📚 Interdisciplinary Statistics Practice for KS3 | KS3 跨学科统计综合题型训练
Welcome to your KS3 Edexcel statistics workout. This article blends data handling, averages, charts, and probability with real-world themes from science, geography, and everyday life. Each section poses a challenge that requires you to collect, present, and interpret data, then draw conclusions – just like a statistician working across disciplines. By the end, you’ll feel confident tackling mixed-topic questions and seeing how numbers tell stories beyond the maths classroom.
欢迎来到 KS3 Edexcel 统计专项训练。这篇文章将数据处理、平均数、图表和概率与科学、地理以及日常生活中的真实主题融合在一起。每一节都提出一个挑战，要求你收集、展示和解读数据，然后得出结论——就像一位跨学科工作的统计学家。当你完成全部内容时，你将能够自信地应对混合主题题目，并理解数字如何在数学课堂之外讲述故事。
1. Heart Rate and Exercise | 心率与运动
A group of Year 8 pupils measured their resting heart rates (beats per minute) before and after a two-minute step-up exercise. Here are the results for five pupils: Before – 72, 68, 75, 70, 73. After – 98, 105, 101, 95, 99. Calculate the mean and range for each set. What does the difference in means tell you about the effect of exercise on heart rate?
一组八年级学生测量了他们安静时的心率（每分钟心跳次数）以及完成两分钟台阶运动后的心率。五名学生的数据如下：运动前——72, 68, 75, 70, 73。运动后——98, 105, 101, 95, 99。计算每组的平均数和极差。平均数的差异告诉你运动对心率有什么影响？
The mean increased by 28 beats per minute, showing that exercise raises heart rate. The range also grew slightly, indicating more variation in response after exercise.
运动前平均： (72+68+75+70+73) ÷ 5 = 358 ÷ 5 = 71.6。极差：75 - 68 = 7。运动后平均： (98+105+101+95+99) ÷ 5 = 498 ÷ 5 = 99.6。极差：105 - 95 = 10。平均数增加了每分钟28次，表明运动会使心率升高。极差也略微增大，说明运动后个体反应的差异更明显。
2. Recycling Weights and the Median | 回收重量与中位数
The eco-club collected waste paper and cardboard every week for 9 weeks. The weights in kilograms were: 12, 8, 15, 10, 18, 9, 14, 30, 11. There is an outlier: the 30 kg week was after a school fair. Find the median and mode, then explain which average better represents a typical week and why the outlier matters.
环保社团连续九周收集废纸和纸板。每周的重量（千克）为：12, 8, 15, 10, 18, 9, 14, 30, 11。其中有一个异常值：30千克那一周是在学校游园会之后。找出中位数和众数，然后解释哪个平均数更能代表典型的一周，以及为什么异常值很重要。
) ÷ 9 = 127 ÷ 9 ≈ 14.1 kg. The median of 12 kg is less affected by the outlier of 30 kg, so it gives a fairer idea of a typical week. The outlier pulls the mean up and might mislead someone about usual collection amounts.
将数据排序：8, 9, 10, 11, 12, 14, 15, 18, 30。中位数（第5个值）= 12千克。众数：没有重复数值，所以没有众数。平均数为 (8+9+10+11+12+14+15+18+30) ÷ 9 = 127 ÷ 9 ≈ 14.1 千克。中位数12千克受异常值30千克的影响较小，因此能更公平地反映典型一周的回收量。异常值拉高了平均数，可能会让人们对通常的回收量产生误解。
3. Geography Temperatures and Dual Bar Charts | 地理气温与复式条形图
A weather station recorded the average monthly temperatures for two cities, Coastville and Inlandtown, over three months. The data are: January – Coastville 5°C, Inlandtown 1°C; February – Coastville 6°C, Inlandtown 3°C; March – Coastville 9°C, Inlandtown 8°C. Draw a dual bar chart, then write two comparisons. Suggest one geographical reason for the differences.
一座气象站记录了两个城市——海岸镇和内陆城——三个月的平均月气温。数据如下：一月——海岸镇5°C，内陆城1°C；二月——海岸镇6°C，内陆城3°C；三月——海岸镇9°C，内陆城8°C。绘制一幅复式条形图，然后写出两处对比。提出一个导致差异的地理原因。
On your dual bar chart, group months on the horizontal axis and use two differently coloured bars. Comparison 1: Coastville is consistently warmer than Inlandtown across all three months. Comparison 2: The temperature gap closes from 4°C in January to only 1°C in March. Geographical reason: Coastville is near the sea, which retains heat in winter and keeps winters milder; Inlandtown is inland, so it cools faster.
在你的复式条形图中，横轴显示月份，并使用两种不同颜色的条形。对比1：在所有三个月中，海岸镇的气温始终高于内陆城。对比2：温差从一月的4°C缩小到三月的仅1°C。地理原因：海岸镇靠近海洋，海洋冬季保留热量，使得冬季较为温和；内陆城地处内陆，因此降温更快。
4. Probability from a Two-Way Table | 双向表中的概率
70 students were asked whether they prefer swimming or cycling. The results are summarised in a two-way table:
| Swimming | Cycling | Total | |
|---|---|---|---|
| Boys | 18 | 22 | 40 |
| Girls | 20 | 10 | 30 |
| Total | 38 | 32 | 70 |
Use the table to find the probability that a randomly chosen student is a girl who prefers cycling. Then find the probability that a student prefers swimming given that they are a boy. Present your answers as fractions in simplest form.
70名学生被问及他们更喜欢游泳还是骑行。结果总结在下面的双向表中。利用表格计算随机选取一名学生恰好是喜欢骑行的女生的概率。然后计算在已知是男生的情况下，该学生更喜欢游泳的概率。答案用最简分数表示。
ity (swimming given boy) = number of boys who prefer swimming / total boys = 18/40 = 9/20.
概率（女生且喜欢骑行）= 喜欢骑行的女生人数 / 总人数 = 10/70 = 1/7。概率（在男生中喜欢游泳）= 喜欢游泳的男生人数 / 男生总人数 = 18/40 = 9/20。
5. Science Experiment: Plant Growth and Line Graphs | 科学实验：植物生长与折线图
Two bean plants were measured over five weeks. Plant A (with fertiliser) heights in cm: 2, 5, 9, 14, 20. Plant B (no fertiliser): 2, 4, 6, 8, 10. Plot both sets on the same line graph, using time (weeks) on the x-axis. Describe the trend for each plant. Use the data to argue whether fertiliser made a difference.
两株豆苗在五周内被测量高度。植物A（施加肥料）的高度（厘米）：2, 5, 9, 14, 20。植物B（无肥料）：2, 4, 6, 8, 10。将两组数据绘制在同一幅折线图上，x轴表示时间（周）。描述每株植物的变化趋势。利用数据论证肥料是否产生了作用。
The line graph should show both series. Plant A' final heights and the increasing rate suggest fertiliser boosted growth. Statistical reasoning: you could compute the mean weekly growth rates: Plant A (18 cm total increase ÷ 4 weeks = 4.5 cm/week), Plant B (8 ÷ 4 = 2 cm/week).
折线图应同时展示两组数据。植物A的生长呈加速趋势，每周坡度更陡。植物B以大约每周2厘米的速度稳定增长。到第五周时，植物A高出10厘米。最终高度差异以及不断增大的生长速度表明肥料促进了生长。统计推理：你可以计算平均每周生长速率：植物A（总增长18厘米 ÷ 4周 = 4.5厘米/周），植物B（8 ÷ 4 = 2厘米/周）。
6. Pie Chart of Household Energy Use | 家庭能源使用的饼图
An energy survey recorded the percentage of electricity used by different appliances in a typical home: Heating 40%, Kitchen appliances 25%, Lighting 15%, Entertainment 12%, Others 8%. Draw a pie chart to represent this data. Calculate the angle for each sector. What energy-saving advice could you give based on the largest sector?
一项能源调查记录了一个典型家庭中不同电器用电量的百分比：取暖40%，厨房电器25%，照明15%，娱乐设备12%，其他8%。绘制一幅饼图来表示这些数据。计算每个扇形的角度。根据占比最大的部分，你能给出什么节能建议？
Angles: Heating = 40% = 28.8° (round to 29°). Construct your pie chart with these angles, label sectors and percentages. Energy-saving tip: Since heating uses the largest share, lowering the thermostat by 1°C or improving insulation could significantly reduce electricity consumption.
角度：取暖 = 360° 的40% = 0.4 × 360° = 144°。厨房 = 0.25 × 360° = 90°。照明 = 0.15 × 360° = 54°。娱乐 = 0.12 × 360° = 43.2°（四舍五入到43°）。其他 = 0.08 × 360° = 28.8°（四舍五入到29°）。用这些角度构建饼图，标注各扇区和百分比。节能建议：因为取暖用电占比最大，把恒温器调低1°C或改善隔热可以显著减少电力消耗。
7. Book Survey and Comparative Mean | 阅读调查与平均数比较
A library surveyed the number of books borrowed by Year 7 and Year 9 students in one month. Year 7 sample: 3, 5, 2, 6, 4, 7, 3, 5, 4, 6. Year 9 sample: 8, 10, 7, 9, 11, 12, 8, 9, 10, 6. Find the mean for each year group. What does the difference suggest about reading habits? Identify one potential source of bias if data were collected only from students visiting the library.
图书馆调查了七年级和九年级学生在一个月内的借书数量。七年级样本：3, 5, 2, 6, 4, 7, 3, 5, 4, 6。九年级样本：8, 10, 7, 9, 11, 12, 8, 9, 10, 6。计算每个年级的平均数。平均数的差异说明了阅读习惯的什么情况？如果数据仅从到访图书馆的学生中收集，请指出一个潜在的偏差来源。
ce as many books on average, perhaps because they have longer texts or more independent reading required by their curriculum. Bias: the sample excludes students who do not visit the library, who might borrow fewer books; thus the results may overestimate borrowing rates for the whole year group.
七年级平均：总和 = 3+5+2+6+4+7+3+5+4+6 = 45, 平均 = 45 ÷ 10 = 4.5。九年级平均：总和 = 8+10+7+9+11+12+8+9+10+6 = 90, 平均 = 9.0。九年级学生平均借书量是七年级的两倍，这可能是因为他们的读物更长，或者课程要求更多独立阅读。偏差：样本没有包括那些不到访图书馆的学生，他们可能借书较少；因此结果可能高估了整个年级的借阅率。
8. Scatter Graph: Height and Arm Span | 散点图：身高与臂展
In a PE class, students measured their height and arm span (cm). The first five pairs are: (145, 143), (152, 150), (160, 162), (138, 137), (168, 167). Plot these on a scatter graph with height on the x-axis. Describe the correlation. Estimate the arm span for a student 155 cm tall. Is it reasonable to predict arm span for a height of 200 cm? Explain.
在一节体育课上，学生们测量了身高和臂展（厘米）。前五对数据为：(145, 143), (152, 150), (160, 162), (138, 137), (168, 167)。以身高为x轴绘制散点图。描述相关性。估计一名身高155厘米的学生的臂展。对于身高200厘米的人预测臂展是否合理？请解释。
The points show strong positive correlation – as height increases, arm span tends to increase. Draw a line of best fit. For 155 cm on the x-axis, read upwards to your line and across to the y-axis; this gives an estimate around 154–156 cm. Prediction for 200 cm is less reliable because it lies far outside the data range (extrapolation); the relationship may not stay linear, and you have no data to support that estimate.
散点显示出强正相关——身高增加时，臂展也倾向于增加。画一条最佳拟合线。在x轴上从155厘米处向上作垂线与直线相交，再水平读出y轴数值，估计大约在154–156厘米范围内。对于200厘米的预测不太可靠，因为这远远超出了数据范围（外推）；关系可能不再保持线性，而且没有数据支持这一估计。
9. Combined Probability in a Game | 游戏中的组合概率
A fair six-sided dice is rolled and a fair coin is flipped at the same time. Use a sample space diagram to list all possible outcomes. Find the probability of getting a head and an even number. Then find the probability of getting a tail or a number greater than 4. Write your answers as decimals.
同时掷一枚均匀的六面骰子和抛一枚均匀的硬币。使用样本空间图列出所有可能的结果。求得到正面和一个偶数的概率。然后求得到反面或者大于4的数字的概率。答案用小数表示。
s H2, H4, H6 → 3/12 = 0.25. Probability (Tail or number >4): Tail outcomes: 6 outcomes (T1– distinct outcomes: T1,T2,T3,T4,T5,T6,H5,H6 → 8 outcomes. Probability = 8/12 = 2/3 ≈ 0.667.
样本空间包含12个结果：H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6。概率（正面且偶数）= 结果H2, H4, H6 → 3/12 = 0.25。概率（反面或数字大于4）：反面结果：6个（T1至T6）。数字大于4：出现在硬币两面，即H5, H6, T5, T6（4个结果）。并集：计数所有不同的结果：T1,T2,T3,T4,T5,T6,H5,H6 → 8个结果。概率 = 8/12 = 2/3 ≈ 0.667。
10. Comparing Distributions with Box Plots | 用箱形图比较分布
The science department tested two brands of batteries by recording how many hours they lasted in identical torches. Brand X hours: 8, 9, 10, 10, 11, 12, 14. Brand Y: 7, 8, 9, 9, 10, 13, 15. Construct a box plot for each, using the five-number summary. Compare the medians and interquartile ranges. Which battery would you recommend and why?
科学部对两种品牌的电池进行了测试，记录它们在相同的手电筒中持续使用的小时数。品牌X的小时数：8, 9, 10, 10, 11, 12, 14。品牌Y：7, 8, 9, 9, 10, 13, 15。利用五数概括法为每种品牌绘制箱形图。比较它们的中位数和四分位距。你会推荐哪种电池？为什么？
n (10 vs 9) and a smaller IQR, meaning it is more consistent and typically lasts longer. Brand Y has greater spread, so some last very long but others are shorter. Recommendation: Brand X for reliability.
品牌X：最小值=8，Q1=9，中位数=10，Q3=12，最大值=14。IQR = 12-9=3。品牌Y：最小值=7，Q1=8，中位数=9，Q3=13，最大值=15。IQR = 13-8=5。品牌X的中位数更高（10对9），且四分位距更小，意味着它更加稳定，通常续航时间更长。品牌Y的分布更分散，虽然有一些电池续航很久，但其他的较短。推荐：选择品牌X以获得可靠性。
11. Designing a Questionnaire for a Canteen Survey | 设计食堂调查问卷
The school wants to improve the canteen menu. Write three statistical questions that could be included in a pupil questionnaire. One must generate categorical data, one discrete data, and one continuous data. For each question, state the type of chart you would use to display the results and explain how you would avoid bias in the wording.
学校希望改善食堂菜单。写出三个可以包含在学生问卷中的统计问题。其中一个必须生成分类数据，一个生成离散数据，另一个生成连续数据。对于每个问题，说明你将使用哪种图表来展示结果，并解释如何在措辞上避免偏差。
Question 1 (categorical): 'What is your favourite type of main dish? (Pasta / Pizza / Salad / Curry / Other)' – Display with a bar chart or pie chart. Avoid bias by not implying one option is better, and include an 'Other' category. Question 2 (discrete): 'How many days per week do you buy food from the canteen? (0,1,2,3,4,5)' – Display with a frequency diagram or bar chart. Avoid bias by using neutral wording and a full range of options. Question 3 (continuous): 'About how many minutes do you spend queuing at the canteen each visit? (Collect as a number)' – Display with a histogram (grouped if needed). Avoid bias by not suggesting a 'normal' time in the question.
问题1（分类数据）：“你最喜欢哪类主菜？（意大利面/披萨/沙拉/咖喱/其他）”——使用条形图或饼图展示。通过不暗示某个选项更好，并加入“其他”类别来避免偏差。问题2（离散数据）：“你每周有多少天在食堂购买食物？（0,1,2,3,4,5）”——使用频数图或条形图展示。使用中性措辞并提供完整选项范围以避免偏差。问题3（连续数据）：“你每次在食堂大约排队多少分钟？（收集具体数字）”——使用直方图（必要时分组）展示。通过不在问题中暗示“正常”时间的方式来避免偏差。
12. Evaluating a Statistical Claim from Media | 评价媒体中的统计论断
A fitness app claims: 'Users who complete our 30-day challenge lose an average of 3 kg.' The small print states that only 40% of users finished the challenge and reported their results. Explain why the average might be misleading. How could you design a fairer investigation? Think about missing data, sample size, and comparison groups.
一款健身应用声称：“完成我们30天挑战的用户平均减重3公斤。”小字说明只有40%的用户完成了挑战并报告了结果。解释为什么这个平均数可能具有误导性。你如何设计一个更公平的调查？考虑缺失数据、样本大小和对照组。
The average only includes those who finished and reported, ignoring the 60% who dropped out or didn't report; these users may have lost less weight or even gained weight, which would lower the true average. This is 'non-response bias'. A fairer investigation would track all participants from the start, measure weight change for everyone (including dropouts), use a control group who do not do the challenge, and calculate the mean difference between groups with a larger, random sample.
这个平均数只包含了完成并报告的用户，忽略了60%退出或未报告的用户；这些用户可能减重更少，甚至体重增加，从而拉低真实的平均值。这是“无应答偏差”。更公平的调查应从开始就追踪所有参与者，测量每个人（包括退出者）的体重变化，设立一个不参与挑战的对照组，并在更大规模的随机样本中计算组间平均差异。
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