.2 秒以内。通过标准化，我们可以确定随机选择一名队员达标的概率，将体育分析与正态概率计算相结合。
Z = (X - μ)/σ = (11.2 - 11.8)/0.5 = -1.20
tables, P(Z < -1.20) = 1 - P(Z < 1.20) ≈ 1 - 0.8849 = 0.1151. Hence there is about an 11.5% ed. P(X ≥ 2) = 1 - P(0) - P(1). P(0) = 0.8849²⁰ ≈ 0.090, P(1) = 20 × 0.1151 × 0.8849¹⁹ ≈ 0.234, so probability ≈ 0.676. This interdisciplinary approach demonstrates how normal and binomial models interact in personnel selection.
查标准正态表，P(Z < -1.20) ≈ 0.1151。因此运动员达标概率约为 11.5%。如果教练需要从 20 次独立试验中至少 2 人达标，可使用二项分布 B(20, 0.1151)。P(X ≥ 2) = 1 - P(0) - P(1) ≈ 0.676。此跨学科方法展示了正态与二项模型如何在人员选拔中相互作用。
6. Business: Stratified Sampling and Confidence Intervals | 商业：分层抽样与置信区间
A marketing firm surveys customer satisfaction across three product lines. The population sizes are: Basic (2500 users), Premium (1500), and Elite (1000). A stratified sample of 500 customers is to be selected. Calculate the sample sizes per stratum and then, given a sample proportion of satisfied customers, construct a 95% confidence interval for the true proportion in the Premium line. This exemplifies business decisions relying on proper sampling theory.
一家市场调研公司对三个产品线的顾客满意度进行调查。总体规模为：基础款 2500 人，高级款 1500 人，精英款 1000 人。按比例抽取 500 名顾客的分层样本。计算每层的抽样人数，然后根据高级款中满意顾客的样本比例，构建真实比例的 95% 置信区间。这展示了依赖正确抽样理论的商业决策。
1 = 250; Premium 150; Elite 50. In the Premium sample of 150, suppose 108 are satisfied. p̂ = 108/150 = 0.72. Standard error = √[p̂(1-p̂)/n] = √[0.72×0.28/150] ≈ 0.0367. For 95% nfidence, Z = 1.96. Interval: 0.72 ± 1.96×0.0367 = (0.648, 0.792). Thus the firm can be 95% confident that between 64.8% and 79.2% of all Premium users are satisfied, enabling targeted improvement strategies.
总人口 = 5000，抽样比 = 0.1。各层样本数：基础 250，高级 150，精英 50。假设 150 名高级用户样本中有 108 人满意，p̂ = 0.72，标准误 ≈ 0.0367，95% 置信区间：0.72 ± 1.96×0.0367 = (0.648, 0.792)。因此公司可以有 95% 的把握认为所有高级用户满意度在 64.8% 到 79.2% 之间，从而制定有针对性的改进策略。
7. Health Sciences: Chi-Squared Test of Independence | 健康科学：卡方独立性检验
Health researchers investigate whether smoking status is independent of a lung capacity index (low/high). The observed frequencies are tabulated. A chi-squared test at the 5% level reveals if there is a significant association, linking epidemiological studies with categorical data analysis.
健康研究人员调查吸烟状况是否与肺活量指数（低/高）独立。将观测频数列入联表。在 5% 显著性水平下的卡方检验可揭示是否存在显著关联，将流行病学研究与分类数据分析联系起来。
| Low capacity | High capacity | Total | |
|---|---|---|---|
| Smoker | 45 | 30 | 75 |
| Non-smoker | 28 | 47 | 75 |
| Total | 73 | 77 | 150 |
ic χ² = Σ (O - E)²/E = (45-36.5)²/36.5 + (30-38.5)²/38.5 + (28-36.5)²/36.5 + (47-38.5)²/38.5 ≈ 1.98 + 1.88 + 1.98 + 1.88 = 7.72. Degrees of freedom = (2-1)(2-1)=1; critical value at 5% city index. This trains students to interpret such results for health policy implications.
期望频数：吸烟/低 = 36.5 等。χ² = Σ (O - E)²/E ≈ 7.72。自由度 1，5% 临界值 3.841。由于 7.72 > 3.841，拒绝原假设。有证据表明吸烟与肺活量指数之间存在关联。这训练学生解读此类结果对卫生政策的意义。
8. Environmental Science: Histograms and Frequency Density | 环境科学：直方图与频率密度
An environmental agency records the concentration of a pollutant (µg/m³) in 80 air samples. The data are grouped into unequal class widths. Constructing a histogram with frequency density on the vertical axis allows accurate visual representation, a skill OCR examiners prioritise when testing data presentation in scientific contexts.
一家环保机构记录了 80 个空气样本中某污染物的浓度 (µg/m³)。数据按不等组距分组。通过以频率密度为纵轴绘制直方图，可以实现精确的可视化呈现，这是 OCR 考官在测试科学背景下数据呈现时优先考查的技能。
| Concentration (µg/m³) | Frequency | Class width | Frequency density |
|---|---|---|---|
| 0 - 10 | 5 | 10 | 0.5 |
| 10 - 30 | 18 | 20 | 0.9 |
| 30 - 50 | 30 | 20 | 1.5 |
| 50 - 80 | 22 | 30 | 0.73 |
| 80 - 100 | 5 | 20 | 0.25 |
s width. Drawing bars with heights 0.5, 0.9, 1.5, 0.73, 0.25 over the respective intervals gives a histogram where the area of each bar is proportional to frequency. The distribution appears positively skewed, indicating most pollution readings are low but occasional spikes occur. Environmental scientists can use this to set emission thresholds or issue health alerts when concentrations exceed 80 µg/m³.
频率密度 = 频数 / 组距。绘制高度分别为 0.5、0.9、1.5、0.73、0.25 的柱状图，每个直方的面积与频数成正比。分布呈正偏态，表明大多数污染读数较低但偶尔有高值出现。环境科学家可利用这些信息设定排放阈值或在浓度超过 80 µg/m³ 时发布健康警报。
9. Psychology: Sign Test for Paired Data | 心理学：配对数据的符号检验
A psychologist measures reaction times (ms) before and after a stress-reduction course for 10 participants. To test whether the course reduces reaction time, a sign test is appropriate for this paired-design experiment. This is a non-parametric method explicitly required by the OCR specification.
一位心理学家测量了 10 名参与者在减压课程前后的反应时间（毫秒）。为了检验课程是否缩短了反应时间，符号检验适用于这一配对设计实验。这是 OCR 规格明确要求的一种非参数方法。
| Participant | Before | After | Sign (After - Before) |
|---|---|---|---|
| 1 | 320 | 305 | - |
| 2 | 290 | 288 | - |
| 3 | 350 | 340 | - |
| 4 | 280 | 282 | + |
| 5 | 310 | 300 | - |
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