KS3 Cambridge Statistics: Mock Test Paper Analysis | KS3 剑桥统计：单元测试模拟卷解析

📚 KS3 Cambridge Statistics: Mock Test Paper Analysis | KS3 剑桥统计：单元测试模拟卷解析

This mock test paper analysis is designed to help KS3 Cambridge students master key statistical concepts. By working through real exam-style questions, you will reinforce your understanding of mean, median, mode, range, interpreting charts, probability, comparing data sets and identifying bias. Each question is broken down step by step to highlight common mistakes and effective problem-solving strategies. Use this revision resource to build confidence and accuracy ahead of your unit test.

本模拟测试卷解析旨在帮助 KS3 剑桥学生掌握核心统计概念。通过练习真实考试风格的题目，你将巩固对平均数、中位数、众数、极差、图表解读、概率、数据集比较以及识别偏差等知识的理解。每道题都逐步拆解，突出常见错误和有效解题策略。请利用这份复习资料，在单元测试前建立信心、提升解题准确性。

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1. Mean, Median, Mode & Range | 平均数、中位数、众数和极差

Question 1: The heights (in cm) of six students are: 152, 158, 163, 158, 149, 160. Calculate the mean, median, mode and range.

题目1：六名学生的身高（厘米）为：152, 158, 163, 158, 149, 160。计算平均数、中位数、众数和极差。

Solution:

解答：

Mean: Add all values: 152 + 158 + 163 + 158 + 149 + 160 = 940. Divide by 6: 940 ÷ 6 ≈ 156.7 cm (to 1 decimal place).

平均数：将所有值相加：152 + 158 + 163 + 158 + 149 + 160 = 940。除以 6：940 ÷ 6 ≈ 156.7 厘米（保留一位小数）。

Median: Order the data: 149, 152, 158, 158, 160, 163. With six numbers, the median is the average of the 3rd and 4th values: (158 + 158) ÷ 2 = 158 cm.

中位数：将数据从小到大排列：149, 152, 158, 158, 160, 163。当数据个数为偶数时，中位数为中间两个数的平均值：(158 + 158) ÷ 2 = 158 厘米。

Mode: The most frequent value is 158 cm (appears twice).

众数：出现次数最多的值是 158 厘米（出现两次）。

Range: Highest value – lowest value = 163 – 149 = 14 cm.

极差：最大值减最小值 = 163 – 149 = 14 厘米。

Always remember to order the data for median, and check for repeated values when finding the mode. The mean can be affected by extreme values, while median and mode are more robust.

求中位数时始终要先将数据排序；找众数时要检查重复值。平均数可能受极端值影响，中位数和众数则更稳健。

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2. Interpreting Bar Charts | 条形图解读

Question 2: The bar chart shows the number of goals scored by four football teams: Team A (6 goals), Team B (2 goals), Team C (5 goals), Team D (3 goals). How many goals were scored in total? Which team scored the most? What fraction of total goals did Team C score?

题目2：条形图显示了四支足球队的进球数：A队（6球）、B队（2球）、C队（5球）、D队（3球）。总进球数是多少？哪队进球最多？C队进球数占总数的几分之几？

Total goals = 6 + 2 + 5 + 3 = 16. Team A scored the most. Team C scored 5 goals, so the fraction is 5/16. This cannot be simplified further.

总进球数 = 6 + 2 + 5 + 3 = 16。A队进球最多。C队进了5球，所以分数为 5/16，已是最简形式。

When reading bar charts, carefully check the scale on the vertical axis. Use a ruler to ensure accurate reading if the chart is printed. In exam questions, you may be asked to compare categories or compute proportions.

解读条形图时，要仔细检查纵轴的刻度。如果图表是打印版，可用直尺辅助精确读数。考试中常要求比较类别或计算比例。

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3. Pie Charts: Angle and Frequency | 饼图：角度与频数

Question 3: A pie chart represents 180 students’ favourite snacks. The sector for ‘Fruit’ has an angle of 80°. How many students chose fruit? What percentage prefer fruit?

题目3：一个饼图表示180名学生对零食的喜好。标记为“水果”的扇区角度为80°。有多少名学生选择了水果？喜欢水果的百分比是多少？

The entire circle is 360°, representing 180 students. Each degree corresponds to 180 ÷ 360 = 0.5 students. So, 80° represents 80 × 0.5 = 40 students. Percentage = (40 ÷ 180) × 100 ≈ 22.2% (to 1 decimal place).

整个圆为360°，代表180名学生。每度对应 180 ÷ 360 = 0.5 名学生。因此，80°对应 80 × 0.5 = 40 名学生。百分比 = (40 ÷ 180) × 100 ≈ 22.2%（保留一位小数）。

Alternatively, use proportion: (Angle/360) × Total frequency = (80/360) × 180 = 40. Distance between sectors does not matter; only the angle determines the frequency.

也可用比例法：(角度/360) × 总频数 = (80/360) × 180 = 40。扇区之间的距离无关紧要，只有角度决定频数。

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4. Scatter Graphs and Correlation | 散点图与相关性

Question 4: A scatter graph plots hours of revision against exam score. Points rise from bottom-left to top-right. Describe the correlation. Can you conclude that more revision causes higher scores?

题目4：散点图绘制了复习时间与考试成绩的关系，各点从左下方到右上方上升。描述其相关性。能否得出更多复习导致更高成绩的结论？

The graph shows a positive correlation: as hours increase, scores tend to increase. However, correlation does not imply causation; there may be other factors such as prior knowledge or sleep. The scatter graph only indicates an association, not a cause-and-effect relationship.

该图显示正相关：随着复习时间增加，成绩也倾向于提高。但是，相关性不意味着因果关系；可能存在其他因素，如原有知识或睡眠。散点图仅表明关联，并非因果关系。

Outliers can weaken correlation. A line of best fit could be drawn to make predictions, but extrapolation beyond the data range is unreliable.

异常值会削弱相关性。可绘制最佳拟合线进行预测，但超出数据范围的推断不可靠。

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5. Mean from a Frequency Table | 频数表求均值

Question 5: The table shows the number of pets owned by children.

Number of pets	0	1	2	3
Frequency	4	7	5	4

Calculate the mean number of pets.

计算宠物数量的平均数。

Create a new row for ‘Pets × Frequency’: 0×4=0, 1×7=7, 2×5=10, 3×4=12. Sum of these products = 0+7+10+12 = 29. Total frequency = 4+7+5+4 = 20. Mean = 29 ÷ 20 = 1.45 pets.

新增一行计算“宠物数 × 频数”：0×4=0，1×7=7，2×5=10，3×4=12。这些乘积的总和 = 0+7+10+12 = 29。总频数 = 4+7+5+4 = 20。平均数 = 29 ÷ 20 = 1.45 只宠物。

Always multiply each value by its frequency before summing. This method avoids long lists of individual data points.

求总和前务必将每个值与其频数相乘。这种方法可避免罗列冗长的单个数据点。

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6. Experimental vs Theoretical Probability | 实验概率与理论概率

Question 6: A fair coin is flipped 50 times, landing on heads 28 times. What is the experimental probability of heads? How does it compare with the theoretical probability?

题目6：一枚均匀硬币抛掷50次，得到28次正面。正面的实验概率是多少？与理论概率相比如何？

Experimental probability = Number of successful trials / Total trials = 28/50 = 0.56 or 56%. Theoretical probability for a fair coin is 0.5 (50%). The experimental probability is slightly higher, which is normal in small sample sizes due to chance variation. With more trials, the experimental probability should get closer to 0.5.

实验概率 = 成功次数 / 总试验次数 = 28/50 = 0.56，即56%。均匀硬币的理论概率为0.5（50%）。实验概率略高，在小样本中由于随机波动是正常的。随着试验次数增加，实验概率将趋近于0.5。

This demonstrates the Law of Large Numbers. Always express probabilities as fractions, decimals or percentages as specified.

这体现了大数定律。始终按题目要求将概率表示为分数、小数或百分数。

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7. Sample Space and Probability | 样本空间与概率

Question 7: Two fair six-sided dice are rolled. List all outcomes where the sum is 7. Hence, find the probability of rolling a sum of 7.

题目7：同时掷两个均匀的六面骰子。列出所有和为7的可能结果，并由此求出掷出和为7的概率。

Total possible outcomes = 6 × 6 = 36. Pairs summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) – six outcomes. Probability = 6/36 = 1/6.

总可能结果数 = 6 × 6 = 36。和为7的组合：(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — 共六种结果。概率 = 6/36 = 1/6。

Using a systematic list prevents missing combinations. A two-way table can also be helpful to visualize the sample space.

使用系统性列表可避免遗漏组合。双向表也有助于直观展示样本空间。

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8. Comparing Data Sets Using Averages and Range | 运用平均数和极差比较数据集

Question 8: Class 7P scored a mean of 62% with a range of 30% on a test. Class 7Q scored a mean of 64% with a range of 12%. Compare the performance and consistency of the two classes.

题目8：7P班测验平均分62%，极差30%；7Q班平均分64%，极差12%。比较两个班的表现和一致性。

The mean score of 7Q is slightly higher (64% vs 62%), indicating slightly better performance on average. However, the range for 7Q is much smaller (12% vs 30%), meaning scores are more consistent and less spread out. 7P has a wider range, suggesting larger variation in student achievement. So, 7Q is more consistent, though the average difference is small.

7Q班的平均分略高（64% 对 62%），表明平均表现稍好。然而，7Q班的极差小得多（12% 对 30%），意味着成绩更集中、一致性更高。7P班极差较大，说明学生成绩差异大。因此，7Q班成绩更均衡，尽管平均分差异不大。

Use both average and measure of spread (range) for a full comparison. Note that range only uses extremes, so it doesn’t show how data is distributed in between.

全面比较时需同时使用平均数和离散度量（极差）。注意极差仅使用极值，无法显示中间数据的分布情况。

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9. Identifying Bias in Data Collection | 识别数据收集中的偏差

Question 9: A survey asks: ‘Don’t you agree that our school meals are delicious?’ Criticise this question and rewrite it to be unbiased.

题目9：一项调查问道：“你难道不觉得我们学校的午餐很美味吗？”批评该问题并改写成无偏问题。

This is a leading question because it suggests a desired answer (‘are delicious’) and pushes respondents toward agreement. It also uses negative phrasing, causing confusion. An unbiased version could be: ‘How would you rate the taste of our school meals?’ with options: Very good, Good, Neutral, Poor, Very poor. This collects honest opinions without steering respondents.

这是一个诱导性问题，因为它暗示了期望的答案（“很美味”），并推动受访者表示同意。它还使用了否定措辞，容易引起混淆。无偏版本可以是：“您如何评价我校午餐的口味？”选项包括：非常好、好、一般、差、非常差。这样可以收集真实意见，不引导受访者。

Bias can also arise from small or unrepresentative samples. Always check question wording and sampling method.

偏差也可能来自样本量小或样本不具代表性。务必检查问题措辞和抽样方法。

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10. Stem-and-Leaf Diagrams: Median and Range | 茎叶图：中位数与极差

Question 10: The stem-and-leaf diagram shows the ages of people at a concert.

2 | 1 3 5 7
3 | 0 2 4 4 6 9
4 | 1 5

Here 2 | 1 means 21 years. Find the median age and range.

此处 2 | 1 表示 21 岁。求年龄中位数和极差。

List all ages in order: 21, 23, 25, 27, 30, 32, 34, 34, 36, 39, 41, 45. There are 12 values. Median is the average of 6th and 7th values: (32 + 34) ÷ 2 = 33 years. Range = 45 – 21 = 24 years.

按顺序列出所有年龄：21, 23, 25, 27, 30, 32, 34, 34, 36, 39, 41, 45。共12个数据。中位数为第6和第7个数的平均值：(32 + 34) ÷ 2 = 33 岁。极差 = 45 – 21 = 24 岁。

The stem-and-leaf diagram keeps data ordered and shows distribution shape. Remember the key (2|1=21) is essential for interpretation.

茎叶图保持数据有序且显示分布形态。切记图例（2|1=21）对于解读至关重要。

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Published by TutorHao | Statistics Revision Series | aleveler.com

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