📚 Year 10 CCEA Statistics: In-Depth Analysis of Past Exam Questions | Year 10 CCEA 统计：历年真题深度解析
The CCEA Year 10 Statistics course builds a strong foundation in data handling, probability, and statistical reasoning. Analysing past exam questions is one of the most effective ways to understand the exam structure, identify common question types, and reinforce key concepts. This article provides a comprehensive breakdown of frequently tested topics, with worked examples and examiner insights to help you excel.
CCEA Year 10 统计课程为数据处理、概率和统计推理打下坚实基础。分析历年真题是理解考试结构、识别常见题型并巩固关键概念最有效的方法之一。本文对常考专题进行全面解析，配合典型例题和考官视角的点评，助你取得优异成绩。
1. Sampling Methods and Data Collection | 抽样方法与数据收集
A classic past-paper question asks: 'A school with 1200 students wants to select a sample of 60 to investigate attitudes toward homework. Describe how to obtain (a) a simple random sample and (b) a stratified sample by year group.' Understanding these methods is vital, as examiners look for precise descriptions using random number generators and proportional allocation.
一道经典真题问：“一所拥有1200名学生的学校希望选取60名学生作为样本，调查对家庭作业的态度。描述如何获得 (a) 简单随机样本 和 (b) 按年级分层样本。” 理解这些方法至关重要，因为考官要求精确描述随机数生成器使用和比例分配。
For a simple random sample, every student must have an equal chance of selection. Assign each student a unique number from 1 to 1200, then use a random number generator or random number tables to pick 60 distinct numbers; ignore repeats. For a stratified sample, first divide students into year groups (e.g., Year 8 to Year 12), calculate the proportion in each group relative to 1200, and then multiply that proportion by 60 to get the sample size for that stratum. Randomly select that number from each year group.
对于简单随机样本，每个学生必须有等同的选中机会。为每名学生分配1至1200的唯一编号，然后用随机数生成器或随机数字表选出60个不同的号码，忽略重复。对于分层样本，首先将学生按年级分组（例如 Year 8 至 Year 12），计算每组人数占1200的比例，再将比例乘以60得到该层样本量，最后从每个年级组中随机抽取相应人数。
Examiners frequently penalise vague language like 'pick names out of a hat' without describing equal chance, or failing to mention that stratified sampling reflects the population structure. Always specify the tool (e.g., random number generator) and state that stratification ensures each year group is fairly represented.
考官常因用语模糊而扣分，比如只说“从帽子里抽名字”却不描述等可能性，或未提及分层抽样反映总体结构。务必明确工具（如随机数生成器），并说明分层确保了每个年级组都被公平代表。
2. Data Presentation: Bar Charts and Pie Charts | 数据展示：条形图与饼图
pical exam problem provides a frequency table, for example: Favourite sport of 60 students – ector angles using (frequency / total) × 360. For Football: (24/60) × 360 = 144°. Tennis: 72°, Swimming: 108°, Other: 36°. Always check that angles sum to 360°.
典型考题提供频数表，例如：60名学生最喜爱的运动——足球24，网球12，游泳18，其他6。(a) 要求绘制饼图，(b) 解释饼图为何合适。你必须用 (频数 / 总数) × 360 计算扇形角度。足球：(24/60) × 360 = 144°；网球：72°；游泳：108°；其他：36°。始终检查角度总和是否为360°。
When drawing, label each sector clearly or provide a key, and use a protractor accurately. Compound bar charts also appear: the question might give data on boys' and girls' sport choices and ask for a comparative bar chart. Marks are awarded for correct axes, equal bar widths, consistent scales, and a legend. A common error is drawing a bar chart without spaces between bars for categorical data – remember, discrete categories require equal gaps.
绘制时清晰标注各扇形或提供图例，并准确使用量角器。复合条形图也常出现：题目可能给出男生女生运动选择数据，要求画对比条形图。得分点在于正确的坐标轴、等宽的条形、一致的比例尺和图例。常见错误是分类数据的条形之间不留空隙——记住，离散类别需要等距间隔。
3. Frequency Distributions and Cumulative Frequency | 频率分布与累积频率
A past question gave the heights of 80 plants: 0-10 cm (5), 10-20 (12), 20-30 (20), 30-40 (18), 40-50 (15), 50-60 (10). Students had to complete a cumulative frequency table, plot a cumulative frequency curve, and estimate the median and interquartile range. Start by adding a cumulative frequency column: 5, 17, 37, 55, 70, 80. Plot the upper class boundary against cumulative frequency: (10,5), (20,17), (30,37), (40,55), (50,70), (60,80). Join points with a smooth curve.
一道真题给出80株植物的高度：0-10厘米 (5株)，10-20 (12)，20-30 (20)，30-40 (18)，40-50 (15)，50-60 (10)。学生需补全累积频数表，绘制累积频数曲线，并估算中位数和四分位距。先添加累积频数列：5, 17, 37, 55, 70, 80。将上组界对累积频数描点：(10,5)，(20,17)，(30,37)，(40,55)，(50,70)，(60,80)。用光滑曲线连接各点。
lf of 80) on the cumulative frequency axis to the curve, then drop down to the height axis – 7 - 22.5 = 14.5 cm. Examiners expect sensitivity to the scale when reading off, and answers within a tolerance range are accepted.
找中位数时，从累积频数轴上40（80的一半）画水平线与曲线相交，再向下读取高度轴——约29厘米。下四分位数从20处（40/2）横读得约22.5厘米；上四分位数从60处读得约37厘米。四分位距 = 37 - 22.5 = 14.5厘米。考官期望读数与坐标尺度吻合，并接受一定容差范围内的答案。
4. Measures of Central Tendency: Mean, Median, Mode | 集中趋势：平均数、中位数、众数
Consider a data set from a past paper: test scores of 8 students – 45) ÷ 8 = 153 ÷ 8 = 19.125. The median, with data already ordered, is the average of the 4th and 5th values: (15+16) ÷ 2 = 15.5. The mode is 14 (appears twice).
回顾一道真题数据：8名学生的测验分数——12, 14, 14, 15, 16, 17, 20, 45。计算三种平均数，并判断哪一个最能代表典型分数。平均数 = (12+14+14+15+16+17+20+45) ÷ 8 = 153 ÷ 8 = 19.125。数据已排序，中位数是第4与第5个值的平均：(15+16) ÷ 2 = 15.5。众数为14（出现两次）。
pulls the mean upward, making it less representative of the majority who scored around 14–17. Hence the median or mode is more representative. The exam often asks: 'Give a reason for your choice.' quency data, mean is estimated using midpoints: ∑(frequency × midpoint) ÷ total frequency.
极端值45拉高了平均数，使其不能代表大多数14–17分的同学，因此中位数或众数更具代表性。考题常问：“给出你的选择理由。” 优秀答案会解释异常值的影响。对于分组频数数据，平均数用组中值估算：∑(频数 × 组中值) ÷ 总频数。
5. Measures of Dispersion: Range and Quartiles | 离散度：范围与四分位数
as crucial as central tendency. A past question gave: waiting times (minutes) at a clinic – ange, lower quartile (Q1), upper quartile (Q3), and interquartile range (IQR). Range = max – min = 30 – edian 14 → Q1 = 14. Upper half (last 5): 18, 21, 23, 25, 30 has median 23 → Q3 = 23. IQR = 23 – 14 = 9 minutes.
离散度与集中趋势同等重要。一道真题给出：诊所候诊时间（分钟）——10, 12, 14, 15, 17, 18, 21, 23, 25, 30。求范围、下四分位数(Q1)、上四分位数(Q3)和四分位距(IQR)。范围 = 最大值 – 最小值 = 30 – 10 = 20分钟。对于四分位数，10个数的中位数在第5个(17)和第6个(18)之间 → 17.5。低半区（前5个）：10, 12, 14, 15, 17的中位数为14 → Q1 = 14。高半区（后5个）：18, 21, 23, 25, 30的中位数为23 → Q3 = 23。IQR = 23 – 14 = 9分钟。
Some CCEA questions use (n+1) ÷ 2 positioning for quartiles within a cumulative frequency context. However, for a simple list, splitting at the median works. The IQR is preferred over range because it is not affected by extreme values. An exam tip: always list data in order and clearly identify the boundary between the lower and upper halves.
部分CCEA题目在累积频率背景下使用 (n+1) ÷ 2 定位四分位数。但对于简单数列，在中位数处分割即可。IQR比范围更受青睐，因为它不受极端值影响。考试技巧：始终将数据排序，并明确标示低半区与高半区的分界。
6. Box Plots | 箱线图
er provided the five-number summaries for Maths and English test scores (out of 50): Maths – min=12, Q1=20, median=28, Q3=35, max=42; English – t the boxes. For Maths, the box extends from 20 to 35 with the median line at 28; whiskers to 12 and 42. For English, box 24–38, median at 30, whiskers to 15 and 45.
箱线图是用于比较分布的经典题型。一份真题提供了数学与英语测验分数（满分50）的五数概括：数学 – 最小值=12，Q1=20，中位数=28，Q3=35，最大值=42；英语 – 最小值=15，Q1=24，中位数=30，Q3=38，最大值=45。要求在同一尺度上画出两个箱线图，并做两项比较。先画坐标轴，再绘制箱体。数学的箱体从20到35，中位线在28处；胡须线延伸至12和42。英语箱体24–38，中位线30，胡须至15和45。
Comparisons should refer to median and spread (IQR or range). For example: 'On average, English scores were higher, as the median (30) is above the Maths median (28).' ' lightly smaller than that of Maths (30), but the IQR of English (14) is greater than Maths's IQR (15), indicating more variation in the middle 50% for Maths.' quick check: Maths IQR = 35-20=15, English IQR=38-24=14. Good comparisons earn full marks.
比较应涉及中位数和离散程度（IQR或范围）。例如：“平均而言，英语分数更高，因为英语中位数(30)高于数学中位数(28)。”“英语分数的范围(30)与数学(30)相同，但数学的IQR(15)大于英语的IQR(14)，说明数学中间50%的变异更大。”精准的比较能获满分。
7. Scatter Graphs and Correlation | 散点图与相关
Past paper data: outdoor temperature (°C) and number of ice creams sold at a kiosk. Temperature: 18, 22, 25, 28, 15, 20, 21, 27, 30, 17; Sales: 45, 60, 78, 85, 40, 55, 63, 83, 92, 48. Plot the points on a scatter graph. The pattern shows a clear upward trend – as temperature increases, sales increase. Describe correlation as 'strong positive correlation'. You might also be asked to estimate a value from the graph or to draw a line of best fit in a later part.
真题数据：室外温度(°C)与售货亭冰淇淋销售量。温度：18, 22, 25, 28, 15, 20, 21, 27, 30, 17；销量：45, 60, 78, 85, 40, 55, 63, 83, 92, 48。在散点图上描点。图形呈现明显上升趋势——温度升高，销量增加。将相关性描述为“强正相关”。后续题目可能要求根据图形估算数值或画最佳拟合线。
When describing correlation, use standard phrases: 'positive' (as one variable rises, the other tends to rise), 'negative' (one rises, the other falls), or 'no correlation'. Strength can be given as 'strong', 'moderate', or 'weak'. An examiner expects you to relate the context – here, warmer weather leads to more ice cream sales. Do not just say 'positive'; explain what it means in the situation.
描述相关时使用标准表述：“正相关”（一变量升高，另一变量趋于升高）、“负相关”或“无相关”。强度可用“强”、“中等”或“弱”。考官期望联系上下文——此处，天气变暖导致冰淇淋销量增加。不要只说“正相关”，而要解释在场景中的含义。
8. Lines of Best Fit and Prediction | 回归线与预测
Continuing the ice cream scatter graph, the next part asks to draw a line of best fit and predict sales at 23°C. The line should have roughly equal numbers of points above and below, following the general trend. From the line, at 23°C, read vertically to the line and across – around 67 sales. State that this is an estimate. A follow-up question might ask about the reliability of predicting sales at 35°C. Since 35°C is beyond the data range (extrapolation), the estimate is unreliable because we do not know if the linear pattern continues.
接续冰淇淋散点图，下一部分要求画最佳拟合线并预测23°C时的销量。直线应使上下两侧点数大致相等，并跟随总体趋势。从23°C处垂直向上与线相交，再横向读销量——约为67。说明这是一个估算值。后续可能问预测35°C时销量的可靠性。由于35°C超出数据范围（外推），该估算不可靠，因为我们不知道线性模式是否延续。
Even interpolation (within the data range) has some uncertainty, but extrapolation carries greater risk. The exam response should mention that the line of best fit is only a model and actual values may differ. Using the line to read off values: draw construction lines on the graph to show the process and earn method marks.
即使是内插（数据范围内）也有一定不确定性，但外推风险更大。考试答案应提及最佳拟合线仅是一个模型，实际值可能不同。用直线读取数值时，在图上画辅助线以展示过程，可获方法分。
9. Basic Probability | 概率基础
A standard question: 'A fair six-sided die is rolled. What is the probability of rolling a factor of 6?' Factors of 6 are 1, 2, 3, 6 – four outcomes. Probability = 4/6 = 2/3. Another typical part: ' he die is rolled 300 times, how many times would you expect to get a number greater than 4?' ers >4 are 5 and 6, probability = 2/6 = 1/3. Expected frequency = (1/3) × 300 = 100 times.
标准题型：“投掷一枚均匀六面骰子，掷出6的因数的概率是多少？”6的因数为1, 2, 3, 6——四种结果。概率 = 4/6 = 2/3。另一典型问法：“若投掷骰子300次，期望得到大于4的数的次数是多少？”大于4的数为5和6，概率 = 2/6 = 1/3。期望频数 = (1/3) × 300 = 100次。
Students often confuse 'expected number' with 'exact occurrence' ; the expected value is a long-run average. Questions on relative frequency give data from an experiment: ' n is dropped 200 times and lands point up 124 times. Estimate the probability of point up.' s are conducted, relative frequency tends to stabilise around the theoretical probability.
学生常混淆“期望次数”与“确切发生次数”；期望值是一个长期平均值。有关相对频率的题目给出实验结果：“一枚图钉抛落200次，有124次钉尖向上。估计钉尖向上的概率。”答案：124/200 = 0.62。随着试验次数增加，相对频率会趋于稳定在理论概率附近。
10. Probability Tree Diagrams | 概率树形图
A classic past paper problem: ' l possible outcomes and their probabilities. Find the probability that both balls are red.' d)=4/7, P(Blue)=3/7. If first was blue, remaining 5 red, 2 blue → P(Red)=5/7, P(Blue)=2/7.
经典真题：“袋中有5个红球和3个蓝球。无放回地抽取两球。画出树形图表示所有可能结果及其概率。求两球均为红色的概率。”第一次抽取：P(红) = 5/8，P(蓝) = 3/8。第二次抽取概率改变：若第一次取红，剩余4红3蓝 → P(红)=4/7，P(蓝)=3/7；若第一次取蓝，剩余5红2蓝 → P(红)=5/7，P(蓝)=2/7。
ability is along the branch Red-Red: (5/8) × (4/7) = 20/56 = 5/14. A follow-up may ask for 'at least one blue'. It is often easier to find the complement: 1 – P(both red) = 1 – 5/14 = 9/14. Always reduce fractions to simplest form
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