📚 Year 9 AQA Statistics: Interdisciplinary Mixed Question Practice | 跨学科综合题型训练
Statistics is the science of collecting, organising, and interpreting data. In the AQA Year 9 curriculum, you must be able to apply statistical methods across a wide range of subjects – from science experiments and sports scores to business sales and environmental studies. This article provides a set of interdisciplinary mixed questions that mirror real-life situations, allowing you to practise everything from drawing stem-and-leaf diagrams to interpreting scatter graphs and calculating probabilities. Each section models the thought process and the key skills you need to master for your assessments.
统计是收集、整理和解读数据的科学。在 AQA 九年级课程中，你必须能够在广泛的学科中应用统计方法 —— 从科学实验和体育比分到商业销售和环境研究。本文提供了一系列跨学科综合题型，模拟真实生活情境，让你练习从绘制茎叶图到解读散点图以及计算概率的所有技能。每一节都示范了你在评估中需要掌握的关键思维过程和技能。
1. Collecting Data in Science Experiments | 科学实验中的数据收集
When investigating how light intensity affects plant growth, you must design a data collection table. Decide on your independent variable (light intensity measured in lux) and dependent variable (plant height in cm). Ensure you control variables such as water volume, soil type, and temperature. The data you gather should be recorded clearly, including units. For example, you might measure the height of 10 plants under low light and 10 plants under high light over 14 days.
在研究光照强度如何影响植物生长时，你必须设计一个数据收集表。确定你的自变量（以勒克斯为单位的光照强度）和因变量（以厘米为单位的植物高度）。确保你控制变量，例如水量、土壤类型和温度。你收集的数据应清晰记录，包括单位。例如，你可能在 14 天内测量 10 株低光照和 10 株高光照下的植物高度。
In an exam, a typical question might ask: ‘Describe two things that must be kept the same to make the experiment fair.’ The answer would include keeping the type of seed identical and giving the same amount of water daily. Another question could give a table of results and ask you to identify any anomalies – a value that does not fit the pattern – and suggest how to deal with it, usually by repeating the measurement.
在考试中，一个典型的问题可能是：“描述必须保持相同的两件事，以使实验公平。”答案将包括保持种子类型相同以及每天浇灌等量的水。另一个问题可能给出一个结果表，并要求你识别任何异常值 —— 一个不符合模式的值 —— 并建议如何处理它，通常是通过重复测量。
2. Stem-and-Leaf Diagrams for Sports Scores | 体育比分的茎叶图
Consider the points scored by a basketball team in 15 matches:
23, 34, 41, 29, 36, 52, 61, 45, 38, 49, 55, 28, 33, 47, 58.
A stem-and-leaf diagram organises this data. The stem represents the tens digit, and the leaf shows the units digit. Always provide a key, such as 2|3 represents 23 points.
考虑一支篮球队在 15 场比赛中的得分：23、34、41、29、36、52、61、45、38、49、55、28、33、47、58。茎叶图可以整理这些数据。茎代表十位数，叶代表个位数。始终提供一个图例，例如 2|3 表示 23 分。
| Stem | Leaf |
| 2 | 3, 8, 9 |
| 3 | 3, 4, 6, 8 |
| 4 | 1, 5, 7, 9 |
| 5 | 2, 5, 8 |
| 6 | 1 |
From the diagram, you can quickly find the median (the middle value) and the range. Ordering the leaves, we have 15 values. The median is the 8th value: 38 points. The range = highest – lowest = 61 – 23 = 38 points. The mode, the most frequent score, is 49 as it appears twice? Actually check: 49 appears only once, 23, 28, 29, 33, 34, 36, 38, 41, 45, 47, 49, 52, 55, 58, 61 – each appears once, so no mode. But if there were a repeated leaf, the mode is easy to spot.
从图中你可以快速找到中位数（中间值）和全距。将叶子排序后，我们有 15 个值。中位数是第 8 个值：38 分。全距 = 最大值 – 最小值 = 61 – 23 = 38 分。众数，即出现最频繁的分数，是 49 吗？实际上检查一下：49 只出现一次，23、28、29、33、34、36、38、41、45、47、49、52、55、58、61 —— 每个出现一次，所以没有众数。但如果有一个重复的叶值，众数就很容易发现。
3. Calculating Averages and Range from Geography Data | 从地理数据计算平均数与全距
The annual rainfall (in mm) for six cities is: London 557, Cairo 25, Mumbai 2200, Sydney 1217, Tokyo 1528, Nairobi 869. Calculate the mean, median, mode, and range. The mean = (557 + 25 + 2200 + 1217 + 1528 + 869) ÷ 6. The sum is 6396, so mean = 1066 mm. To find the median, order the data: 25, 557, 869, 1217, 1528, 2200. With an even number of values, the median is the average of the 3rd and 4th: (869 + 1217) ÷ 2 = 1043 mm. There is no mode as each value is unique. Range = 2200 – 25 = 2175 mm.
六个城市的年降雨量（毫米）为：伦敦 557，开罗 25，孟买 2200，悉尼 1217，东京 1528，内罗毕 869。计算平均数、中位数、众数和全距。平均数 = (557 + 25 + 2200 + 1217 + 1528 + 869) ÷ 6。总和为 6396，因此平均数为 1066 毫米。要找到中位数，将数据排序：25，557，869，1217，1528，2200。由于数值个数为偶数，中位数是第 3 和第 4 个值的平均数：(869 + 1217) ÷ 2 = 1043 毫米。由于每个值都唯一，没有众数。全距 = 2200 – 25 = 2175 毫米。
Geography questions often ask: ‘Explain why the mean is not a good representation of rainfall in Cairo.’ The mean is pulled up by extremely wet cities, so the median better represents the typical rainfall. Also, you might be given rainfall data in a bar chart and asked which average is affected by an outlier. Always justify your choice of average based on the context.
地理问题经常会问：“解释为什么平均数不能很好地代表开罗的降雨量。”平均数被极湿的城市拉高，因此中位数更能代表典型降雨量。此外，你可能会看到一个条形图中的降雨量数据，并被问到哪个平均数受异常值影响。始终根据上下文证明你选择的平均数是合理的。
4. Scatter Graphs and Correlation in Social Studies | 社会科学中的散点图与相关
A class collected data on the number of hours spent on social media per week and the score on a concentration test (out of 20). The paired data: (2, 18), (5, 16), (8, 14), (3, 19), (10, 11), (6, 15), (12, 9), (1, 20), (9, 13), (7, 14). Plot these on a scatter graph with ‘Social media hours’ on the x-axis and ‘Concentration score’ on the y-axis.
一个班级收集了每周花在社交媒体上的小时数和专注力测试得分（满分 20）的数据。配对数据为：(2, 18), (5, 16), (8, 14), (3, 19), (10, 11), (6, 15), (12, 9), (1, 20), (9, 13), (7, 14)。将这些点绘制在散点图上，x 轴为“社交媒体小时数”，y 轴为“专注力得分”。
The plotted points show a clear negative correlation: as social media use increases, concentration scores tend to decrease. You can draw a line of best fit that passes through the middle of the points. Do not connect the dots! The line should have roughly equal numbers of points above and below it. Use the line to estimate the score for someone who uses social media for 4 hours; read across from 4 on the x-axis up to the line, then across to the y-axis giving about 17.5.
绘制的点显示出明显的负相关：随着社交媒体使用时间增加，专注力得分趋于下降。你可以画一条通过各点中间的最佳拟合线。不要连接各点！这条线应该大致具有相同数量的点在其上方和下方。使用这条线估计一个使用社交媒体 4 小时的人的得分；从 x 轴上的 4 向上读到直线，然后水平读至 y 轴，得出大约 17.5。
Beware of extrapolation: predicting outside the data range. If you try to predict for 20 hours, the line may not be reliable because you have no data there. In an exam, you might be asked to describe the relationship (e.g., ‘strong negative correlation’) and point out any outliers that do not fit the pattern.
注意外推：预测数据范围之外的值。如果你试图预测 20 小时，这条线可能不可靠，因为你在那里没有数据。在考试中，你可能会被要求描述关系（例如，“强负相关”）并指出任何不符合规律的异常值。
5. Pie Charts and Percentages for Business Sales | 商业销售的饼图与百分比
A small café recorded the percentage of sales from four items: coffee 45%, sandwiches 30%, cakes 15%, and tea 10%. To draw a pie chart, convert each percentage to an angle by multiplying by 3.6°. Coffee: 45 × 3.6 = 162°; sandwiches: 108°; cakes: 54°; tea: 36°. Check that the angles sum to 360°. Draw a circle and use a protractor to mark each sector.
一家小咖啡馆记录了四种商品的销售额百分比：咖啡 45%，三明治 30%，蛋糕 15%，茶 10%。要绘制饼图，将每个百分比乘以 3.6° 转换为角度。咖啡：45 × 3.6 = 162°；三明治：108°；蛋糕：54°；茶：36°。检查角度总和为 360°。画一个圆，用量角器标出每个扇形。
In an interdisciplinary business context, you might be given a table of actual sales figures instead of percentages. First calculate the total sales: if the café sold 240 items in a day, each category frequency is used. For example, coffee sold 108 items (45% of 240), sandwiches 72, cakes 36, tea 24. Then calculate the angle for coffee: (108/240) × 360 = 162°, same as before. Questions often ask: ‘What fraction of total sales were cakes?’ Answer: 36/240 = 3/20. Also, be able to compare two pie charts with different totals and explain that a larger slice does not necessarily mean a larger number unless you know the total.
在跨学科的商业情境中，你可能会得到一个实际销售数据的表，而不是百分比。首先计算总销售：如果咖啡馆一天销售了 240 件商品，每个类别的频数分别为：咖啡 108 件（240 的 45%），三明治 72 件，蛋糕 36 件，茶 24 件。然后计算咖啡的角度：(108/240) × 360 = 162°，与之前相同。问题经常会问：“蛋糕占总销售的几分之几？”答案：36/240 = 3/20。此外，要能比较两个总数不同的饼图，并解释一个较大的扇形不一定意味着更大的数量，除非你知道总数。
6. Grouped Frequency Tables and Histograms for Health Data | 健康数据的分组频数表与直方图
Pulse rates (beats per minute) of 30 students were recorded:
56, 62, 71, 75, 68, 80, 82, 59, 64, 73, 78, 85, 90, 66, 69, 74, 77, 81, 88, 92, 63, 72, 76, 79, 84, 87, 67, 70, 83, 86.
Create a grouped frequency table using intervals 50–59, 60–69, 70–79, 80–89, 90–99. Tally the frequencies:
记录了 30 名学生的脉搏率（每分钟心跳次数）：56、62、71、75、68、80、82、59、64、73、78、85、90、66、69、74、77、81、88、92、63、72、76、79、84、87、67、70、83、86。使用区间 50–59、60–69、70–79、80–89、90–99 创建一个分组频数表。统计频数：
| Pulse rate (bpm) | Frequency |
| 50–59 | 2 |
| 60–69 | 8 |
| 70–79 | 10 |
| 80–89 | 8 |
| 90–99 | 2 |
To draw a histogram, note that for continuous data the bars touch. The class boundaries should be carefully considered: 50 ≤ rate < 60, 60 ≤ rate < 70, and so on. The height of each bar represents the frequency, and the width is the class interval. In an exam, you might need to estimate the mean from a grouped table: find the midpoint of each class (e.g., 54.5, 64.5, 74.5, 84.5, 94.5), multiply by the frequency, sum these products, and divide by total frequency. The estimated mean = (54.5×2 + 64.5×8 + 74.5×10 + 84.5×8 + 94.5×2) ÷ 30 = (109 + 516 + 745 + 676 + 189) ÷ 30 = 2235 ÷ 30 = 74.5 bpm.
要绘制直方图，注意对于连续数据，条形相互接触。组界限应仔细考虑：50 ≤ 心率 < 60，60 ≤ 心率 < 70，依此类推。每个条形的高度代表频数，宽度是组距。在考试中，你可能需要从分组表格中估计平均数：找到每个组的组中值（例如，54.5、64.5、74.5、84.5、94.5），乘以频数，将这些乘积相加，再除以总频数。估计的平均数 = (54.5×2 + 64.5×8 + 74.5×10 + 84.5×8 + 94.5×2) ÷ 30 = (109 + 516 + 745 + 676 + 189) ÷ 30 = 2235 ÷ 30 = 74.5 bpm。
Health-related questions often ask which interval contains the median. The median is the (30+1)/2 = 15.5th value, so the median class is 70–79. This links statistics to physical education concepts.
与健康相关的问题经常问哪个区间包含中位数。中位数是第 (30+1)/2 = 15.5 个值，因此中位数组是 70–79。这将统计与体育概念联系起来。
7. Box Plots Comparing Temperatures in Different Regions | 箱线图比较不同地区气温
Monthly average maximum temperature (°C) for London: 8, 8, 11, 14, 17, 20, 22, 22, 19, 15, 11, 9.
Order the data: 8, 8, 9, 11, 11, 14, 15, 17, 19, 20, 22, 22.
Five-number summary: minimum = 8, Q₁ (lower quartile) = median of lower half (8,8,9,11,11,14) = (9+11)/2 = 10, median = (14+15)/2 = 14.5, Q₃ (upper quartile) = median of upper half (17,19,20,22,22) = 20, maximum = 22. Draw a box plot with a scale.
伦敦月平均最高气温（℃）：8、8、11、14、17、20、22、22、19、15、11、9。排序数据：8、8、9、11、11、14、15、17、19、20、22、22。五数概括：最小值 = 8，Q₁（下四分位数）= 下半部分的中位数 (8,8,9,11,11,14) = (9+11)/2 = 10，中位数 = (14+15)/2 = 14.5，Q₃（上四分位数）= 上半部分的中位数 (17,19,20,22,22) = 20，最大值 = 22。使用比例尺绘制箱线图。
Now do the same for Sydney: (Jan–Dec): 26, 26, 25, 22, 19, 17, 16, 18, 20, 22, 24, 25. Ordered: 16,17,18,19,20,22,22,24,25,25,26,26. Min = 16, Q₁ = (18+19)/2 = 18.5, median = (22+22)/2 = 22, Q₃ = (25+25)/2 = 25, max = 26. Comparing the box plots shows Sydney is warmer overall with a smaller interquartile range (IQR = Q₃ – Q₁ = 6.5°C) compared to London (IQR = 20 – 10 = 10°C), meaning London’s temperatures are more spread out.
现在对悉尼做同样的处理：(1月–12月)：26、26、25、22、19、17、16、18、20、22、24、25。排序后：16,17,18,19,20,22,22,24,25,25,26,26。最小值 = 16，Q₁ = (18+19)/2 = 18.5，中位数 = (22+22)/2 = 22，Q₃ = (25+25)/2 = 25，最大值 = 26。将箱线图进行比较，可以看出悉尼总体更温暖，四分位距 (IQR = Q₃ – Q₁ = 6.5°C) 比伦敦 (IQR = 20 – 10 = 10°C) 更小，这意味着伦敦的温度更分散。
Geography questions may ask: ‘What percentage of months in London have a maximum temperature above 20°C?’ From the data, only two months (July and August, 22°C each) are above 20°C, so 2/12 ≈ 16.7%. This type of question links box plots to proportional reasoning.
地理问题可能会问：“伦敦有百分之几的月份最高气温高于 20°C？”根据数据，只有两个月（七月和八月，各 22°C）高于 20°C，所以比例是 2/12 ≈ 16.7%。这类问题将箱线图与比例推理联系起来。
8. Time Series Graphs for Environmental Data | 环境数据的时间序列图
The amount of plastic waste recycled by a school each term (kg) from 2021 to 2024 is recorded:
Spring 21: 45, Summer 21: 52, Autumn 21: 48, Spring 22: 55, Summer 22: 60, Autumn 22: 58, Spring 23: 63, Summer 23: 70, Autumn 23: 68, Spring 24: 72, Summer 24: 78.
Plot these as points on a time series graph with time on the x-axis. Join the points with straight lines to show the trend.
记录了 2021 年至 2024 年一所学校每学期回收的塑料废物量（千克）：2021 年春：45，2021 年夏：52，2021 年秋：48，2022 年春：55，2022 年夏：60，2022 年秋：58，2023 年春：63，2023 年夏：70，2023 年秋：68，2024 年春：72，2024 年夏：78。将这些数据作为点绘制在时间序列图上，时间在 x 轴上。用直线连接各点以显示趋势。
The graph reveals an overall upward trend, though there are slight seasonal variations (e.g., a drop in Autumn 2021). You can comment on the long-term increase, which suggests that recycling efforts are improving. In exams, you might be asked to predict the amount for Autumn 2024 using the trend line. Extend the line and read the approximate value, say 76 kg. However, always be cautious about seasonal effects; Autumn values tend to dip slightly.
该图显示出总体上升趋势，尽管有轻微的季节性变化（例如 2021 年秋季下降）。你可以评论长期增长，这表明回收工作正在改善。在考试中，你可能会被要求使用趋势线预测 2024 年秋季的数量。延长该线并读出近似值，比如说 76 千克。然而，总要当心季节效应；秋季数值往往会略有下降。
Time series graphs are common in environmental science and geography. You must label axes clearly, use an appropriate scale, and understand the difference between a trend and a seasonal fluctuation. A question might ask: ‘Why might the amount drop in Autumn 2021?’ Possible reasons: school holidays reducing collection, or a specific one-off event.
时间序列图在环境科学和地理中很常见。你必须清楚地标注坐标轴，使用合适的比例尺，并理解趋势与季节性波动之间的区别。一个问题可能会问：“为什么 2021 年秋季的数量可能下降？”可能的原因：学校假期减少了收集量，或某个一次性事件。
9. Probability from Genetic Crosses | 遗传杂交中的概率
In biology, you study how traits are inherited. Consider pea plants where the allele for tall stem (T) is dominant over short stem (t). Two heterozygous plants (Tt) are crossed. Use a Punnett square (a sample space diagram) to find the probability that an offspring is short. The possible combinations from each parent’s gametes are T and t.
在生物学中，你研究性状是如何遗传的。考虑茎高性状，其中高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。两个杂合植株 (Tt) 杂交。使用庞纳特方格（样本空间图）求出一个后代为矮茎的概率。每个亲本配子的可能组合是 T 和 t。
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
The grid shows four equally likely outcomes: TT, Tt, Tt, tt. Only the genotype tt gives a short stem because it has both recessive alleles. Therefore, P(short) = 1/4. P(tall) = 3/4. This can also be expressed as a ratio 3:1. This is a classic application of probability in genetics.
这个表格显示了四个等可能的结果：TT、Tt、Tt、tt。只有基因型 tt 会产生矮茎，因为它具有两个隐性等位基因。因此，P(矮) = 1/4。P(高) = 3/4。这也可以表示为 3:1 的比例。这是概率在遗传学中的经典应用。
Exam questions may ask: ‘If the plants produce 20 offspring, about how many would you expect to be tall?’ Expected number = 3/4 × 20 = 15. You could also be asked to explain the difference between experimental probability and theoretical probability if the actual results were 12 tall and 8 short, discussing the role of sample size.
考试问题可能会问：“如果这些植株产生 20 个后代，你预计大约有多少个是高茎？”期望值 = 3/4 × 20 = 15。如果实际结果是 12 高 8 矮，你也可能被要求解释实验概率和理论概率之间的差异，并讨论样本量的作用。
10. Interpreting Dual Bar Charts in Geography | 地理中的复式条形图解读
A dual bar chart compares the percentage of the population living in urban areas in two countries, Nigeria and UK, from 1980 to 2020 at ten-year intervals. The approximate data:
1980: UK 78%, Nigeria 22%; 1990: UK 80%, Nigeria 30%; 2000: UK 82%, Nigeria 35%; 2010: UK 83%, Nigeria 45%; 2020: UK 84%, Nigeria 52%.
一个复式条形图比较了尼日利亚和英国两个国家从 1980 年到 2020 年每十年城市人口百分比。近似数据为：1980 年：英国 78%，尼日利亚 22%；1990 年：英国 80%，尼日利亚 30%；2000 年：英国 82%，尼日利亚 35%；2010 年：英国 83%，尼日利亚 45%；2020 年：英国 84%，尼日利亚 52%。
When the bars are drawn side by side, you can easily compare the urbanisation levels. The UK has a high and slowly increasing urban percentage, whereas Nigeria’s urban population has grown rapidly, more than doubling from 22% to 52%. The gap between the two countries has narrowed from 56% in 1980 to 32% in 2020.
当条形并排绘制时，你可以轻松地比较城市化水平。英国的城市人口比例很高且缓慢增长，而尼日利亚的城市人口增长迅速，从 22% 增加到 52%，翻了一番多。两国之间的差距从 1980 年的 56% 缩小
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