📚 Year 10 Cambridge Statistics: Interdisciplinary Exam-Style Practice | 剑桥十年级统计：跨学科综合题型训练
Statistics is not just a collection of isolated formulas; it is the language of data that connects biology, business, social science, finance and physics. This article presents a set of exam‑style, cross‑curricular problems tailored for Year 10 Cambridge Statistics. Each section blends real‑world context with core statistical techniques, helping you build fluency in interpreting data, calculating probabilities and drawing valid conclusions – exactly the skills assessed in IGCSE and O Level examinations.
统计不仅仅是孤立的公式合集，它是连接生物学、商业、社会科学、金融和物理学的数据语言。本文为剑桥十年级统计量身定制了一套跨学科考试题型训练。每个章节都将真实情境与核心统计技术融合，帮助你在解读数据、计算概率和得出有效结论方面形成扎实的能力——这些正是 IGCSE 与 O Level 考试评估的关键技能。
1. Descriptive Statistics in Biology: Mean, Median and Mode from Raw Data | 生物学中的描述统计：原始数据的均值、中位数与众数
A biology student measured the resting heart rates (beats per minute) of eight classmates: 72, 68, 75, 70, 69, 74, 68, 71. Find the mean, median and mode of this sample. Comment on which measure best represents the typical heart rate.
一名生物专业的学生测量了八位同学静息心率（次/分）：72, 68, 75, 70, 69, 74, 68, 71。计算样本的均值、中位数和众数，并评价哪一个度量最能代表典型心率。
The mean is calculated by summing all values and dividing by n = 8. Sum = 72+68+75+70+69+74+68+71 = 567. Hence x̄ = 567/8 = 70.875 bpm. To find the median, order the data: 68, 68, 69, 70, 71, 72, 74, 75. With an even number of observations, the median is the average of the 4th and 5th values: (70 + 71)/2 = 70.5 bpm. The mode is 68, which appears twice. Since the data are fairly symmetric and no extreme outliers exist, the mean is a reliable measure of central tendency. However, the median is close to the mean and is less affected by any possible recording error, making both suitable for a biological report.
均值通过求和除以 n=8 得到。总和 = 567，因此 x̄ = 567/8 = 70.875 次/分。将数据排序：68, 68, 69, 70, 71, 72, 74, 75。由于观测个数为偶数，中位数为第4和第5个值的平均：(70+71)/2 = 70.5 次/分。众数为68，出现两次。数据分布较为对称且无明显异常值，因此均值是可靠的集中趋势度量。中位数与均值接近且不易受可能的记录误差影响，两者均适用于生物学报告。
2. Grouped Frequency Tables and Histograms in Social Studies | 社会科学中的分组频率表与直方图
A school surveyed 40 students about daily screen time (hours). The results were grouped as follows: 0 ≤ t < 1: 4 students; 1 ≤ t < 2: 10; 2 ≤ t < 3: 12; 3 ≤ t < 4: 8; 4 ≤ t < 5: 4; 5 ≤ t < 6: 2. Construct a grouped frequency table and draw a histogram. Estimate the percentage of students who spend less than 3 hours on screens.
某学校对40名学生每日屏幕时间（小时）进行了调查。结果分组为：0 ≤ t < 1: 4人；1 ≤ t < 2: 10人；2 ≤ t < 3: 12人；3 ≤ t < 4: 8人；4 ≤ t < 5: 4人；5 ≤ t < 6: 2人。构建分组频率表并绘制直方图。估算每天屏幕时间少于3小时的学生比例。
| Screen time, t (hours) | Frequency |
|---|---|
| 0 ≤ t < 1 | 4 |
| 1 ≤ t < 2 | 10 |
| 2 ≤ t < 3 | 12 |
| 3 ≤ t < 4 | 8 |
| 4 ≤ t < 5 | 4 |
| 5 ≤ t < 6 | 2 |
In a histogram, the area of each bar represents frequency. Since all class intervals are equal (width 1 hour), the height is directly proportional to frequency. To estimate the percentage with screen time less than 3 hours, add the frequencies for the first three classes: 4 + 10 + 12 = 26 students. The percentage is (26/40) × 100% = 65%. Thus, 65% of surveyed students spend under 3 hours daily on screens, a finding that could inform school digital wellness policies.
在直方图中，每个矩形的面积代表频率。由于所有组距宽度均为1小时，矩形高度直接与频率成正比。为估算屏幕时间少于3小时的比例，将前三组的频数相加：4+10+12=26人。百分比为(26/40) × 100% = 65%。因此，65%的被调查学生每日屏幕时间不足3小时，该发现可为学校的数字健康政策提供参考。
3. Cumulative Frequency and Quartiles in Environmental Science | 环境科学中的累积频率与四分位数
An environmental scientist recorded the daily rainfall (mm) in a region over 30 days. The cumulative frequency graph shows that the 30th percentile is 2.4 mm, the lower quartile is 3.8 mm, the median is 6.5 mm, and the upper quartile is 9.2 mm. Interpret the quartiles and calculate the interquartile range. What does the IQR reveal about rainfall variability?
一名环境科学家记录了某地区30天的日降雨量（毫米）。累积频率图显示第30百分位数为2.4 mm，下四分位数为3.8 mm，中位数为6.5 mm，上四分位数为9.2 mm。解读这些四分位数并计算四分位距。IQR 揭示了降雨量的什么变化特征？
The lower quartile (Q₁ = 3.8 mm) indicates that 25% of the days had rainfall at or below 3.8 mm. The median (Q₂ = 6.5 mm) shows that half the days recorded 6.5 mm or less. The upper quartile (Q₃ = 9.2 mm) tells us that 75% of days received at most 9.2 mm of rain. The interquartile range IQR = Q₃ – Q₁ = 9.2 – 3.8 = 5.4 mm. The IQR measures the spread of the middle 50% of the data. A moderate IQR relative to the median suggests that most daily rainfall amounts cluster between 3.8 mm and 9.2 mm, although extreme dry or wet days could still occur outside this interval.
下四分位数（Q₁ = 3.8 mm）表明有25%的日子降雨量≤3.8 mm。中位数（Q₂ = 6.5 mm）意味着半数日子降雨量不超过6.5 mm。上四分位数（Q₃ = 9.2 mm）告知75%的日子降雨量最高为9.2 mm。四分位距 IQR = Q₃ – Q₁ = 9.2 – 3.8 = 5.4 mm。IQR 衡量中间50%数据的离散程度。与中位数相比，适中的 IQR 说明大多数日降雨量集中在3.8 mm至9.2 mm之间，但该区间外仍可能出现极端干燥或暴雨天气。
4. Interpreting Bar Charts and Pie Charts in Business | 商业中的条形图和饼图解读
A small bakery recorded quarterly sales: Q1: £12,000; Q2: £15,000; Q3: £18,000; Q4: £25,000. Present this data on a bar chart and construct a pie chart, calculating the angle for each quarter. Identify which quarter contributed most to the annual revenue.
一家小面包店记录了季度销售额：第一季度 £12,000；第二季度 £15,000；第三季度 £18,000；第四季度 £25,000。用条形图呈现数据并绘制饼图，计算每个季度的圆心角。指出哪个季度对年收入贡献最大。
The total annual revenue is £12,000 + £15,000 + £18,000 + £25,000 = £70,000. For a pie chart, each sector angle = (Quarter sales / Total) × 360°. Angles: Q1 = (12,000/70,000)×360° ≈ 61.7°; Q2 = (15,000/70,000)×360° ≈ 77.1°; Q3 = (18,000/70,000)×360° ≈ 92.6°; Q4 = (25,000/70,000)×360° ≈ 128.6°. The bar chart uses labelled axes with Sales (£) on the vertical axis and Quarter on the horizontal axis, drawing bars of corresponding heights. Q4, with 25,000, is clearly the largest contributor, accounting for more than a third of yearly sales, likely driven by holiday demand.
年总收入为 £12,000 + £15,000 + £18,000 + £25,000 = £70,000。饼图中，每个扇形的圆心角 = (季度销售额/总销售额) × 360°。各角度：Q1 ≈ 61.7°，Q2 ≈ 77.1°，Q3 ≈ 92.6°，Q4 ≈ 128.6°。条形图以销售额（£）为纵轴、季度为横轴，绘制对应高度的条形。第四季度以25,000英镑遥遥领先，贡献了年度销售额的三分之一以上，很可能是受节假日需求推动。
5. Probability and Tree Diagrams in Genetics | 遗传学中的概率与树状图
In pea plants, the allele for tall stems (T) is dominant over short stems (t). Two heterozygous tall plants (Tt) are crossed. Draw a tree diagram to model the possible genotypes of the offspring and calculate the probability that an offspring is short. Hence find the expected number of short plants in a sample of 200 offspring.
在豌豆植株中，高茎等位基因（T）对矮茎等位基因（t）为显性。两株杂合高茎植株（Tt）杂交。画出树状图展示子代可能的基因型，并计算子代为矮茎的概率。进而求出在200株子代中矮茎植株的期望数量。
Each parent can pass on either T or t with equal probability (½). The tree diagram has two branches from the first parent, each then splitting for the second. The possible offspring genotypes are: TT (½ × ½ = ¼), Tt (½ × ½ = ¼ from T then t, plus ½ × ½ = ¼ from t then T, total ½), and tt (½ × ½ = ¼). A plant is short only with genotype tt. So P(short) = ¼. With 200 offspring, the expected number of short plants is 200 × ¼ = 50. This prediction aligns with Mendelian ratios, demonstrating how probability supports experimental genetics.
每个亲本以等概率（½）传递 T 或 t。树状图先显示第一个亲本的两个分支，每支再对应第二个亲本的两种可能。子代可能的基因型为：TT（½ × ½ = ¼）、Tt（½ × ½ 两次，共 ½）和 tt（¼）。只有基因型为 tt 的植株表型为矮茎。因此 P(矮茎) = ¼。若子代共200株，矮茎期望数量为 200 × ¼ = 50。该预测符合孟德尔比例，表明概率如何支撑实验遗传学。
6. Scatter Graphs and Correlation in Sports Science | 体育科学中的散点图与相关性
A coach recorded the weekly training hours (x) and the competition score (y) for ten athletes. The data pairs: (4, 62), (6, 68), (5, 65), (8, 78), (7, 72), (3, 58), (9, 82), (6, 70), (7, 75), (5, 63). Plot a scatter graph, describe the correlation and suggest a line of best fit. Use your line to predict the score for an athlete training 7.5 hours per week.
一位教练记录10名运动员的每周训练时数（x）和比赛得分（y）。数据对为：(4,62), (6,68), (5,65), (8,78), (7,72), (3,58), (9,82), (6,70), (7,75), (5,63)。绘制散点图，描述相关性并画出最佳拟合线。利用该线预测一名每周训练7.5小时的运动员的比赛得分。
The scatter graph shows a clear upward trend: as training hours increase, competition scores tend to rise. The correlation is strong positive. A line of best fit can be drawn passing roughly through the centroid of the points, with equation estimated as y ≈ 4x + 46 (where y is score and x is hours). For x = 7.5, predicted score ≈ 4(7.5) + 46 = 76. This is interpolation since 7.5 lies within the data range. Coaches should note that correlation does not imply causation, but the relationship supports designing training plans based on desired performance levels.
散点图显示明显的上升趋势：训练时数增加，比赛得分趋于提升，相关性为强正相关。最佳拟合线可大致穿过数据点重心，其方程估算为 y ≈ 4x + 46（y为得分，x为小时）。当 x = 7.5 时，预测得分 ≈ 4×7.5 + 46 = 76。7.5小时位于数据范围内，因此为内插预测。教练须注意相关性并不意味着因果性，但这种关系有助于根据目标表现设计训练计划。
7. Probability Distributions and Expected Value in Finance | 金融中的概率分布与期望值
An investor evaluates a start-up project with three possible outcomes: Profit £50,000 with probability 0.3; Profit £10,000 with probability 0.5; Loss £20,000 with probability 0.2. Create a probability distribution table and calculate the expected profit. Should the investor proceed if the minimum acceptable expected profit is £15,000?
一位投资者评估一个初创项目，有三种可能结果：获利50,000英镑（概率0.3）；获利10,000英镑（概率0.5）；亏损20,000英镑（概率0.2）。制作概率分布表并计算期望利润。若投资者可接受的最低期望利润为15,000英镑，应否继续投资？
| Outcome (£) | Probability, P(X=x) |
|---|---|
| 50,000 | 0.3 |
| 10,000 | 0.5 |
| -20,000 | 0.2 |
Expected profit E(X) = Σ[xᵢ · P(X=xᵢ)] = (50,000 × 0.3) + (10,000 × 0.5) + (–20,000 × 0.2) = 15,000 + 5,000 – 4,000 = £16,000. Since the expected profit (£16,000) exceeds the threshold of £15,000, the project meets the investment criterion. However, the investor must still consider risk; the 20% chance of a £20,000 loss may be unacceptable if capital is limited. Expected value provides a rational starting point for financial decisions.
期望利润 E(X) = Σ[xᵢ · P(X=xᵢ)] = (50,000×0.3) + (10,000×0.5) + (–20,000×0.2) = 15,000 + 5,000 – 4,000 = £16,000。由于期望利润（16,000英镑）高于门槛值15,000英镑，项目符合投资标准。但投资者仍需考虑风险：20%的亏损概率可能因资本有限而无法接受。期望值为金融决策提供了理性出发点。
8. Sampling Methods and Bias in Market Research | 市场调研中的抽样方法与偏差
A marketing team wants to estimate the average monthly spending of teenagers on streaming services. They plan to collect data by posting a survey link on a gaming forum. Identify the sampling method, describe the potential bias, and suggest a more representative sampling strategy.
一个营销团队希望估计青少年在流媒体服务上的月均支出。他们计划在某游戏论坛发布问卷链接。识别此抽样方法，描述潜在的偏差，并提出一种更具代表性的抽样策略。
The method described is a voluntary response sample, where individuals self-select to participate. It is likely to produce bias because forum users are not representative of all teenagers: they may be more tech-oriented, have different spending habits, and those with strong opinions might be more likely to respond. A better approach is stratified random sampling: divide the teenage population into strata based on age, gender and region, then randomly select participants from each group in proportion to their size. This ensures that diverse subgroups are included and the sample more accurately reflects the target population.
所描述的方法是自愿响应样本，个体自行选择参与。这很可能产生偏差，因为论坛用户并不能代表所有青少年：他们可能更偏向技术型，消费习惯不同，且观点强烈者更倾向参与。更好的方法是分层随机抽样：将青少年总体按年龄、性别和地区分层，然后按比例从每层随机选取参与者。这能确保纳入多元子群体，样本更准确地反映目标总体。
9. Interquartile Range and Outliers in Chemistry Experiment Data | 化学实验数据中的四分位距与异常值
A student repeated a titration six times and obtained volumes (cm³): 24.8, 25.2, 24.9, 27.1, 25.0, 24.9. Determine the median, quartiles and IQR. Using the 1.5 × IQR rule, decide if 27.1 cm³ is an outlier. What should the student do with outliers in a lab report?
一名学生重复滴定实验六次，得到体积（cm³）：24.8, 25.2, 24.9, 27.1, 25.0, 24.9。确定中位数、四分位数和 IQR。利用 1.5 × IQR 法则判断 27.1 cm³ 是否为异常值。学生在实验报告中应如何处理异常值？
Order the data: 24.8, 24.9, 24.9, 25.0, 25.2, 27.1. Median (Q₂) = average of 3rd and 4th: (24.9 + 25.0)/2 = 24.95 cm³. Lower quartile Q₁ = median of lower half (24.8, 24.9, 24.9) = 24.9 cm³. Upper quartile Q₃ = median of upper half (25.0, 25.2, 27.1) = 25.2 cm³. IQR = 25.2 – 24.9 = 0.3 cm³. Outlier boundaries: lower fence = Q₁ – 1.5×IQR = 24.9 – 0.45 = 24.45; upper fence = Q₃ + 1.5×IQR = 25.2 + 0.45 = 25.65. Any value outside [24.45, 25.65] is an outlier. 27.1 > 25.65, so it is an outlier. In a lab report, the student should check for procedural errors, record the anomaly, and possibly repeat the experiment. The outlier should not be automatically discarded; instead, present the data with and without the outlier for transparency.
排序数据：24.8, 24.9, 24.9, 25.0, 25.2, 27.1。中位数 Q₂ = 24.95 cm³。下四分位数 Q₁ = 24.9 cm³；上四分位数 Q₃ = 25.2 cm³。IQR = 0.3 cm³。异常值界限：下界 = Q₁ – 1.5×IQR = 24.9 – 0.45 = 24.45；上界 = Q₃ + 1.5×IQR = 25.2 + 0.45 = 25.65。27.1 > 25.65，因此为异常值。实验报告中学生应检查操作错误，记录异常，必要时重做实验。不应随意丢弃异常值，而应同时展示包含和剔除该值的数据，以保持透明度。
10. Real‑life Integrated Problem: Travel Survey Analysis | 真实情境综合题：旅行调查分析
The local council surveyed 200 commuters about their daily travel time (minutes). Summary statistics: mean 34.2 min, median 28 min, range 90 min, standard deviation 12.5 min. A histogram of travel times is positively skewed. The council also collected data on mode of transport: bus (80 people), car (70), bicycle (30), walk (20). (a) Explain why the median is lower than the mean. (b) Calculate the percentage who use bus and represent it in a pie chart. (c) Discuss a possible reason for the high variability in travel times and suggest a follow‑up stratified sample to investigate congestion.
地方议会调查了200名通勤者的每日通勤时间（分钟）。汇总统计：均值34.2分钟，中位数28分钟，极差90分钟，标准差12.5分钟。旅行时间直方图呈正偏态。议会还收集了交通方式数据：公交车（80人），私家车（70人），自行车（30人），步行（20人）。(a) 解释为什么中位数低于均值。(b) 计算使用公交车的百分比并在饼图中表示。(c) 讨论旅行时间高变异性的一个可能原因，并建议一个分层抽样后续调查以研究拥堵情况。
(a) The positively skewed distribution means a few commuters have extremely long travel times, pulling the mean upward, while the median resists extreme values. Thus the median (28 min) is lower than the mean (34.2 min). (b) Percentage using bus = (80/200)×100% = 40%. In a pie chart, the bus sector angle = 40% × 360° = 144°. Similarly, car: 126°, bicycle: 54°, walk: 36°. (c) High variability may arise because the sample includes both inner‑city residents with short commutes and suburban commuters traveling far. A follow‑up study could use stratified sampling by residential zone (urban, suburban, rural) to compare average travel times and identify where congestion interventions are most needed. This integrated analysis illustrates how descriptive statistics, graphical representation and sampling design combine to
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