📚 KS3 OCR Statistics: Interdisciplinary Mixed Question Practice | KS3 OCR 统计：跨学科综合题型训练
Statistics is the one topic that follows you into every classroom. You record plant growth in science, plot rainfall in geography, analyse reaction times in PE, and compare turnout figures in history. This guide builds a complete set of interdisciplinary mixed questions for the OCR KS3 curriculum, so you practise collecting data, choosing the right chart, calculating averages and spread, handling grouped data, judging measurement error, interpreting probability in real contexts, and writing answers that earn full marks.
统计学是唯一会跟着你走进每一间教室的知识。你在科学课上记录植物生长，在地理课上绘制降雨量，在体育课上分析反应时间，在历史课上比较投票率。本指南为 OCR KS3 课程搭建了一整套跨学科综合题型，让你练习收集数据、选择合适的图表、计算平均数与离散程度、处理分组数据、判断测量误差、在真实情境中解读概率，并写出能拿满分的答案。
1. Why Statistics Travels Across Subjects | 为什么统计学贯穿各学科
Statistics is a tool subject rather than a content subject. Geography needs it to describe population and climate; science needs it to decide whether a result is real or just noise; history and citizenship need it to compare social data fairly. In the OCR KS3 specification, the assessed skills are always the same four steps, whatever the context looks like.
统计学是工具学科，而不是内容学科。地理需要它描述人口与气候；科学需要它判断一个结果是真实存在还是只是噪声；历史与公民课需要它公平地比较社会数据。在 OCR KS3 的考试说明中，无论情境如何变化，被考查的技能始终是同样的四个步骤。
Collect: design a table, choose suitable units, record data accurately.
收集：设计表格、选择合适的单位、准确记录数据。
Process: group data, calculate averages, measures of spread and relative frequency.
处理：对数据分组，计算平均数、离散程度度量和相对频数。
Represent: pick a chart that matches the type of data and label it fully.
呈现：选择与数据类型匹配的图表，并完整标注。
Interpret: describe the pattern, compare groups and comment on reliability.
解读：描述规律、比较各组，并评价数据的可靠性。
Mixed questions simply swap the context while keeping those four steps. If you can name the step you are on, you rarely get lost, even when the question mentions enzymes, migration or cricket scores.
综合题型只是更换情境，却保留这四个步骤。只要你能说清自己正处在哪一步，即使题目提到酶、人口迁移或板球比分，你也很少会迷失方向。
2. Collecting and Organising Data in Science Experiments | 收集和整理科学实验中的数据
In any science investigation, the first step is to decide what data to gather and how to record it clearly. A well-structured table with headings and units prevents mistakes later on. For example, when measuring the bounce height of a ball dropped from different heights, you might use a table with columns "Drop height (cm)", "Bounce 1 (cm)", "Bounce 2 (cm)" and "Mean bounce (cm)".
在任何科学探究中，第一步都是决定收集哪些数据以及如何清晰地记录。一张标题明确并带有单位的表格可以避免之后的错误。例如，测量从不同高度落下的球的反弹高度时，你可以使用包含 "下落高度 (cm)"、"第一次反弹 (cm)"、"第二次反弹 (cm)"、"平均反弹 (cm)" 这几栏的表格。
The three rules for a good data table are simple: the independent variable goes in the first column, all readings are in the same units, and repeated readings sit side by side so they can be averaged. Write the unit in the heading, not next to every number.
一张好的数据表有三条简单规则：自变量放第一栏；所有读数的单位一致；重复读数并排放置，方便取平均。单位写在标题里，而不是写在每个数字旁边。
| Drop height (cm) | Bounce 1 (cm) | Bounce 2 (cm) | Mean bounce (cm) |
|---|---|---|---|
| 20 | 13 | 15 | 14 |
| 40 | 27 | 27 | 27 |
| 60 | 41 | 43 | 42 |
| 80 | 55 | 54 | 54.5 |
Tally charts are a quick way to record categorical data, such as the colour of flowers in a field. You can group tallies in fives using a diagonal stroke through four vertical lines. This makes counting the total frequency much faster and reduces counting errors.
计数表是记录类别数据的快捷方式，例如田野中花朵的颜色。你可以用四竖一横的划线法每五个一组来标记。这样能让总数统计快得多，也减少计数错误。
3. Grouped Frequency Tables and Class Intervals | 分组频数表与组距
Once data is collected, you often need to organise large sets of numbers into a grouped frequency table. Choose equal class intervals, such as 0-9, 10-19, 20-29, so that patterns appear without the data becoming unmanageable. Unequal intervals make comparisons between groups unfair and lose marks.
收集完数据后，你常常需要把大量数字整理成分组频数表。选择相等的组距，如 0-9、10-19、20-29，既能揭示模式，又不会让数据变得难以处理。组距不相等会让组间比较不公平，也会丢分。
Two technical points matter in every grouped table. First, intervals such as 0-9 and 10-19 must not overlap: a value of 10 belongs only to the second group. Second, write the class mid-point for each group, because it is needed later to estimate the mean.
每张分组表都有两个技术要点。第一，像 0-9 与 10-19 这样的区间不能重叠：数值 10 只属于第二组。第二，为每组写出组中值，因为后面估算平均数时要用到它。
Class mid-point = (lower boundary + upper boundary) ÷ 2
29) ÷ 2 = 24.5. Always check that the class widths really are equal before you calculate.
对区间 20-29，组中值是 (20 + 29) ÷ 2 = 24.5。计算之前一定要检查各组的组距是否真的相等。
4. Displaying Data: Charts for Geography and Science | 展示数据：地理和科学中的图表
Bar charts are ideal when you want to compare distinct categories, such as the number of plant species found in four different habitats, or the population of five counties. The bars are separated by gaps because the categories are not continuous.
条形图非常适合比较互不相同的类别，比如四个不同栖息地中发现的植物物种数，或五个郡的人口。柱子之间留有空隙，因为这些类别不是连续的。
For data measured over time, use a line graph. Monthly temperature or population change plotted as a line lets the reader see a trend, which a bar chart hides. Continuous grouped data, such as heights of 200 students, is best shown as a histogram, where the bars touch because the variable is continuous.
对于随时间测量的数据，要用折线图。把每月气温或人口变化画成折线，读者能看到趋势，而条形图会掩盖这一点。连续的分组数据，例如 200 名学生的身高，最好用直方图展示，柱子相互接触，因为变量是连续的。
When two variables might be linked, a scatter graph is the right choice. Plot one variable on each axis and look for a positive, negative or no correlation. This is exactly the graph scientists use for height against shoe size, or temperature against enzyme activity.
当两个变量可能存在关联时，散点图是正确的选择。两个轴各放一个变量，观察是正相关、负相关还是无相关。这正是科学家用来画身高与鞋码、或温度与酶活性关系的那种图。
A pie chart shows parts of a single whole, and works only if the frequencies add up to the total.
饼图展示同一个整体的各部分，只有在频数之和等于总数时才适用。
Every graph needs a title, both axes labelled with units, and an even scale.
每张图都需要标题、两个带单位的坐标轴标签，以及均匀的刻度。
Never break the scale axis to make a difference look bigger; it misleads the reader.
不要为了放大差异而断开坐标轴刻度；这会误导读者。
5. Averages and Spread: Mean, Median, Mode and Range | 平均数与离散程度：平均数、中位数、众数和极差
Three averages describe the centre of a data set, and each answers a slightly different question. The mean uses every value, the median is not affected by extreme values, and the mode is the only average that works for categorical data such as favourite sport.
三种平均数描述数据集的中心，各自回答略有不同的问题。平均数用到了每一个数值；中位数不受极端值影响；众数是唯一适用于类别数据（例如最喜欢的运动）的平均数。
Mean = Σx ÷ n Range = highest value − lowest value
h and 5th values: (27 + 41) ÷ 2 = 34 cm. The mode is 27, and the range is 55 − 13 = 42 cm.
以反弹数据 13、15、27、27、41、43、54、55 为例。总和是 275，共 8 个数值，所以平均数是 34.375 cm。中位数是第 4 与第 5 个数值的平均数：(27 + 41) ÷ 2 = 34 cm。众数是 27，极差是 55 − 13 = 42 cm。
Always report spread as well as an average. Two classes can share the same mean of 60 marks, yet one may have a range of 10 while the other has a range of 55. The second class is far less consistent, and that conclusion is worth a mark in every subject.
报告平均数时一定要同时报告离散程度。两个班级的平均分可能都是 60 分，但一个班的极差是 10，另一个班的极差是 55。第二个班级的稳定性差得多，而这个结论在任何学科都值一分。
6. Estimating the Mean from Grouped Data | 由分组数据估算平均数
When data has already been grouped, the individual values are lost, so you can only estimate the mean. The method is to assume every value in a group sits at the class mid-point, then multiply each mid-point by its frequency.
当数据已经被分组时，具体的单个数值已经丢失，所以你只能估算平均数。方法是假设每组中所有数值都位于组中值处，然后把每个组中值乘以它的频数。
Estimated mean = Σ(f × m) ÷ Σf
Here f is the frequency of a group and m is its mid-point. The symbol Σ means "add up all of them". You should also identify the modal class, which is simply the group with the highest frequency, because that is what a grouped table can tell you about the mode.
这里 f 是某一组的频数，m 是它的组中值。符号 Σ 表示 "把它们全部相加"。你还应找出众数所在组，也就是频数最高的那一组，因为分组表能告诉你的众数信息就只有这个。
| Time (minutes) | Mid-point m | Frequency f | f × m |
|---|---|---|---|
| 0-9 | 4.5 | 6 | 27 |
| 10-19 | 14.5 | 10 | 145 |
| 20-29 | 24.5 | 4 | 98 |
giving an estimated mean of 270 ÷ 20 = 13.5 minutes. The modal class is 10-19 minutes. Say "estimated" in your answer: it shows the examiner that you know grouped data has lost information.
总计 Σf = 20，Σ(f × m) = 270，因此估算平均数为 270 ÷ 20 = 13.5 分钟。众数所在组是 10-19 分钟。答案中要写 "估算"，这能向阅卷人表明你知道分组数据已经丢失了信息。
7. Measurement, Error and Significant Figures in Science | 科学中的测量、误差与有效数字
nt and averaging reduces random error, which is why the bounce experiment in Section 2 takes two readings at each height.
任何测量都不精确。如果你用刻度尺读到最近的厘米，每个读数都带有约 ± 0.5 cm 的不确定度。重复测量并取平均可以减少随机误差，这正是第 2 节反弹实验在每一高度取两次读数的原因。
A reading that is very different from the others, such as 41 cm when every other value is near 27 cm, is an outlier. Check whether it was caused by a mistake in the method; if it was, discard it and explain why. Do not remove a value merely because it looks inconvenient.
与其它数值差别很大的读数，例如当其它值都接近 27 cm 时出现的 41 cm，就是异常值。要检查它是否由方法上的错误造成；如果是，就剔除它并说明理由。不要仅仅因为某个数值看起来碍事就把它去掉。
Report a mean to roughly the same precision as the readings, for example 14 cm rather than 14.333 cm.
报告平均数时，精度应与原始读数大致相同，例如写 14 cm 而不是 14.333 cm。
Never quote more significant figures than your instrument can justify.
有效数字不要多于你的仪器所能支撑的程度。
State units in every answer; a number without a unit is incomplete.
每个答案都要写单位；没有单位的数值是不完整的。
8. Probability in Real Contexts | 真实情境中的概率
Theoretical probability assumes equally likely outcomes, so the probability of rolling a 4 on a fair dice is 1 ÷ 6. Experimental probability, also called relative frequency, is calculated from data you actually collected.
理论概率假定所有结果等可能，因此掷一枚均匀骰子得到 4 的概率是 1 ÷ 6。实验概率也称为相对频数，是由你实际收集到的数据计算出来的。
e outcomes ÷ total outcomes Relative frequency = frequency ÷ number of trials
Genetics predicts a theoretical value of 0.75 for the dominant purple trait, so the experimental result is close but not identical. As the number of trials increases, the experimental value generally moves closer to the theoretical one.
如果在科学课上种出的 80 株豌豆中有 58 株开紫花，相对频数就是 58 ÷ 80 = 0.725。遗传学预测显性紫花性状的理论值为 0.75，所以实验结果接近但并不完全相同。随着试验次数增加，实验值通常会越来越接近理论值。
lusive outcomes is exactly 1, which is how you find the probability of an event not happening: P(not A) = 1 − P(A).
概率始终满足 0 ≤ P ≤ 1。如果你算出 1.2 或负数，说明某处出错了。所有互斥结果的概率之和恰好为 1，这也正是求某事件不发生概率的方法：P(不发生 A) = 1 − P(A)。
9. Surveys, Sampling and Bias | 调查、抽样与偏差
When you cannot question or measure everyone, you take a sample from the population. A good sample must be large enough to be reliable and selected at random, so that every member of the population has an equal chance of being chosen.
当你无法询问或测量每一个人时，就要从总体中抽取样本。好的样本必须足够大以保证可靠性，并且是随机抽取的，使总体中每个个体被选中的机会相同。
Bias creeps in through the way the sample is chosen or through the wording of a question. Asking students queuing for a football match whether they enjoy sport will overestimate the answer, because the sample is not representative. Leading questions such as "Do you agree that homework should be reduced?" push the respondent towards one answer.
偏差会通过样本选取方式或问题措辞悄悄混入。在足球比赛排队的人群中询问学生是否喜欢运动，会高估答案，因为这个样本不具代表性。像 "你是否同意应该减少作业？" 这样的诱导性问题会把受访者推向某个答案。
Bad sampling method: the sample is unrepresentative, for example only boys in Year 9.
糟糕的抽样方法：样本不具代表性，例如只调查九年级男生。
Bad question wording: it is leading, ambiguous, or offers overlapping response boxes.
糟糕的问题措辞：带有诱导性、含糊不清，或者选项区间相互重叠。
10. Interdisciplinary Mixed Question Set | 跨学科综合题组
Work through the four questions below without a calculator where possible, and write down the method as well as the answer. Each question deliberately mixes skills from more than one subject area.
请尽可能不用计算器完成下面四道题，并且把方法与答案一起写下来。每道题都刻意混合了不止一个学科领域的技能。
Question 1 (Science and Physics). A ball is dropped from 20 cm, 40 cm, 60 cm and 80 cm. The mean bounce heights are 14 cm, 27 cm, 42 cm and 54.5 cm. Calculate the range of the mean bounce heights and describe the relationship between drop height and bounce height.
第 1 题（科学与物理）。一个球分别从 20 cm、40 cm、60 cm、80 cm 的高度落下，平均反弹高度分别为 14 cm、27 cm、42 cm 和 54.5 cm。计算平均反弹高度的极差，并描述下落高度与反弹高度之间的关系。
Question 2 (Geography and Climate). Five daily maximum temperatures in January were 4 °C, 6 °C, 3 °C, 5 °C and 2 °C. Five daily maxima in July were 18 °C, 21 °C, 19 °C, 24 °C and 23 °C. Calculate the mean and range for each month and compare the two climates.
第 2 题（地理与气候）。一月份五个日最高气温为 4 °C、6 °C、3 °C、5 °C 和 2 °C。七月份五个日最高气温为 18 °C、21 °C、19 °C、24 °C 和 23 °C。分别计算每个月的平均数和极差，并比较这两种气候。
Question 3 (Biology and Probability). In a genetics experiment, 80 pea plants are grown: 58 have purple flowers and 22 have white flowers. Calculate the relative frequency of purple flowers and compare it with the theoretical value of 0.75 found in Section 8.
第 3 题（生物与概率）。在一项遗传实验中种了 80 株豌豆：58 株开紫花，22 株开白花。计算紫花的相对频数，并与第 8 节给出的理论值 0.75 进行比较。
Question 4 (Design and Statistics). A student wants to estimate how much screen time Year 9 students have each day. She asks 15 students in the school library at lunchtime. Give one reason why her sample may be biased and suggest a better method.
第 4 题（设计与统计）。一名学生想估算九年级学生每天使用屏幕的时长。她在午餐时间询问了学校图书馆里的 15 名学生。请给出她的样本可能带有偏差的一个理由，并提出一个更好的方法。
11. Worked Solutions and Marking Points | 详解与评分要点
5 cm. The relationship is positive and roughly linear: as the drop height increases, the mean bounce height increases, at approximately 0.7 cm of bounce for every 1 cm of drop height.
第 1 题。极差为 54.5 − 14 = 40.5 cm。关系是正相关且大致为线性：随着下落高度增加，平均反弹高度也增加，大致每下落 1 cm 反弹约 0.7 cm。
he mean is 20 ÷ 5 = 4 °C, and the range is 6 − 2 = 4 °C. July: the sum is 105, so the mean
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