📚 In-Depth Analysis of Past Papers for Year 8 CAIE Statistics | CAIE Year 8 统计历年真题深度解析
The Cambridge Lower Secondary Checkpoint Mathematics (Stage 8) Statistics paper tests students' ability to collect, represent, interpret and analyse data. Working through past exam questions is one of the most effective ways to prepare. This article provides an in-depth breakdown of typical Year 8 statistics problems, highlighting key techniques and common mistakes.
剑桥初中 Checkpoint 数学（阶段8）统计部分考查学生收集、展示、解读和分析数据的能力。练习历年真题是最有效的备考方法之一。本文将深度解析 Year 8 统计典型题目，强调关键解题技巧和常见错误。
1. Mean, Median, Mode and Range | 平均数、中位数、众数和极差
Past Paper Example: The ages (in years) of 11 children in a club are: 9, 7, 8, 10, 7, 8, 9, 10, 11, 8, 7. Find the mean, median, mode and range.
真题示例：一个俱乐部中 11 名儿童的年龄（岁）为：9, 7, 8, 10, 7, 8, 9, 10, 11, 8, 7。求平均数、中位数、众数和极差。
94 ÷ 11 ≈ 8.55 (rounded to two decimal places). Always show the division step clearly to gain full marks.
计算平均数时，先将所有数字相加：9 + 7 + 8 + 10 + 7 + 8 + 9 + 10 + 11 + 8 + 7 = 94。共有 11 个数，所以平均数 = 94 ÷ 11 ≈ 8.55（保留两位小数）。清晰地展示除法步骤可确保获得满分。
For the median, arrange the numbers in ascending order: 7, 7, 7, 8, 8, 8, 9, 9, 10, 10, 11. The median is the 6th value, which is 8. A common mistake is to pick the middle number without arranging the data first.
求中位数时，将数字从小到大排序：7, 7, 7, 8, 8, 8, 9, 9, 10, 10, 11。中位数是第 6 个值，即 8。常见的错误是在未排序的情况下直接选取中间的数。
The mode is the value that appears most frequently. Here both 7 and 8 appear three times, so the data set is bimodal with modes 7 and 8. If all numbers appear only once, there would be no mode.
众数是出现次数最多的数值。本题中 7 和 8 均出现三次，因此数据是双峰的，众数为 7 和 8。如果所有数字都只出现一次，则没有众数。
t and smallest values: 11 - 7 = 4. Remember the range is a single number, not an interval.
极差是最大值与最小值的差：11 - 7 = 4。注意极差是一个数字，而不是一个区间。
2. Bar Charts and Pictograms | 条形图和象形图
Past Paper Example: The pictogram shows the number of books read by four students. A picture of a book represents 4 books. Ali has 3½ symbols, Bella has 2 symbols, Chen has 4 symbols, and Dana has 2½ symbols. How many books did Ali read? How many more books did Chen read than Bella?
真题示例：象形图显示了四名学生阅读的书籍数量。一个书本图标表示 4 本书。Ali 有 3½ 个图标，Bella 有 2 个，Chen 有 4 个，Dana 有 2½ 个。Ali 读了多少本书？Chen 比 Bella 多读了几本书？
books. Remember that a half symbol represents half of 4, which is 2 books. Therefore, Ali' × 4) + 2 = 14. Always check the key carefully; some questions use different scales.
将图标数量乘以图示比例。Ali：3½ × 4 = 14 本书。注意，半个图标代表 4 的一半，即 2 本书。因此 Ali 的总数 = (3 × 4) + 2 = 14。务必仔细检查图例，有的题目会使用不同的比例。
ks, so Chen read 8 more books than Bella. Write the answer in the context of the question.
Bella 读了 2 × 4 = 8 本书。Chen 读了 4 × 4 = 16 本书。差值为 16 - 8 = 8 本书，因此 Chen 比 Bella 多读了 8 本。答题时要结合题目情境给出答案。
3. Pie Charts and Proportional Reasoning | 饼图与比例推理
Past Paper Example: In a survey of 30 students about their favourite fruit, 10 chose apple, 12 chose banana and 8 chose orange. Calculate the angle for each sector in a pie chart.
真题示例：一项关于 30 名学生最喜欢的水果的调查中，10 人选择苹果，12 人选择香蕉，8 人选择橙子。计算饼图中每个扇形的角度。
anana: (12 ÷ 30) × 360° = 144°. For orange: (8 ÷ 30) × 360° = 96°. Always check that the angles sum to 360°: 120 + 144 + 96 = 360°.
总频数为 30。每个扇形的角度 = (频数 ÷ 总数) × 360°。苹果：(10 ÷ 30) × 360° = 120°。香蕉：(12 ÷ 30) × 360° = 144°。橙子：(8 ÷ 30) × 360° = 96°。务必检查角度总和是否为 360°：120 + 144 + 96 = 360°。
total. For instance, if the banana sector is 144° in a pie chart of 30 students, number choosing banana = (144 ÷ 360) × 30 = 12.
一个常见错误是忘记乘以 360。如果题目给出角度和总人数，求频数则用 (角度 ÷ 360) × 总数。例如，若饼图中香蕉扇形的角度为 144°，总人数为 30，则选择香蕉的人数 = (144 ÷ 360) × 30 = 12。
4. Stem-and-Leaf Diagrams | 茎叶图
Past Paper Example: The stem-and-leaf diagram shows the test scores of 10 students. Stem: 2 | Leaf: 3 5 7; Stem: 3 | Leaf: 0 1 4 6; Stem: 4 | Leaf: 2 8 9. Key: 2|3 means 23. Find the median and range.
真题示例：茎叶图显示了 10 名学生的测验分数。茎 2，叶：3 5 7；茎 3，叶：0 1 4 6；茎 4，叶：2 8 9。图例：2|3 表示 23。求中位数和极差。
he average of the 5th and 6th values: (31 + 34) ÷ 2 = 32.5. The range is 49 - 23 = 26.
首先，根据图例列出所有数据值：23, 25, 27, 30, 31, 34, 36, 42, 48, 49。共 10 个值。中位数为第 5 和第 6 个值的平均数：(31 + 34) ÷ 2 = 32.5。极差为 49 - 23 = 26。
Many candidates forget to write the key on their diagram if they are asked to draw one, which results in lost marks. Always include a key. Also, check that the leaves are ordered from smallest to largest to avoid confusion.
许多考生在被要求绘制茎叶图时忘记书写图例，这会导致失分。务必包含图例。同时，确保叶从最小到最大排列，避免混淆。
5. Scatter Graphs and Correlation | 散点图与相关性
Past Paper Example: The scatter graph plots the time spent revising (hours) against the test score (out of 50) for 8 students. The points roughly rise from bottom left to top right. Describe the correlation and estimate the test score for a student who revised for 4.5 hours.
真题示例：散点图描绘了 8 名学生的复习时间（小时）与测验分数（满分 50）的关系。各点大致从左下方向右上方升高。描述相关性并估计一位复习了 4.5 小时的学生的测验分数。
As the points move upwards as we go right, there is a positive correlation. The relationship is fairly strong. To estimate, draw a line of best fit that passes through the middle of the points. At 4.5 hours, read the value on the line, which might be around 42 marks.
当点随着向右移动而上升时，呈正相关。这种关系相当强。要进行估计，需画一条穿过各点中间的最佳拟合线。在 4.5 小时处，读取线上的值，大约为 42 分。
Never join the points dot-to-dot; a line of best fit should be a single straight line. Also, avoid extending the line far beyond the data (extrapolation), because the trend may not continue. If no line is drawn, the estimate is unreliable.
切勿逐点连线；最佳拟合线应为一条单一的直线。同时，避免将线延伸到数据范围之外太远（外推），因为趋势可能不再持续。若不画线，估计结果便不可靠。
6. Frequency Tables and Grouped Data | 频率表与分组数据
Past Paper Example: The table shows the heights of 25 plants: 140 ≤ h < 150 (frequency 5), 150 ≤ h < 160 (8), 160 ≤ h < 170 (10), 170 ≤ h < 180 (2). Calculate an estimate of the mean height.
真题示例：下表显示了 25 株植物的高度：140 ≤ h < 150（频数 5），150 ≤ h < 160（8），160 ≤ h < 170（10），170 ≤ h < 180（2）。计算平均高度的估计值。
1650, 175×2 = 350. Sum of (midpoint × frequency) = 725 + 1240 + 1650 + 350 = 3965. Total frequency = 25. Estimated mean = 3965 ÷ 25 = 158.6 cm.
找出每个区间的组中值：145, 155, 165, 175。将每个组中值乘以对应的频数：145×5 = 725，155×8 = 1240，165×10 = 1650，175×2 = 350。组中值 × 频数的总和 = 725 + 1240 + 1650 + 350 = 3965。总频数 = 25。估计平均值 = 3965 ÷ 25 = 158.6 厘米。
This is an estimate because we do not know the exact heights; we assume they are evenly spread within the interval. A common mistake is to use the class boundaries (e.g., 140 and 150) instead of the midpoint.
这是一个估计值，因为我们不知道确切的高度；我们假设它们在区间内均匀分布。一个常见错误是用区间边界值（如 140 和 150）而非组中值进行计算。
7. Probability Scale and Simple Events | 概率尺度与简单事件
Past Paper Example: A bag contains 5 red marbles, 3 blue marbles and 2 green marbles. One marble is taken at random. Write down the probability that it is (a) red, (b) not red, (c) yellow. Mark these probabilities on a probability scale.
真题示例：一个袋子里有 5 个红色弹珠、3 个蓝色弹珠和 2 个绿色弹珠。随机取出一个弹珠。写出它是 (a) 红色，(b) 不是红色，(c) 黄色 的概率。在概率尺度上标出这些概率。
les = 5 + 3 + 2 = 10. P(red) = 5/10 = 1/2 or 0.5. P(not red) = 1 - P(red) = 1/2. The event 'not red' w marbles. On a scale from 0 to 1, mark 0 (impossible), 0.5 (even chance) and 1 (certain).
弹珠总数 = 5 + 3 + 2 = 10。P(红色) = 5/10 = 1/2 或 0.5。P(不是红色) = 1 - P(红色) = 1/2。事件“不是红色”意味着蓝色或绿色。P(黄色) = 0，因为没有黄色弹珠。在 0 到 1 的尺度上，标记 0（不可能）、0.5（均等机会）和 1（必然）。
Always simplify fractions where possible and ensure probabilities lie between 0 and 1 inclusive. Writing a probability as '1/2' is acceptable; '0.5' is also correct. Never write a probability as a ratio like 5:10.
尽可能简化分数，并确保概率在 0 和 1 之间（含端点）。将概率写作“1/2”是可以的；“0.5”同样正确。切勿将概率写成比例形式，如 5:10。
8. Two-Way Tables | 双向表
Past Paper Example: 40 students choose one sport: football or basketball. 18 are boys, the rest are girls. 25 students chose football. 14 boys chose football. Complete the two-way table and find the probability that a randomly chosen student is a girl who chose basketball.
真题示例：40 名学生选择一项运动：足球或篮球。其中 18 名是男生，其余是女生。25 人选择了足球。14 名男生选择了足球。完成双向表，并求随机选择一名学生是选择篮球的女生的概率。
| Football | Basketball | Total | |
|---|---|---|---|
| Boys | 14 | ? | 18 |
| Girls | ? | ? | ? |
| Total | 25 | ? | 40 |
s who chose basketball = 22 - 11 = 11. Total basketball = 4 + 11 = 15 (also 40 - 25 = 15).
逐步填写表格。选择篮球的男生 = 18 - 14 = 4。女生总数 = 40 - 18 = 22。选择足球的女生 = 25 - 14 = 11。选择篮球的女生 = 22 - 11 = 11。篮球总人数 = 4 + 11 = 15（也可以通过 40 - 25 = 15 得到）。
a fraction. Always use the completed table to check that all row and column totals match.
概率（女生且篮球）= 选择篮球的女生人数 / 总数 = 11/40。可保留分数形式。利用填好的表格检查所有行和列的总和是否匹配。
9. Comparing Data Sets Using Averages | 使用平均数比较数据集
Past Paper Example: Class A had a mean score of 70% with a range of 20%. Class B had a mean score of 75% with a range of 30%. Compare the performance and consistency of the two classes.
真题示例：A 班的平均分数为 70%，极差为 20%；B 班的平均分数为 75%，极差为 30%。比较两个班的表现和稳定性。
The mean tells us about the average performance. Class B has a higher mean (75 > 70), so on average, Class B performed better. The range measures spread: a smaller range means more consistent results. Class A has a smaller range (20 < 30), so the marks in Class A are less spread out, indicating more consistent performance.
平均数反映了平均表现。B 班的平均分更高（75 > 70），因此平均而言 B 班表现更好。极差衡量离散程度：极差越小表示成绩越一致。A 班的极差更小（20 < 30），所以 A 班的分数分布更集中，表明表现更稳定。
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