📚 Cross-disciplinary Integrated Practice for WJEC GCSE Statistics | GCSE WJEC 统计跨学科综合题型训练
GCSE Statistics is never just about numbers – it is about making sense of the world. The WJEC specification regularly tests your ability to apply statistical methods in real-life, cross-disciplinary contexts. From biology to business, geography to sport, this article walks you through a range of integrated practice questions that will sharpen your skills and boost your exam confidence.
GCSE 统计远不止是数字——它关乎理解世界。WJEC 大纲经常考查你在真实的跨学科情境中运用统计方法的能力。从生物到商业、从地理到体育，这篇文章将带你练习一系列综合题型，帮助你磨练技能、提升考试信心。
1. Biology & Box Plots | 生物与箱线图
A botanist measures the heights of 15 seedlings (cm) after 4 weeks under two different light conditions. The data are: Low light: 12, 14, 15, 13, 16, 11, 15, 14, 13, 15, 14, 12, 13, 15, 14. High light: 18, 20, 19, 22, 21, 19, 18, 20, 22, 21, 20, 19, 18, 22, 21. Use box plots to compare the distributions.
一位植物学家测量了两种光照条件下 15 株幼苗 4 周后的高度（厘米）。数据如下：低光照：12, 14, 15, 13, 16, 11, 15, 14, 13, 15, 14, 12, 13, 15, 14。高光照：18, 20, 19, 22, 21, 19, 18, 20, 22, 21, 20, 19, 18, 22, 21。请使用箱线图比较分布。
For Low light, order the data: 11, 12, 12, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16. Median (8th value) = 14. Lower quartile Q₁ (4th value) = 13. Upper quartile Q₃ (12th value) = 15. Minimum = 11, maximum = 16. Interquartile range (IQR) = 15 – 13 = 2.
对于低光照组，排序：11, 12, 12, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16。中位数（第 8 个值）= 14。下四分位数 Q₁（第 4 个值）= 13。上四分位数 Q₃（第 12 个值）= 15。最小值 = 11，最大值 = 16。四分位距 IQR = 15 – 13 = 2。
For High light, ordered: 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22. Median = 20. Q₁ = 19. Q₃ = 21. Minimum = 18, maximum = 22. IQR = 2.
高光照组排序：18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22。中位数 = 20。Q₁ = 19。Q₃ = 21。最小值 = 18，最大值 = 22。IQR = 2。
The box plots show that high light produces consistently taller seedlings, with both medians and quartiles shifted up by about 5–6 cm. The spreads (IQRs) are identical, but the whole distribution is translated upwards.
箱线图显示高光照条件下幼苗普遍更高，中位数和四分位数均上移约 5–6 厘米。离散程度（IQR）相同，但整个分布向上平移。
Comparison: Median(Low) = 14 cm, Median(High) = 20 cm; IQR both = 2 cm.
2. Geography & Population Pyramids | 地理与人口金字塔
A geography student collects age–gender data for a small town. Males: 0–14: 1200, 15–64: 3400, 65+: 900. Females: 0–14: 1150, 15–64: 3600, 65+: 1100. Construct a population pyramid and comment on the dependency ratio.
一名地理专业的学生收集了某小镇的年龄–性别数据。男性：0–14 岁：1200，15–64 岁：3400，65 岁以上：900。女性：0–14 岁：1150，15–64 岁：3600，65 岁以上：1100。绘制人口金字塔并评论抚养比。
To draw the pyramid, use horizontal bars: left for males, right for females. Age groups on the vertical axis. The bar lengths represent frequency. For 0–14, male bar length ∝ 1200, female ∝ 1150. The pyramid shows a relatively narrow base, indicating lower birth rates, and a wider working-age bulge. The dependency ratio = [(young + old) / working-age] × 100.
绘制金字塔时使用水平条形图：左侧为男性，右侧为女性。纵轴为年龄组。条形长度代表频数。0–14 岁组，男性条形长度∝1200，女性∝1150。该金字塔底部相对较窄，表明出生率较低，劳动年龄人口较宽。抚养比 = [(少儿 + 老年) / 劳动年龄] × 100。
Calculation: Young dependents = 1200 + 1150 = 2350. Old dependents = 900 + 1100 = 2000. Total dependents = 4350. Working-age = 3400 + 3600 = 7000. Dependency ratio = (4350 / 7000) × 100 = 62.1%. This means for every 100 working-age people, there are about 62 dependents.
计算：少儿抚养人口 = 1200 + 1150 = 2350。老年抚养人口 = 900 + 1100 = 2000。总抚养人口 = 4350。劳动年龄人口 = 3400 + 3600 = 7000。抚养比 = (4350 / 7000) × 100 = 62.1%。这意味着每 100 名劳动年龄人口对应约 62 名被抚养人口。
3. Physics & Standard Deviation | 物理与标准差
In an experiment, a student times the period of a pendulum (seconds) over 10 trials: 1.95, 2.02, 1.98, 2.05, 2.00, 1.97, 2.03, 1.99, 2.01, 2.04. Calculate the mean and standard deviation to assess precision.
在一项实验中，一名学生测量了单摆的周期（秒），共 10 次：1.95, 2.02, 1.98, 2.05, 2.00, 1.97, 2.03, 1.99, 2.01, 2.04。计算平均值和标准差以评估精密度。
Mean, x̄ = (1.95 + 2.02 + 1.98 + 2.05 + 2.00 + 1.97 + 2.03 + 1.99 + 2.01 + 2.04) ÷ 10 = 20.04 ÷ 10 = 2.004 s.
平均值 x̄ = (1.95 + 2.02 + 1.98 + 2.05 + 2.00 + 1.97 + 2.03 + 1.99 + 2.01 + 2.04) ÷ 10 = 20.04 ÷ 10 = 2.004 s。
Now compute deviations and squared deviations. E.g., (1.95 – 2.004) = -0.054; square = 0.002916. Sum of squared deviations = 0.002916 + 0.000256 + 0.000576 + 0.002116 + 0.000016 + 0.001156 + 0.000676 + 0.000196 + 0.000036 + 0.001296 = 0.00924. Variance (sample) = 0.00924 ÷ (10 – 1) = 0.0010267. Standard deviation s = √0.0010267 ≈ 0.0320 s. The small standard deviation relative to the mean indicates high precision.
然后计算偏差和平方偏差。例如 (1.95 – 2.004) = -0.054；平方 = 0.002916。平方偏差之和 = 0.002916 + 0.000256 + 0.000576 + 0.002116 + 0.000016 + 0.001156 + 0.000676 + 0.000196 + 0.000036 + 0.001296 = 0.00924。方差（样本）= 0.00924 ÷ (10 – 1) = 0.0010267。标准差 s = √0.0010267 ≈ 0.0320 s。标准差相对于平均值很小，说明精密度高。
4. Business & Moving Averages | 商业与移动平均
A company records quarterly sales (£1000s) over two years. Q1: 23, Q2: 35, Q3: 42, Q4: 28, Q5: 25, Q6: 38, Q7: 46, Q8: 30. Calculate a four-point moving average to identify the trend.
一家公司记录了两年的季度销售额（千英镑）。Q1: 23, Q2: 35, Q3: 42, Q4: 28, Q5: 25, Q6: 38, Q7: 46, Q8: 30。计算四点移动平均以确定趋势。
Four-point moving averages are centred between quarters. First average: (23+35+42+28)/4 = 32.0. Second: (35+42+28+25)/4 = 32.5. Third: (42+28+25+38)/4 = 33.25. Fourth: (28+25+38+46)/4 = 34.25. Fifth: (25+38+46+30)/4 = 34.75. To centre, we need to average each pair. Centred moving averages: (32.0+32.5)/2 = 32.25 (aligned with original Q3). Next: (32.5+33.25)/2 = 32.875 (Q4). (33.25+34.25)/2 = 33.75 (Q5). (34.25+34.75)/2 = 34.5 (Q6). The trend shows a gradual increase.
四点移动平均置于季度之间。第一个平均：(23+35+42+28)/4 = 32.0。第二个：(35+42+28+25)/4 = 32.5。第三个：(42+28+25+38)/4 = 33.25。第四个：(28+25+38+46)/4 = 34.25。第五个：(25+38+46+30)/4 = 34.75。为使其对应原季度，需对每对取平均。中心移动平均：(32.0+32.5)/2 = 32.25（对应原 Q3）。(32.5+33.25)/2 = 32.875（Q4）。(33.25+34.25)/2 = 33.75（Q5）。(34.25+34.75)/2 = 34.5（Q6）。趋势呈现逐步上升。
| Quarter | Sales | 4-pt MA | Centred MA |
| Q3 | 42 | — | 32.25 |
| Q4 | 28 | — | 32.875 |
| Q5 | 25 | — | 33.75 |
| Q6 | 38 | — | 34.5 |
5. Sports Science & Scatter Graphs | 体育科学与散点图
A coach records the number of hours of training per week and the 100 m sprint time (s) for 8 athletes. Data: (5, 12.1), (7, 11.8), (4, 12.5), (8, 11.2), (6, 12.0), (9, 10.9), (3, 13.0), (10, 10.5). Draw a scatter graph and describe the correlation.
一位教练记录了 8 名运动员每周训练时数与 100 米短跑时间（秒）。数据：(5, 12.1), (7, 11.8), (4, 12.5), (8, 11.2), (6, 12.0), (9, 10.9), (3, 13.0), (10, 10.5)。绘制散点图并描述相关性。
Plotting the points with training hours on the x-axis and sprint time on the y-axis shows a clear negative correlation: as training hours increase, sprint time decreases. The product-moment correlation coefficient can be calculated, but visual inspection suggests a strong negative linear relationship. This implies that more training is associated with faster sprint times.
以训练时数为 x 轴，短跑时间为 y 轴描点，显示出明显的负相关：训练时数增加，短跑时间下降。可计算积矩相关系数，但目视检查即可知存在强负线性关系。这意味着训练越多，短跑时间越短。
To estimate the line of best fit, we can find the mean point: x̄ = (5+7+4+8+6+9+3+10)/8 = 52/8 = 6.5 hours; ȳ = (12.1+11.8+12.5+11.2+12.0+10.9+13.0+10.5)/8 = 94.0/8 = 11.75 s. The line passes through (6.5, 11.75). Using two points on the line, we can find the equation. For example, using (4, 12.5) and (9, 10.9): slope = (10.9 – 12.5) / (9 – 4) = -1.6/5 = -0.32. Equation: time = 13.78 – 0.32 × hours (approx). For each extra training hour, sprint time drops by about 0.32 s.
为估计最佳拟合线，计算均值点：x̄ = (5+7+4+8+6+9+3+10)/8 = 52/8 = 6.5 小时；ȳ = (12.1+11.8+12.5+11.2+12.0+10.9+13.0+10.5)/8 = 94.0/8 = 11.75 秒。直线经过 (6.5, 11.75)。利用线上两点求方程。例如用 (4, 12.5) 和 (9, 10.9)：斜率 = (10.9 – 12.5) / (9 – 4) = -1.6/5 = -0.32。方程：时间 = 13.78 – 0.32 × 时数（近似）。每多训练一小时，短跑时间约减少 0.32 秒。
6. Environmental Science & Air Quality Index | 环境科学与空气质量指数
An environmental agency uses a composite index to report daily air quality. The index combines PM₂.₅ (weight 0.5), NO₂ (weight 0.3) and O₃ (weight 0.2). On a particular day, normalized sub-indices are: PM₂.₅ = 82, NO₂ = 65, O₃ = 44. Calculate the overall Air Quality Index (AQI) and interpret.
一家环境机构使用综合指数报告每日空气质量。该指数结合了 PM₂.₅（权重 0.5）、NO₂（权重 0.3）和 O₃（权重 0.2）。某日的标准化分指数为：PM₂.₅ = 82，NO₂ = 65，O₃ = 44。计算总体空气质量指数 (AQI) 并解释。
AQI = (0.5 × 82) + (0.3 × 65) + (0.2 × 44) = 41 + 19.5 + 8.8 = 69.3. According to standard bands, 0–50 is Good, 51–100 is Moderate, so 69.3 falls into the Moderate category. This means air quality is acceptable; however, there may be a risk for some people who are unusually sensitive to air pollution.
AQI = (0.5 × 82) + (0.3 × 65) + (0.2 × 44) = 41 + 19.5 + 8.8 = 69.3。根据标准分级，0–50 为优，51–100 为良，因此 69.3 属于良级别。这意味着空气质量是可接受的；但对于极少数异常敏感人群可能存在风险。
The weighted index approach is typical in GCSE Statistics when dealing with composite indicators. Always check that the weights sum to 1. Here, 0.5+0.3+0.2 = 1. If the weights are given as ratios, first convert them to proportions.
加权指数方法是 GCSE 统计中处理综合指标的典型做法。始终检查权重之和是否为 1。这里 0.5+0.3+0.2 = 1。若权重以比值给出，需先转化为比例。
7. Psychology & Questionnaire Bias | 心理学与问卷偏差
A psychology student designs a survey to study screen time and sleep quality among teenagers. The questionnaire asks: ‘How many hours do you spend on screens, and do you agree that screen time ruins your sleep?’ Identify and explain potential sources of bias.
一名心理学专业学生设计了一份问卷，研究青少年屏幕时间与睡眠质量。问卷问道：“你每天使用屏幕多少小时，你是否同意屏幕时间毁了你的睡眠？”请指出并解释可能的偏差来源。
The wording of the second question is leading; it assumes screen time ruins sleep, which may influence respondents to agree. This is response bias. The sampling method also matters: if the survey is only distributed online, it may miss teenagers without internet access, causing coverage bias. If only friends are asked, it is a convenience sample and not representative. The timing of the survey (e.g., during exams) could also affect responses.
第二个问题的措辞具有引导性；它预设屏幕时间会毁掉睡眠，这可能会影响受访者同意该说法。这属于回答偏差。抽样方法也很重要：如果问卷仅在网上分发，可能会漏掉没有网络的青少年，导致覆盖偏差。如果仅询问朋友，则是便利样本，不具代表性。调查时间（如考试期间）也可能影响回答。
To improve, use neutral language: ‘How many hours of screen time do you have daily, and on a scale of 1–5 how would you rate your sleep quality?’ Use stratified sampling to ensure different age groups and genders are proportionally represented. Pilot the questionnaire to identify ambiguous questions.
改进方法是使用中性语言：“你每天屏幕时间是多少小时？请用 1–5 分评价你的睡眠质量。”使用分层抽样以确保不同年龄组和性别按比例代表。对问卷进行预测试，找出有歧义的问题。
8. Economics & Inflation Index | 经济学与通货膨胀指数
An economist tracks the price of a basket of goods over three years. Base year (Year 1): basket cost = £200. Year 2: £214. Year 3: £226. Calculate the simple price index for each year (base = 100) and the annual inflation rates.
一位经济学家追踪了一篮子商品三年的价格。基年（第 1 年）：篮子成本 = 200 英镑。第 2 年：214 英镑。第 3 年：226 英镑。计算每年的简单价格指数（基期 = 100）及年通货膨胀率。
Index Year 1 = (200/200)×100 = 100. Index Year 2 = (214/200)×100 = 107. Index Year 3 = (226/200)×100 = 113. The inflation rate from Year 1 to Year 2 = (107-100)/100 ×100% = 7%. From Year 2 to Year 3 = (113-107)/107 ×100% ≈ 5.6%. Notice the percentage change uses the previous year as the base, not the original base year.
第 1 年指数 = (200/200)×100 = 100。第 2 年指数 = (214/200)×100 = 107。第 3 年指数 = (226/200)×100 = 113。第 1 年至第 2 年通货膨胀率 = (107-100)/100 ×100% = 7%。第 2 年至第 3 年 = (113-107)/107 ×100% ≈ 5.6%。注意变化率以前一年为基准，而非原基年。
This type of index number calculation is fundamental in GCSE Statistics. You may also need to re-base the index if the base period changes. For example, if Year 2 becomes the new base, the Year 3 index = (113/107)×100 ≈ 105.6. Always state which base year you are using.
此类指数计算是 GCSE 统计的基础。若基期改变，可能还需要重定基数。例如，若第 2 年成为新基期，则第 3 年指数 = (113/107)×100 ≈ 105.6。务必说明所使用的基年。
9. Medicine & Probability Trees | 医学与概率树
A medical test for a virus has a sensitivity of 94% (true positive rate) and a specificity of 98% (true negative rate). The virus prevalence in the population is 0.5%. If a person tests positive, what is the probability they actually have the virus? Use a tree diagram.
某病毒检测的灵敏度为 94%（真阳性率），特异度为 98%（真阴性率）。该病毒在人群中的患病率为 0.5%。若某人检测呈阳性，其真正感染病毒的概率是多少？用树形图求解。
Let V = has virus, V’ = no virus. P(V) = 0.005, P(V’) = 0.995. Positive test given virus: P(+|V) = 0.94; negative given virus: P(-|V) = 0.06. Positive given no virus: P(+|V’) = 1 – 0.98 = 0.02; negative given no virus: P(-|V’) = 0.98. Probability of testing positive overall: P(+) = P(V ∩ +) + P(V’ ∩ +) = (0.005×0.94) + (0.995×0.02) = 0.0047 + 0.0199 = 0.0246. Then, P(V|+) = P(V ∩ +) / P(+) = 0.0047 / 0.0246 ≈ 0.1911 or 19.1%.
设 V = 感染病毒，V’ = 未感染。P(V) = 0.005，P(V’) = 0.995。感染病毒时阳性：P(+|V) = 0.94；阴性：P(-|V) = 0.06。未感染时阳性：P(+|V’) = 1 – 0.98 = 0.02；阴性：P(-|V’) = 0.98。整体检测阳性的概率：P(+) = P(V ∩ +) + P(V’ ∩ +) = (0.005×0.94) + (0.995×0.02) = 0.0047 + 0.0199 = 0.0246。于是 P(V|+) = P(V ∩ +) / P(+) = 0.0047 / 0.0246 ≈ 0.1911 或 19.1%。
Despite the high specificity and sensitivity, the low prevalence means that a positive result has a surprisingly low positive predictive value of about 19%. This counter-intuitive result is important in medical statistics and highlights the need for full understanding of conditional probability.
尽管特异度和灵敏度都很高，但低患病率导致阳性结果的阳性预测值仅有约 19%，出人意料。这一反直觉的结果在医学统计中很重要，也凸显了全面理解条件概率的必要性。
10. Mixed Cross-Disciplinary Challenge | 跨学科综合挑战
A multi-part question: In a health study, data on weekly exercise hours and BMI are collected from 30 adults. (a) Summarise the exercise data with a stem-and-leaf diagram. (b) Draw a histogram for BMI using unequal class widths. (c) Calculate Spearman’s rank correlation coefficient to test for association. (d) Critically evaluate the reliability of the study.
一道多部分题目：在一项健康研究中，收集了 30 名成年人的每周锻炼时数和 BMI 数据。(a) 用茎叶图汇总锻炼数据。(b) 用不等宽组距绘制 BMI 直方图。(c) 计算斯皮尔曼等级相关系数以检验关联。(d) 批判性评估研究的可靠性。
For (a), sort the exercise data and split stems by tens, leaves by units. For (b), choose appropriate class boundaries so that the area of each bar is proportional to frequency. Frequency density = frequency ÷ class width. For (c), rank both variables separately, find d (difference in ranks) for each adult, compute ∑d², and use the formula rₛ = 1 – (6∑d²)/(n(n²-1)). For (d), discuss sampling method, possible confounding variables (diet, genetics), and whether correlation implies causation.
对于 (a)，将锻炼数据排序，以十位数为茎、个位数为叶。对于 (b)，选择合适的组界，确保每个长条面积与频数成正比。频数密度 = 频数 ÷ 组距。对于 (c)，分别对两个变量排序，计算每个人的秩差 d，求 ∑d²，再用公式 rₛ = 1 – (6∑d²)/(n(n²-1))。对于 (d)，讨论抽样方法、可能的混杂变量（饮食、遗传）以及相关性是否意味着因果关系。
In trial data, suppose the exercise hours (stem: leaf) are 0: 5 7; 1: 0 2 3 5; 2: 0 1 4 8 9; 3: 2 5; 4: 1; 5: 0. The distribution is positively skewed. The Spearman’s rank for hypothetical paired BMI data could yield rₛ = -0.68, indicating a moderately strong negative association (more exercise, lower BMI). Evaluation must point out that the sample size (30) is relatively small and the study is observational.
在试验数据中，假设锻炼时数茎叶图（茎: 叶）为 0: 5 7; 1: 0 2 3 5; 2: 0 1 4 8 9; 3: 2 5; 4: 1; 5: 0。分布呈正偏态。对于假定的配对 BMI 数据，斯皮尔曼等级相关系数可能为 rₛ = -0.68，表明中等强度的负关联（锻炼越多，BMI 越低）。评估须指出样本量 (30) 较小，且研究属观察性研究。
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