Year 7 OCR Statistics: Interdisciplinary Mixed Practice | Year 7 OCR 统计：跨学科综合题型训练

📚 Year 7 OCR Statistics: Interdisciplinary Mixed Practice | Year 7 OCR 统计：跨学科综合题型训练

Statistics is not just about numbers in isolation – it is a powerful tool used across subjects like Science, Geography, History, and even Sport. In Year 7 OCR Statistics, you will encounter questions that combine data handling skills with real-world contexts from other disciplines. This article provides a structured mixed practice covering key statistical concepts, graphical interpretation, probability, and cross-curricular problem-solving to help you excel.

统计不仅仅是孤立的数字——它是一种强大的工具，应用于科学、地理、历史甚至体育等学科。在 Year 7 OCR 统计中，你会遇到将数据处理技能与其他学科的真实情境相结合的题目。本文提供结构化的综合训练，涵盖关键统计概念、图表解读、概率和跨学科问题解决，助你取得优异成绩。

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1. Statistics in Science Experiments: Collecting and Analysing Data | 科学实验中的统计：收集与分析数据

In science, you often record measurements and then use statistics to make sense of the results. Imagine a botanist measuring the growth of bean seedlings with different fertilisers. A clear table, a bar chart, and the mean height help to compare which treatment works best.

在科学中，你经常记录测量值，然后利用统计来理解结果。想象一位植物学家测量施用不同肥料后豆苗的生长情况。清晰的表格、条形图和平均高度有助于比较哪种处理效果最好。

Example: Tom tested four fertilisers (A, B, C, and a control with no fertiliser). He planted 10 seeds per pot and after 14 days recorded the average seedling height in centimetres (cm).

例题：Tom 测试了四种肥料（A、B、C 和一组无肥料的对照组）。他每个花盆种下 10 颗种子，14 天后记录幼苗的平均高度（单位：厘米）。

Fertiliser	Height (cm)
None (control)	4.2
A	5.8
B	7.1
C	6.4

To find which fertiliser gave the tallest seedlings, you can calculate the mean of these four values or simply read the table. The bar chart should have fertiliser type on the horizontal axis and height on the vertical axis, with bars labelled clearly.

要找出哪种肥料使幼苗长得最高，你可以计算这四个数值的平均值，或者直接读取表格。条形图的横轴应为肥料种类，纵轴为高度，条形需清晰标注。

Practice question: Draw a bar chart for the data. Calculate the mean height and the range. Why is the control group important?

练习题：为数据绘制条形图。计算平均高度和极差。为什么对照组很重要？

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2. Statistics in Geography: Comparing Climate Data | 地理中的统计：比较气候数据

Geographers use statistics to compare weather patterns in different places. Monthly rainfall data can be displayed in a line graph, making it easy to see wet and dry seasons. Let’s look at two cities: London and Barcelona.

地理学家使用统计数据来比较不同地方的天气模式。月降雨量数据可以用折线图展示，便于看出雨季和旱季。我们来看两个城市：伦敦和巴塞罗那。

Monthly Rainfall (mm)

Month	London	Barcelona
Jan	55	41
Feb	40	29
Mar	37	35
Apr	43	48
May	46	56
Jun	45	33
Jul	42	15
Aug	49	36
Sep	49	68
Oct	69	92
Nov	59	58
Dec	55	43

Plot both sets of data on the same line graph. Use a blue line for London and a red line for Barcelona. Add a key. Then answer: Which city has a drier summer? What is the mean monthly rainfall for London?

将两组数据绘制在同一折线图上。用蓝线表示伦敦，红线表示巴塞罗那，并添加图例。然后回答：哪个城市夏季更干燥？伦敦的月平均降雨量是多少？

By calculating the mean for London (add all 12 values and divide by 12), you get about 49.1 mm. The range is 69 – 37 = 32 mm. Barcelona’s summer months (Jun–Aug) show much lower rainfall, illustrating a Mediterranean climate.

通过计算伦敦的平均值（将 12 个数值相加再除以 12），得到约 49.1 mm。极差为 69 – 37 = 32 mm。巴塞罗那的夏季月份（6–8 月）降雨量低得多，体现了地中海气候的特征。

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3. Statistics in History: Population Changes Over Time | 历史中的统计：人口随时间的变化

Historians use data to understand how populations grow and decline. A line graph showing population every ten years helps to identify trends and relate them to historical events like wars or industrial revolutions.

历史学家利用数据来理解人口的增长与下降。每十年显示一次人口数据的折线图有助于识别趋势，并将其与战争或工业革命等历史事件联系起来。

Example: The estimated population of a town from 1901 to 2001 is given below.

例题：以下是某城镇 1901 年至 2001 年的人口估计值。

Year	Population
1901	8,200
1911	9,500
1921	8,900
1931	10,400
1941	9,800
1951	12,100
1961	14,600
1971	16,000
1981	15,800
1991	17,200
2001	18,500

Draw a time series line graph. Describe the overall trend. Can you spot any dips that might link to World War I or World War II? Calculate the range and the median population over the century.

绘制时间序列折线图。描述总体趋势。你能发现可能与第一次世界大战或第二次世界大战相关的低谷吗？计算极差和这百年间的中位数人口数。

The range is 18,500 – 8,200 = 10,300. To find the median, list the 11 populations in order: the 6th value is 12,100. The dip in 1921 and 1941 reflects wartime impacts.

极差为 18,500 – 8,200 = 10,300。将 11 个人口数值从小到大排列，第 6 个数是中位数 12,100。1921 年和 1941 年的低谷反映了战争的影响。

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4. Statistics in Sport: Analysing Player Performance | 体育中的统计：分析运动员表现

Coaches use averages to pick their best players. Imagine a basketball player’s points scored over 10 matches: 18, 22, 15, 31, 22, 19, 22, 27, 16, 20. We can find the mean, median, and mode to describe typical performance.

教练用平均数来挑选最佳球员。假设一名篮球运动员在 10 场比赛中的得分为：18, 22, 15, 31, 22, 19, 22, 27, 16, 20。我们可以计算均值、中位数和众数来描述典型表现。

First, order the data: 15, 16, 18, 19, 20, 22, 22, 22, 27, 31. The mode is 22 (most frequent). The median is the middle of the ordered list: (20 + 22) ÷ 2 = 21. The mean is the sum (212) divided by 10, giving 21.2.

首先，将数据排序：15, 16, 18, 19, 20, 22, 22, 22, 27, 31。众数是 22（出现最频繁）。中位数是排序列表的中间值：(20 + 22) ÷ 2 = 21。均值是总和 (212) 除以 10，得 21.2。

Which average is most useful? The mean is close to the median, so both represent the player well. However, the mode highlights that 22 points is a common score. If the player had one extreme value (e.g., 50 points), the mean would be pulled up, making the median more reliable.

哪种平均数最有用？均值与中位数接近，两者都能很好地代表该球员。但众数突出显示了 22 分是常见得分。如果球员有一次极端值（例如 50 分），均值会被拉高，此时中位数更可靠。

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5. Probability in Science: Predicting Outcomes | 科学中的概率：预测结果

Probability helps scientists predict how likely an event is. When you toss a fair coin, the theoretical probability of heads is ½. If you toss it 100 times, you might not get exactly 50 heads, but as you toss more, the experimental probability gets closer to ½.

概率帮助科学家预测事件发生的可能性。抛一枚公平硬币时，正面朝上的理论概率是 ½。如果抛 100 次，你可能得不到正好 50 次正面，但随着抛掷次数增加，实验概率会趋近于 ½。

In genetics, Gregor Mendel’s pea plants showed that when crossing a pure purple-flowered plant (PP) with a pure white-flowered plant (pp), all offspring (Pp) had purple flowers. In the next generation, the probability of a white flower was ¼. You can model this with a Punnett square or a simple tree diagram.

在遗传学中，孟德尔的豌豆实验表明，纯种紫花植株 (PP) 与纯种白花植株 (pp) 杂交后，所有子代 (Pp) 都开紫花。在下一代中，开白花的概率为 ¼。你可以用庞纳特方格或简单的树状图来模拟。

Practice: Suppose you roll a fair six-sided die. What is the probability of rolling an even number? Answer: 3/6 = ½. Record 30 rolls and compare your experimental probability with theory.

练习：假设你掷一个公平的六面骰子。掷出偶数的概率是多少？答案：3/6 = ½。记录 30 次投掷结果，并比较实验概率与理论概率。

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6. Interpreting Graphs and Spotting Misleading Charts | 解读图表与发现误导性图表

Data can be presented in ways that exaggerate or hide the truth. A bar chart with a y-axis that does not start at zero can make small differences look huge. Always check the scale and labels.

数据可能以夸大或掩盖真相的方式呈现。纵轴不是从零开始的条形图会使微小的差异看起来巨大。务必检查刻度和标签。

For instance, a bar chart showing the number of books read by two classes might use a y-axis from 25 to 35 instead of 0 to 35. A difference of 2 books could appear ten times larger. A responsible statistician always uses a full scale or clearly marks breaks.

例如，一张显示两个班级阅读书籍数量的条形图，纵轴若设为 25 到 35 而非 0 到 35，2 本书的差异可能会被放大十倍。负责任的统计人员总是使用完整刻度或明确标注断点。

Also, beware of 3D pie charts that distort angles, making some slices look bigger than they really are. Stick to 2D charts for honest comparisons.

此外，要警惕扭曲角度的 3D 饼图，它会使某些扇区看起来比实际更大。坚持使用 2D 图表以进行诚实的比较。

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7. Cross-Curricular Challenge: Integrated Problem Scenario | 跨学科挑战：综合问题情境

Now bring your skills together. A school eco-club wants to build a vegetable garden. They record daily sunshine hours over one week, weekly rainfall, and the growth of lettuce plants (in cm) for three weeks. Use the sample data below to advise them.

现在综合运用你的技能。学校环保俱乐部想建一个菜园。他们记录了一周内每天的日照时数、每周降雨量以及莴苣植株三周内的生长高度（cm）。利用下面的样本数据为他们提供建议。

Sunshine hours (day 1 to 7): 5, 6, 4, 8, 7, 5, 6. Weekly rainfall: Week 1: 12 mm, Week 2: 8 mm, Week 3: 14 mm. Lettuce height start of Week 1: 2 cm, end Week 1: 4 cm, end Week 2: 7 cm, end Week 3: 11 cm.

日照时数（第1天到第7天）：5, 6, 4, 8, 7, 5, 6。每周降雨量：第1周 12 mm，第2周 8 mm，第3周 14 mm。莴苣高度：第1周初 2 cm，第1周末 4 cm，第2周末 7 cm，第3周末 11 cm。

Tasks: (a) Calculate the mean daily sunshine. (b) Draw a line graph of lettuce height over time. (c) Which week had the most growth? (d) If lettuce grows best with average sunshine above 5.5 hours and rainfall under 10 mm, did the weather help? Explain using statistics.

任务：(a) 计算日均日照时数。(b) 绘制莴苣高度随时间变化的折线图。(c) 哪一周生长最多？(d) 如果莴苣在平均日照超过 5.5 小时且降雨量低于 10 mm 时生长最好，天气是否有利？用统计数据说明。

Solutions: Mean sunshine = (5+6+4+8+7+5+6)/7 = 41/7 ≈ 5.86 hours (good). Week 2 had only 8 mm rain but growth from 4 cm to 7 cm is 3 cm; Week 3 had 14 mm rain and growth 4 cm. However, the combination of sunshine and moderate rain in Week 2 was ideal, while Week 3 had more rain but still strong growth, suggesting the plant is robust. The club should monitor and perhaps provide shelter during heavy rain.

解答：平均日照 = (5+6+4+8+7+5+6)/7 = 41/7 ≈ 5.86 小时（有利）。第2周降雨仅 8 mm，但生长量为 3 cm（4 cm 到 7 cm）；第3周降雨 14 mm，生长量 4 cm。然而第2周日照与适度降雨的组合最为理想；第3周雨水较多但仍强劲生长，表明植株健壮。俱乐部应持续

Published by TutorHao | Year 7 统计 Revision Series | aleveler.com

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