Interdisciplinary Problem-Solving Practice for Year 9 Edexcel Statistics | 跨学科综合题型训练

📚 Interdisciplinary Problem-Solving Practice for Year 9 Edexcel Statistics | 跨学科综合题型训练

Year 9 Edexcel Statistics is not just a stand-alone subject – it provides the essential toolkit for interpreting data in biology, geography, business, sports science and beyond. In this revision guide, we work through mixed, real-world problems that require you to select the right graph, calculate averages and spread, and draw valid conclusions. Each section models a dual-language explanation so you can master both the techniques and the subject-specific vocabulary.

九年级爱德思统计绝非孤立的学科——它为解读生物、地理、商业、体育科学等领域的数据提供了核心工具箱。本复习指南将通过跨学科的真实问题，训练你选择合适的图表、计算平均数与离散程度，并得出有效结论。每一节都配有中英双语解析，帮助你同时掌握解题技巧和学科术语。

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1. Why Interdisciplinary Practice Matters | 跨学科训练的重要性

Statistical skills are transferable. A biologist plotting plant growth, a geographer comparing population structures, and a business analyst tracking quarterly sales all rely on the same core methods: calculating averages, constructing charts, and describing distributions. By practising problems that blend subjects, you learn to recognise which statistical tool fits the context most effectively.

统计技能是可迁移的。生物学家绘制植物生长图、地理学家比较人口结构、商业分析师追踪季度销售额，都依赖相同的核心方法：计算平均数、构建图表和描述分布。通过混合学科的问题训练，你将学会识别哪种统计工具最适合具体情景。

In the Edexcel assessment, many questions are set in realistic contexts. Students who have practiced interdisciplinary problems are better at extracting key numbers from text, choosing between a bar chart and a line graph, and writing comparative statements supported by evidence.

在爱德思评估中，许多题目都设置在真实情境里。练习过跨学科解决问题的学生，能更好地从文本中提取关键数据，在条形图与折线图之间做出选择，并写出有证据支持的比较性陈述。

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2. Biology: Analysing Plant Growth | 生物：分析植物生长

A student measures the height of a bean plant every two days. The data are recorded in the table below. We need to find the mean height, plot a line graph, and comment on the trend.

一名学生每两天测量一次豆类植物的高度。数据记录在下表中。我们需要计算平均高度、绘制折线图并评述趋势。

Day	Height (cm)
1	2.0
3	3.2
5	4.5
7	5.8
9	7.0
11	8.3
13	9.5

Step 1 – Calculate the mean height: Add all heights: 2.0 + 3.2 + 4.5 + 5.8 + 7.0 + 8.3 + 9.5 = 40.3 cm. Divide by the number of data points (7).

步骤1 – 计算平均高度： 将所有高度相加：2.0 + 3.2 + 4.5 + 5.8 + 7.0 + 8.3 + 9.5 = 40.3 cm。除以数据点个数 (7)。

Mean = 40.3 ÷ 7 ≈ 5.76 cm

Step 2 – Graph choice: A line graph is best because it shows the change in height over time. Plot Day on the horizontal axis and Height on the vertical axis. The line rises steadily, indicating continuous growth.

步骤2 – 图表选择： 折线图最合适，因为它显示高度随时间的变化。将“天数”放在横轴，“高度”放在纵轴。线条平稳上升，表明持续生长。

Step 3 – Interpret: The plant grows fastest between Day 7 and Day 9, where the slope is steepest. The overall pattern suggests a healthy, increasing trend.

步骤3 – 解读： 植物在第7天到第9天之间生长最快，此段斜率最陡。总体模式表明健康、递增的趋势。

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3. Geography: Population Pyramids & Median Age | 地理：人口金字塔与中位年龄

A geography project investigates the age structure of a town. The frequency table below shows the number of residents in each age group. We will estimate the median age.

某地理课题研究一城镇的年龄结构。下面的频数表显示了各年龄段居民人数。我们将估算中位年龄。

Age group (years)	Frequency	Cumulative frequency
0 – 14	1200	1200
15 – 29	1500	2700
30 – 44	1800	4500
45 – 59	1300	5800
60 – 74	800	6600
75+	400	7000

Total population = 7000. The median position is the 3500^th value. The cumulative frequency column shows that the 30 – 44 group contains the median, because its cumulative frequency reaches 4500 and the previous group stops at 2700.

总人口 = 7000。中位位置是第3500个值。累积频数列显示，30 – 44 岁组包含中位数，因为该组累计频数达到4500，而前一组止于2700。

To estimate the exact median, use linear interpolation within the group. The lower boundary is 30, group width is 15, frequency of the group is 1800, and the number needed into the group is 3500 – 2700 = 800.

要估计精确的中位数，可在组内使用线性插值法。下限为30，组距为15，组频数为1800，进入该组所需数量为 3500 – 2700 = 800。

Median ≈ 30 + (800 / 1800) × 15 = 30 + 6.67 ≈ 36.7 years

This tells us that half the residents are younger than about 36.7 years. In a comparative geography question, you could then contrast this with another region’s median, discussing implications for services like schools or care homes.

这告诉我们，一半居民年龄低于约36.7岁。在一道比较性地理题中，你可以将此与另一个地区的中位数对比，讨论对学校或养老院等服务的影响。

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4. Business: Interpreting Bar Charts & Profit Trends | 商业：解读条形图与利润趋势

A small company’s quarterly profits (in £1000s) are shown below. We will calculate the mean quarterly profit, identify the best and worst quarters, and suggest a suitable chart.

一家小公司的季度利润（单位：千英镑）如下所示。我们将计算平均季度利润、确定最佳和最差季度，并建议合适的图表。

Quarter	Profit (£1000s)
Q1	25
Q2	40
Q3	35
Q4	50

Mean profit: (25 + 40 + 35 + 50) ÷ 4 = 150 ÷ 4 = 37.5. So the average quarterly profit is £37,500.

平均利润： (25 + 40 + 35 + 50) ÷ 4 = 150 ÷ 4 = 37.5。因此平均季度利润为 37,500 英镑。

Best and worst: Q4 is the highest (50), Q1 is the lowest (25). The range is 50 – 25 = 25, showing considerable variability.

最佳与最差： 第四季度最高（50），第一季度最低（25）。全距为 50 – 25 = 25，显示出相当大的波动性。

Chart choice: A bar chart (or vertical bar graph) effectively compares the discrete categories of quarters. The height of each bar represents profit, making it easy to see that profits generally rise towards the end of the year.

图表选择： 条形图（或垂直条形图）能有效比较季度的离散类别。每个条形的高度代表利润，易于看出利润通常在年底上升。

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5. Sports: Comparing Athletes’ Performances with Box Plots | 体育：用箱线图比较运动员表现

Two long jump athletes, A and B, have their best jumps (in metres) over 10 competitions recorded. We use five-number summaries to draw box plots and compare consistency.

两名跳远运动员 A 和 B 在 10 场比赛中的最佳成绩（米）被记录下来。我们使用五数概括绘制箱线图并进行一致性比较。

A: 4.5, 4.7, 4.9, 5.0, 5.2, 5.3, 5.5, 5.6, 5.8, 6.0
B: 4.8, 4.9, 5.0, 5.0, 5.1, 5.2, 5.2, 5.3, 5.4, 5.9

For athlete A, minimum = 4.5, maximum = 6.0. Q1 is at position (10+1)/4 = 2.75, so Q1 = 4.7 + 0.75×(4.9 – 4.7) = 4.85. Median position 5.5 gives median = 5.2 + 0.5×(5.3 – 5.2) = 5.25. Q3 position 8.25 gives Q3 = 5.6 + 0.25×(5.8 – 5.6) = 5.65. IQR = 5.65 – 4.85 = 0.80.

对于运动员 A，最小值 = 4.5，最大值 = 6.0。Q1 位置为 (10+1)/4 = 2.75，因此 Q1 = 4.7 + 0.75×(4.9 – 4.7) = 4.85。中位数位置 5.5 得出中位数 = 5.2 + 0.5×(5.3 –

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