KS3 CCEA Statistics: Mock Unit Test Walkthrough | KS3 CCEA 统计：单元测试模拟卷解析

📚 KS3 CCEA Statistics: Mock Unit Test Walkthrough | KS3 CCEA 统计：单元测试模拟卷解析

Welcome to this detailed walkthrough of a mock unit test for KS3 CCEA Statistics. This paper covers key topics such as averages, charts, probability, and data interpretation. By working through each question carefully, you will strengthen your understanding and build confidence for your real assessment.

欢迎阅读 KS3 CCEA 统计单元测试模拟卷的详细解析。这份试卷涵盖了平均数、图表、概率和数据解读等核心主题。通过仔细攻克每道题目，你将加深理解，为真正的考试建立信心。

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1. Calculating Mean, Median, Mode and Range | 计算平均数、中位数、众数和极差

Question: A student recorded the number of books read by 8 friends: 12, 15, 14, 13, 15, 16, 14, 13. Find the mean, median, mode and range.

问题：一名学生记录了8位朋友阅读的书籍数量：12, 15, 14, 13, 15, 16, 14, 13。请计算平均数、中位数、众数和极差。

Step 1: Sort the data into ascending order: 12, 13, 13, 14, 14, 15, 15, 16. This makes it easier to find the median and mode.

步骤一：将数据按升序排列：12, 13, 13, 14, 14, 15, 15, 16。这样更容易找到中位数和众数。

Step 2: Calculate the mean. Sum = 12 + 13 + 13 + 14 + 14 + 15 + 15 + 16 = 112. Number of values = 8. Mean = 112 ÷ 8 = 14.

步骤二：计算平均数。总和 = 12 + 13 + 13 + 14 + 14 + 15 + 15 + 16 = 112，数据个数 = 8，平均数 = 112 ÷ 8 = 14。

Step 3: Find the median. With 8 ordered values, the median is the average of the 4th and 5th numbers. Both are 14, so median = 14.

步骤三：求中位数。8个有序数据的中位数是第4和第5个数的平均值。两个数都是14，因此中位数 = 14。

Step 4: Identify the mode. The numbers 13, 14 and 15 each appear twice, so there are three modes: 13, 14 and 15. The data is multi-modal.

步骤四：确定众数。13、14和15各出现两次，因此有3个众数：13、14和15。这组数据是多模态的。

Step 5: Calculate the range. Range = maximum – minimum = 16 – 12 = 4.

步骤五：计算极差。极差 = 最大值 – 最小值 = 16 – 12 = 4。

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2. Interpreting Frequency Tables | 解读频数表

Question: The table shows the number of pets owned by 20 families.

问题：下表显示了20个家庭拥有的宠物数量。

Number of pets	Frequency
0	4
1	6
2	5
3	4
4	1

Find the mode and calculate the mean number of pets.

求众数并计算每户平均宠物数量。

Step 1: The mode is the value with the highest frequency. Here, 1 pet has the frequency 6, so mode = 1.

步骤一：众数是频率最高的值。这里，1只宠物的频率为6，因此众数 = 1。

Step 2: To find the mean, multiply each number of pets by its frequency, sum these products, then divide by total frequency. Total frequency = 20.

步骤二：计算平均数，把每个宠物数量乘以对应频率，求这些积的总和，再除以总频率。总频率 = 20。

Step 3: Sum of products = (0×4) + (1×6) + (2×5) + (3×4) + (4×1) = 0 + 6 + 10 + 12 + 4 = 32.

步骤三：乘积之和 = (0×4) + (1×6) + (2×5) + (3×4) + (4×1) = 0 + 6 + 10 + 12 + 4 = 32。

Step 4: Mean = 32 ÷ 20 = 1.6. The average family owns 1.6 pets.

步骤四：平均数 = 32 ÷ 20 = 1.6。平均每个家庭拥有1.6只宠物。

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3. Drawing and Reading Bar Charts | 绘制与读取条形图

Question: A bar chart displays students’ favourite colours: Blue 10, Red 8, Green 5, Yellow 7. Use the chart to answer these questions.

问题：一个条形图显示了学生最喜欢的颜色：蓝色10人，红色8人，绿色5人，黄色7人。请根据图表回答问题。

a) Which colour is the mode? The tallest bar is for Blue, so mode = Blue.

a) 哪种颜色是众数？最高的条形是蓝色，因此众数 = 蓝色。

b) How many students chose Red? The bar for Red reaches 8, so 8 students chose red.

b) 有多少学生选择了红色？代表红色的条形高度为8，因此8名学生选择了红色。

c) What is the total number of students surveyed? Total = 10 + 8 + 5 + 7 = 30 students.

c) 被调查的学生总数是多少？总数 = 10 + 8 + 5 + 7 = 30 名学生。

d) Explain how you would draw this bar chart. The horizontal axis shows colours; the vertical axis is frequency, numbered from 0 to 12. Each bar must be of equal width, with gaps between them, and height exactly representing the frequency.

d) 解释你如何绘制这个条形图。横轴表示颜色，纵轴是频率，标度从0到12。每个条形必须等宽，条与条之间有间隔，高度准确对应频率。

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4. Pie Charts and Angle Calculations | 饼图与角度计算

Question: A survey of 30 students on how they travel to school gave these results: Walking 12, Cycling 9, Bus 6, Car 3. Calculate the angle for each sector in a pie chart.

问题：一项针对30名学生上学交通方式的调查结果如下：步行12人，骑行9人，公交6人，小汽车3人。计算饼图中每个扇形的角度。

Step 1: Remember that the whole pie chart corresponds to 360°. The angle for a category = (frequency ÷ total) × 360°.

步骤一：记住整个饼图对应360°。某个类别的角度 = (该类别频数 ÷ 总数) × 360°。

Step 2: Walking angle = (12 ÷ 30) × 360° = 0.4 × 360° = 144°.

步骤二：步行角度 = (12 ÷ 30) × 360° = 0.4 × 360° = 144°。

Step 3: Cycling angle = (9 ÷ 30) × 360° = 0.3 × 360° = 108°.

步骤三：骑行角度 = (9 ÷ 30) × 360° = 0.3 × 360° = 108°。

Step 4: Bus angle = (6 ÷ 30) × 360° = 0.2 × 360° = 72°.

步骤四：公交角度 = (6 ÷ 30) × 360° = 0.2 × 360° = 72°。

Step 5: Car angle = (3 ÷ 30) × 360° = 0.1 × 360° = 36°.

步骤五：小汽车角度 = (3 ÷ 30) × 360° = 0.1 × 360° = 36°。

Step 6: Check the total: 144° + 108° + 72° + 36° = 360°. The angles are correct; you can now draw the chart with a protractor.

步骤六：检验总和：144° + 108° + 72° + 36° = 360°。角度正确，现在可以用量角器绘制图表。

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5. Probability of a Single Event | 单个事件的概率

Question: A bag contains 3 red, 5 blue and 2 green marbles. One marble is picked at random. Find the probability that it is (a) red, (b) not blue.

问题：一个袋子装有3个红色、5个蓝色和2个绿色弹珠。随机抽取一个。求抽到 (a) 红色弹珠的概率，以及 (b) 不是蓝色弹珠的概率。

Step 1: Total number of marbles = 3 + 5 + 2 = 10.

步骤一：弹珠总数 = 3 + 5 + 2 = 10。

Step 2: P(red) = number of red marbles / total = 3/10.

步骤二：P(红色) = 红色弹珠数 / 总数 = 3/10。

Step 3: Not blue means the marble is either red or green. Number of not blue = 3 + 2 = 5. So P(not blue) = 5/10 = 1/2.

步骤三：不是蓝色即弹珠为红色或绿色。不是蓝色的数量 = 3 + 2 = 5。因此 P(不是蓝色) = 5/10 = 1/2。

Step 4: You can also write probabilities as decimals or percentages: 0.3 and 0.5 or 30% and 50%.

步骤四：你也可以将概率写作小数或百分数：0.3和0.5，或者30%和50%。

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

Question: A fair coin is tossed 50 times and lands heads 22 times. Calculate the experimental probability of getting heads and compare it with the theoretical probability.

问题：一枚均匀硬币抛掷50次，出现正面22次。计算出现正面的实验概率，并与理论概率比较。

Step 1: Experimental probability = number of times heads occurs / total trials = 22/50 = 0.44.

步骤一：实验概率 = 正面出现次数 / 总试验次数 = 22/50 = 0.44。

Step 2: Theoretical probability for a fair coin is 1/2 = 0.5.

步骤二：均匀硬币的理论概率为1/2 = 0.5。

Step 3: The experimental probability (0.44) is slightly less than the theoretical probability (0.5). This difference is expected due to chance; with more trials the experimental value tends to get closer to 0.5.

步骤三：实验概率（0.44）略低于理论概率（0.5）。由于随机性，这种差异是正常的；随着试验次数增加，实验值会趋近于0.5。

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

Question: Two classes took the same test. Class A scores: 12, 15, 14, 16, 18. Class B scores: 10, 12, 20, 14, 14. Compare the two classes’ performance using the mean and range.

问题：两个班级参加了同一测试。A班分数：12, 15, 14, 16, 18。B班分数：10, 12, 20, 14, 14。使用平均数和极差比较两个班的表现。

Step 1: Class A mean = (12+15+14+16+18) ÷ 5 = 75 ÷ 5 = 15. Class B mean = (10+12+20+14+14) ÷ 5 = 70 ÷ 5 = 14.

步骤一：A班平均数 = (12+15+14+16+18) ÷ 5 = 75 ÷ 5 = 15。B班平均数 = (10+12+20+14+14) ÷ 5 = 70 ÷ 5 = 14。

Step 2: On average, Class A scored 1 mark higher than Class B.

步骤二：平均而言，A班比B班高1分。

Step 3: Class A range = 18 – 12 = 6. Class B range = 20 – 10 = 10.

步骤三：A班极差 = 18 – 12 = 6。B班极差 = 20 – 10 = 10。

Step 4: Class A has a smaller range, meaning the scores are more consistent. Class B has a wider spread, indicating one very high score and one very low score.

步骤四：A班的极差更小，说明分数更稳定。B班的分布更广，表明有一个极高和一个极低的分数。

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8. Interpreting Scatter Graphs | 解读散点图

Question: A scatter graph shows the relationship between hours spent revising and test marks for 10 students. The points rise from bottom left to top right. Describe the correlation and estimate the mark for a student who revised for 4 hours if the line of best fit passes through (2, 50) and (6, 80).

问题：一个散点图显示了10名学生复习时间与考试成绩的关系。所有点从左下到右上分布。描述其相关性，并根据通过(2, 50)和(6, 80)的最佳拟合线，估计复习4小时的学生的分数。

Step 1: The points going up to the right indicate a positive correlation: as revision hours increase, marks tend to increase.

步骤一：点向右上方延伸表示正相关：复习时间增加，分数倾向于提高。

Step 2: Use the line of best fit to make an estimate. The line passes through (2, 50) and (6, 80). The slope shows that for every extra hour, the mark rises by (80–50)/(6–2) = 30/4 = 7.5 marks per hour.

步骤二：使用最佳拟合线进行估计。该线经过点(2, 50)和(6, 80)。斜率表明每增加1小时，成绩上升 (80–50) ÷ (6–2) = 30/4 = 7.5 分/小时。

Step 3: From 2 hours to 4 hours is an increase of 2 hours. Therefore, estimated mark at 4 hours = 50 + 2 × 7.5 = 50 + 15 = 65.

步骤三：从2小时到4小时增加了2小时。因此，估计4小时的分数 = 50 + 2 × 7.5 = 50 + 15 = 65。

Step 4: So a student who revises for 4 hours is predicted to score around 65 marks.

步骤四：因此，复习4小时的学生预计得分约65分。

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9. Identifying Misleading Graphs | 识别误导性图表

Question: A bar chart shows sales of two products. The vertical axis starts at 80 instead of 0, making the bar for product A (actual 90) look twice as tall as product B (actual 85). Explain why the graph is misleading and how to fix it.

问题：一个条形图显示了两种产品的销量。纵轴从80而不是0开始，使得产品A（实际90）的条形看起来是产品B（实际85）的两倍高。解释为什么该图形具有误导性以及如何纠正。

Step 1: The vertical scale does not start at zero, so the heights exaggerate small differences. The visual difference does not match the actual numerical difference of only 5 units.

步骤一：纵轴刻度不从零开始，因此高度夸大了微小差异。视觉差异与仅5单位的实际数值差异不相符。

Step 2: To fix the graph, redraw the vertical axis from 0 to 100 with equal intervals. Then the bars will show the true proportion, and the difference will appear much smaller.

步骤二：要纠正该图表，应重新将纵轴从0到100等距绘制。这样条形将显示真实比例，差异会看起来小得多。

Step 3: Always check the scale when reading graphs; a broken axis without a clear label can be misleading.

步骤三：读图时务必检查刻度；没有明确标注的截断轴可能产生误导。

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10. Sample Space and Combined Event Probability | 样本空间与组合事件概率

Question: Two fair six-sided dice are rolled. List the sample space and find the probability that the sum of the numbers is 7.

问题：同时掷两个均匀六面骰子。列出样本空间，并求点数之和为7的概率。

Step 1: The sample space consists of 36 equally likely outcomes, which can be shown in a table with rows 1–6 for the first die and columns 1–6 for the second die.

步骤一：样本空间包含36个等可能结果，可用表格表示，行和列分别为第一个骰子和第二个骰子的1–6点。

Step 2: The sum is 7 for the combinations: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). That gives 6 favourable outcomes.

步骤二：和为7的组合有：(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)。共计6个有利结果。

Step 3: P(sum = 7) = number of favourable outcomes / total outcomes = 6/36 = 1/6.

步骤三：P(和为7) = 有利结果数量 / 总结果数量 = 6/36 = 1/6。

Step 4: You could also represent this as a simple fraction or decimal: 1/6 ≈ 0.167.

步骤四：你也可以用分数或小数表示：1/6 ≈ 0.167。

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Published by TutorHao | Statistics Revision Series | aleveler.com

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