📚 Year 9 WJEC Statistics: Interdisciplinary Problem-Solving Practice | Year 9 WJEC 统计：跨学科综合题型训练
Statistics is not just about numbers in a textbook; it is a universal language that helps us make sense of the world. In Year 9 WJEC Statistics, you are expected to apply statistical techniques to real-life problems drawn from science, geography, business, sports, and beyond. This article provides a comprehensive set of interdisciplinary problem-solving exercises designed to strengthen your ability to transfer statistical skills across subjects. Each section introduces a contextual scenario, highlights the key statistical tools you need, and offers practice questions with worked solutions.
统计不只是课本上的数字，它是一种帮助我们理解世界的通用语言。在 Year 9 WJEC 统计课程中，你需要将统计方法应用到科学、地理、商业、体育等现实问题中。本文提供了一系列跨学科综合题型训练，旨在强化你跨学科迁移统计技能的能力。每一节都引入一个真实情境，指出你需要的关键统计工具，并提供练习题和详细解答。
1. Interpreting Scientific Experiment Data | 解读科学实验数据
In science lessons, you often collect measurements during experiments. For example, a biology class might record the growth of cress seedlings over seven days under different light conditions. You need to organise the raw data, calculate measures of central tendency, and draw conclusions. Key statistical skills include calculating the mean, median, mode, range, and constructing line graphs or bar charts to visualise trends.
在科学课上，你经常会在实验中收集测量数据。比如，生物课可能会记录水芹幼苗在不同光照条件下七天的生长情况。你需要整理原始数据，计算集中趋势指标，并得出结论。关键的统计技能包括计算平均数、中位数、众数、极差，以及绘制折线图或条形图来可视化趋势。
Practice Task: The lengths (mm) of 10 cress seedlings grown in full light after 7 days were: 12, 15, 14, 13, 16, 14, 18, 11, 14, 15. Find the mean, median, mode and range. Which measure best describes the typical growth? Explain why.
练习题：在充足光照下生长了 7 天的 10 株水芹幼苗长度（毫米）为：12, 15, 14, 13, 16, 14, 18, 11, 14, 15。请计算均值、中位数、众数和极差。哪一个指标最能描述典型的生长情况？并解释原因。
2 = 14 mm. Mode = 14 mm. Range = 18 - 11 = 7 mm. The mean (14.2) and median (14) are both representative, but because there are no extreme outliers, the mean gives a good overall average. The mode also confirms that 14 mm is the most frequent value.
解答：均值 = (12+15+14+13+16+14+18+11+14+15) ÷ 10 = 142 ÷ 10 = 14.2 毫米。排序后：11,12,13,14,14,14,15,15,16,18。中位数 = (14+14)/2 = 14 毫米。众数 = 14 毫米。极差 = 18 - 11 = 7 毫米。均值（14.2）和中位数（14）都具有代表性，但由于没有极端异常值，均值能很好地反映总体平均水平。众数也证实 14 毫米是最常见的值。
2. Geographical Data: Population Statistics and Graphs | 地理数据分析：人口统计与图表
Geography frequently uses statistical data to describe population structures. You might be given a table showing the population of five cities and asked to create a comparative bar chart or a pie chart to represent proportions. You need to calculate percentages, draw appropriate charts, and interpret demographic patterns. Understanding how to choose the right graph is a core WJEC skill.
地理学经常用统计数据来描述人口结构。你可能会拿到一个显示五个城市人口的表格，并被要求绘制比较条形图或饼图来表示比例。你需要计算百分比，绘制合适的图表，并解释人口模式。懂得如何选择合适的图表是 WJEC 的核心技能之一。
| City | Population (thousands) |
|---|---|
| Cardiff | 370 |
| Swansea | 240 |
| Newport | 150 |
| Bangor | 60 |
| St Davids | 2 |
Question: Calculate the percentage of the total population that lives in each city. Then decide whether a pie chart or a bar chart would be more suitable to compare city sizes and explain your choice.
问题：计算每个城市人口占总人口的百分比。然后判断饼图和条形图哪个更适合比较城市规模，并解释你的选择。
Total population = 822,000. Cardiff: (370/822)*100 ≈ 45.0%, Swansea: 29.2%, Newport: 18.2%, Bangor: 7.3%, St Davids: 0.2% t easy to compare the absolute sizes of cities; a pie chart with one very small slice (0.2%) would be hard to read.
总人口 = 822,000。卡迪夫：(370/822)*100 ≈ 45.0%，斯旺西：29.2%，纽波特：18.2%，班戈：7.3%，圣戴维兹：0.2%。条形图更合适，因为它更容易比较城市的绝对规模；带有一个极小扇形（0.2%）的饼图会很难阅读。
3. Business and Economics: Statistical Analysis of Profit and Budget | 商业与经济：利润与收支的统计分析
Young entrepreneurs use statistics to manage finances. A typical problem gives monthly sales revenue and costs, asking you to calculate profit, construct a compound bar chart, and find average monthly profit. You might also use scatter graphs to explore the relationship between advertising spend and sales income.
年轻的创业者用统计来管理财务。典型题目会给出每月销售收入和成本，要求你计算利润、绘制复式条形图，并找出月平均利润。你可能还会用到散点图来探究广告支出与销售收入之间的关系。
Example: A school tuck shop recorded revenue and costs for four weeks:
| Week | Revenue (£) | Cost (£) |
|---|---|---|
| 1 | 200 | 120 |
| 2 | 250 | 150 |
| 3 | 180 | 130 |
| 4 | 300 | 160 |
- cost. Find the mean weekly profit and the range of profit. Then draw a dual bar chart showing revenue and cost side by side.
任务：每周利润 = 收入 - 成本。计算平均每周利润和利润的极差。然后绘制收入和成本并排的复式条形图。
Solution: Profits: Week 1: £80, Week 2: £100, Week 3: £50, Week 4: £140. Mean profit = (80+100+50+140) ÷ 4 = £92.50. Range = 140 - 50 = £ ual bar chart, use pairs of bars for each week with a key distinguishing revenue and cost.
解答：利润：第 1 周：80 英镑，第 2 周：100 英镑，第 3 周：50 英镑，第 4 周：140 英镑。平均利润 = (80+100+50+140) ÷ 4 = 92.50 英镑。极差 = 140 - 50 = 90 英镑。绘制复式条形图时，每一周用一对条形，并用图例区分收入和成本。
4. Sports Statistics: Performance Data Analysis | 体育统计：运动表现数据分析
Statistics is widely used in sports to monitor athlete performance. You may be given the times (in seconds) for a sprinter in six 100 m races: 11.2, 10.8, 11.5, 10.9, 11.1, 10.7. Calculate the mean time and discuss consistency using the range. Lower range indicates more consistent performance. You can also represent the data with a stem-and-leaf diagram or a box plot once you learn about quartiles.
统计在体育中被广泛用于监测运动员表现。你可能会拿到一名短跑运动员六次 100 米比赛的成绩（秒）：11.2, 10.8, 11.5, 10.9, 11.1, 10.7。计算平均时间，并用极差讨论成绩的稳定性。极差越小表示表现越稳定。在学习了四分位数之后，你还可以用茎叶图或箱线图来表示数据。
1.5 - 10.7 = 0.8 s. This relatively small range suggests the sprinter is quite consistent.
计算：均值 = (11.2+10.8+11.5+10.9+11.1+10.7) ÷ 6 = 66.2 ÷ 6 ≈ 11.03 秒。排序：10.7, 10.8, 10.9, 11.1, 11.2, 11.5。极差 = 11.5 - 10.7 = 0.8 秒。这个较小的极差表明这位短跑选手相当稳定。
5. Health and Nutrition: Statistical Analysis of Food Composition | 健康与营养：食物成分的统计分析
Health education and food technology classes often involve comparing nutritional content. A table might show the sugar content (g) per 100 g for five breakfast cereals. You can calculate the mean sugar content, find the median, and determine whether any cereal is an outlier using quartiles and the interquartile range (IQR). This combines descriptive statistics with health awareness.
健康教育和食品技术课经常涉及比较营养成分。一张表格可能列出五种早餐麦片每 100 克中的糖含量（克）。你可以计算平均糖含量，找出中位数，并利用四分位数和四分位距（IQR）判断是否有麦片属于异常值。这结合了描述性统计与健康意识。
. IQR = 26 - 6.5 = 19.5 g. Outlier boundaries: Lower = Q1 - 1.5×IQR = 6.5 - 29.25 = -22.75 (no low outlier); Upper = Q3 + 1.5×IQR = 26 + 29.25 = 55.25. No outliers here, but 30 g is noticeably high. Always check the context—30 g sugar is high for a cereal.
给定的糖含量：15, 22, 5, 30, 8。排序：5, 8, 15, 22, 30。中位数 = 15 克。Q1 = 下半部分 (5,8) 的中位数 = 6.5 克。Q3 = 上半部分 (22,30) 的中位数 = 26 克。IQR = 26 - 6.5 = 19.5 克。异常值界限：下限 = Q1 - 1.5×IQR = 6.5 - 29.25 = -22.75（无低异常值）；上限 = Q3 + 1.5×IQR = 26 + 29.25 = 55.25。这里没有异常值，但 30 克明显偏高。一定要结合背景——对麦片来说，30 克糖属于高含量。
6. Environmental Science: Trends in Climate Change Data | 环境科学：气候变化数据趋势
Environmental studies rely heavily on time series data. You could be given the average global temperature anomaly (in °C) for ten consecutive years and asked to plot a time series graph, describe the trend, and perhaps calculate a moving average to smooth out fluctuations. This integrates graphical skills with interpretation of environmental change.
环境研究非常依赖时间序列数据。你可能会拿到连续十年的全球平均气温距平（°C），要求绘制时间序列图，描述变化趋势，或许还要计算移动平均来消除波动。这融合了作图技巧与对环境变化的解读。
8: +0.7, Year 9: +0.9, Year 10: +1.0. Plot these on a line graph with years on the x-axis. Trend: general upward movement, indicating warming. A 3‑ .27, and so on. The moving average line will appear smoother and confirm the upward trend.
示例数据：第 1 年：+0.2，第 2 年：+0.3，第 3 年：+0.1，第 4 年：+0.4，第 5 年：+0.6，第 6 年：+0.5，第 7 年：+0.8，第 8 年：+0.7，第 9 年：+0.9，第 10 年：+1.0。在折线图上以年份为 x 轴绘制这些点。趋势：总体向上移动，表明在变暖。第 3 年的 3 年移动平均为 (0.2+0.3+0.1)/3 = 0.2，第 4 年：(0.3+0.1+0.4)/3 ≈ 0.27，依此类推。移动平均线会显得更平滑，并确认上升趋势。
7. Cross-Disciplinary Practice 1: Error Analysis in Physics Experiments | 综合题型训练一：物理实验误差分析
In physics, repeated measurements of the time period of a pendulum are taken to minimise random error. Statistical methods help quantify precision. Suppose five trials give period measurements: 1.23 s, 1.25 s, 1.22 s, 1.24 s, 1.26 s. Calculate the mean, the absolute deviation of each from the mean, and the mean absolute deviation (MAD) as a simple measure of spread. Explain why a small MAD indicates high precision.
在物理中，对单摆周期进行重复测量是为了减少随机误差。统计方法有助于量化精密度。假设五次实验得到的周期测量值为：1.23 s、1.25 s、1.22 s、1.24 s、1.26 s。计算均值、每个值与均值的绝对偏差，以及作为简易离散程度指标的平均绝对偏差（MAD）。请解释为什么较小的 MAD 表示高精密度。
Solution: Mean = (1.23+1.25+1.22+1.24+1.26) ÷ 5 = 6.20 ÷ 5 = 1.24 s. Deviations: |1.23-1.24|=0.01; |1.25-1.24|=0.01; |1.22-1.24|=0.02; |1.24-1.24|=0; |1.26-1.24| most measurements are very close to the mean, showing high precision and low random error.
解答：均值 = (1.23+1.25+1.22+1.24+1.26) ÷ 5 = 6.20 ÷ 5 = 1.24 s。偏差：|1.23-1.24|=0.01；|1.25-1.24|=0.01；|1.22-1.24|=0.02；|1.24-1.24|=0；|1.26-1.24|=0.02。MAD = (0.01+0.01+0.02+0+0.02) ÷ 5 = 0.06 ÷ 5 = 0.012 s。较小的 MAD 意味着多数测量值都非常接近均值，表明高精密度和低随机误差。
8. Cross-Disciplinary Practice 2: Comparing Multivariate Data in Geography | 综合题型训练二：地理多变量数据比较
Geography field trips often collect multiple variables for different sites, such as soil pH, moisture content, and plant diversity index. You can use a radar chart (also called a spider chart) to compare two or more sites visually. Each axis represents one variable, and the values are plotted on a common scale. This allows you to quickly see which site is more alkaline, has higher moisture, etc. A table of data for two sites might be:
地理实地考察通常会为不同地点收集多个变量，例如土壤 pH 值、含水量和植物多样性指数。你可以使用雷达图（也称蛛网图）来直观比较两个或多个地点。每条轴代表一个变量，数值绘制在一个共同的尺度上。这样你可以迅速看出哪个地点碱性更强、含水量更高等。两个地点的数据表可能如下：
| Variable | Woodland A | Woodland B |
|---|---|---|
| Soil pH | 6.2 | 5.1 |
| Moisture (%) | 45 | 60 |
| Diversity index | 3.2 | 4.5 |
Construct a radar chart with axes scaled appropriately (e.g., pH 4 to 7, moisture 30–70%, diversity 2–5). Plot both sites and write a brief comparison. Woodland A has higher pH (less acidic), but Woodland B is moister and more diverse.
构建一个雷达图，轴的范围要合适（例如 pH 4 到 7，湿度 30–70%，多样性指数 2–5）。绘制两个地点并写出简短的比较。林地 A 的 pH 值较高（酸性较弱），但林地 B 更湿润且多样性更高。
9. Cross-Disciplinary Practice 3: Cost-Benefit Analysis in Economics | 综合题型训练三：经济学成本效益分析
ven analysis using cumulative frequency or calculating expected value from probability × outcome. At Year 9 level, a simple version asks you to use a table of costs and revenues over several months to decide whether a project is financially viable. You can calculate the mean monthly net gain and use a bar chart with positive (profit) and negative (loss) months to assess risk.
经济学和商学常常需要通过比较成本与效益来做决策。一种统计方法是利用累积频率进行盈亏平衡分析，或者通过概率 × 结果来计算期望值。在 Year 9 阶段，简化版题目会要求你利用几个月的成本和收入表格来判断一个项目在经济上是否可行。你可以计算平均每月净收益，并用带正（盈利）负（亏损）月份的条形图来评估风险。
Scenario: A pop-up juice bar has monthly net figures: £+120, £-40, £+150, £+80, £0. Find the mean net gain and the range. Mean = (120 - 40 + 150 + 80 - 20) ÷ 5 = 290 ÷ 5 = £58. Because the mean is positive and the range is 150 - (-40) = £ profit but some volatility. This is a good introduction to the concept of investment risk.
情境：一家快闪果汁吧的每月净收益为：+120 英镑、-40 英镑、+150 英镑、+80 英镑、-20 英镑。求平均净收益和极差。均值 = (120 - 40 + 150 + 80 - 20) ÷ 5 = 290 ÷ 5 = 58 英镑。由于均值为正，极差为 150 - (-40) = 190 英镑，这表明总体盈利但存在一定波动。这是对投资风险概念的一个很好的引入。
10. Putting It All Together: Statistical Thinking Across the Curriculum | 融会贯通：跨课程的统计思维
The examples above show that statistical methods are not confined to maths lessons. Whether you are tracking plant growth in biology, mapping population change in geography, or evaluating a business venture, you need to collect, present, and interpret data accurately. The key is to identify the question, choose the right statistical tool, and communicate your findings clearly. Use this cross-curricular approach to prepare for WJEC assessments and develop a life-long skill.
以上例子表明，统计方法并不局限于数学课。无论你是在生物课中追踪植物的生长，在地理课中绘制人口变化图，还是在评估一个商业项目，你都需要准确地收集、展示和解读数据。关键在于明确问题，选择正确的统计工具，并清晰地传达你的发现。运用这种跨学科的方法来准备 WJEC 测评，并发展一项终身技能。
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