📚 Year 7 CIE Statistics: Unit Test Mock Paper Walkthrough | Year 7 CIE 统计：单元测试模拟卷解析
This walkthrough takes you through a typical Year 7 CIE Statistics unit test mock paper, explaining the thought process behind each question. By working through these problems, you will strengthen your understanding of data types, charts, averages, spread, and basic probability – exactly what you need to excel in your checkpoint assessment.
本模拟卷解析带你逐题分析一份典型的 Year 7 CIE 统计单元测试，解释每道题背后的解题思路。通过这些练习，你将加深对数据类型、图表、平均数、离散程度和基础概率的理解——这正是你在 Checkpoint 评估中取得优异成绩所需掌握的内容。
1. Question 1: Types of Data | 第1题：数据类型
Question: Classify each variable as categorical, discrete numerical, or continuous numerical: (a) eye colour, (b) number of books in a school bag, (c) time taken to run 100 m, (d) favourite subject.
题目：将下列变量分类为分类数据、离散数值数据或连续数值数据：(a) 眼睛颜色，(b) 书包里书本的数量，(c) 跑100米所用的时间，(d) 最喜欢的科目。
Categorical data are non-numerical labels or groups. Discrete numerical data can only take certain values, usually whole numbers (counts). Continuous numerical data can take any value within a range and are often measurements.
分类数据是非数字的标签或组别。离散数值数据只能取特定的值，通常是整数（计数）。连续数值数据可以取某个范围内的任何值，通常来自测量。
(a) Eye colour is categorical because it describes a quality, not a quantity. (b) Number of books is discrete numerical – you count whole books. (c) Time is continuous numerical – it can be measured to any precision, e.g. 12.3 s. (d) Favourite subject is categorical, as it names a preference.
(a) 眼睛颜色是分类数据，因为它描述的是性质而非数量。 (b) 书本的数量是离散数值数据——你数的是整本的书。 (c) 时间是连续数值数据——它可以被测量到任意精度，例如12.3秒。 (d) 最喜欢的科目是分类数据，因为它是在命名一种偏好。
2. Question 2: Tally Charts and Frequency Tables | 第2题：划记表和频数表
Question: The snacks chosen by 25 students are: apple, banana, apple, crisps, banana, chocolate, apple, crisps, apple, banana, chocolate, crisps, apple, apple, banana, apple, crisps, chocolate, banana, apple, apple, crisps, banana, chocolate, apple. Complete the tally and frequency table.
题目：25名学生选择的零食如下：apple, banana, apple, crisps, banana, chocolate, apple, crisps, apple, banana, chocolate, crisps, apple, apple, banana, apple, crisps, chocolate, banana, apple, apple, crisps, banana, chocolate, apple。请完成划记和频数表。
Begin by listing the categories: Apple, Banana, Crisps, Chocolate. For each item, draw a tally mark, grouping in fives. Then count the tallies to obtain the frequency.
首先列出类别：Apple, Banana, Crisps, Chocolate。对每一项数据，画一个划记符号，五个一组进行分组。然后数出划记数量，得到频数。
| Snack | Tally | Frequency |
|---|---|---|
| Apple | ~~||||~~ ||| | 9 |
| Banana | ~~||||~~ | 6 |
| Crisps | ~~||||~~ | 5 |
| Chocolate | ~~||||~~ | 5 |
Always check that the total frequency equals the number of data items: 9 + 6 + 5 + 5 = 25. This confirms the tally is correct.
务必检查总频数是否等于数据总数：9 + 6 + 5 + 5 = 25。这可以确认划记无误。
3. Question 3: Bar Charts and Pictograms | 第3题：条形图和象形图
Question: The bar chart shows the number of medals won by four houses. Use it to answer: (a) Which house won the most medals? (b) How many more medals did Eagle win than Falcon? (c) If Falcon is represented by 8 medal icons on a pictogram and each icon stands for 2 medals, is this correct?
题目：条形图显示了四个学院赢得的奖牌数量。根据图表回答：(a) 哪个学院赢得的奖牌最多？ (b) Eagle 学院比 Falcon 学院多赢了多少枚奖牌？ (c) 若 Falcon 在象形图中用8个奖牌图标表示，且每个图标代表2枚奖牌，这样表示正确吗？
Assume the bar chart heights are: Eagle 20, Falcon 12, Hawk 18, Owl 15. (a) Eagle has the tallest bar, so they won the most medals. (b) Difference: 20 − 12 = 8 medals. (c) Falcon has 12 medals. With each icon representing 2 medals, you would need 12 ÷ 2 = 6 icons. Using 8 icons would represent 16 medals, which is incorrect.
假设条形图的高度为：Eagle 20, Falcon 12, Hawk 18, Owl 15。 (a) Eagle 的柱形最高，所以他们赢得最多奖牌。 (b) 差值：20 − 12 = 8枚奖牌。 (c) Falcon 有12枚奖牌。若每个图标代表2枚奖牌，则需要 12 ÷ 2 = 6个图标。使用8个图标将代表16枚奖牌，因此不正确。
4. Question 4: Pie Charts | 第4题：饼图
Question: A pie chart shows favourite sports among 36 students. Football sector angle = 150°, Tennis = 90°, Basketball = 120°. (a) Which sport is the most popular? (b) How many students chose Tennis? (c) What fraction of students chose Basketball? Give your answer in simplest form.
题目：一个饼图显示了36名学生最喜欢的运动。足球对应的扇形角 = 150°，网球 = 90°，篮球 = 120°。 (a) 哪项运动最受欢迎？ (b) 有多少名学生选择了网球？ (c) 选择篮球的学生所占比例是多少？请化成最简分数。
(a) The largest angle is for Football (150°), so it is the most popular. (b) Total angle 360° represents 36 students. Angle per student = 360° ÷ 36 = 10° per student. Tennis has 90°, so number = 90 ÷ 10 = 9 students. (c) Basketball angle 120°, so fraction = 120/360 = 1/3 in simplest form.
(a) 最大角度是足球的150°，因此足球最受欢迎。 (b) 整个圆周角360°代表36名学生。每名学生所占角度 = 360° ÷ 36 = 10°。网球角度90°，所以人数 = 90 ÷ 10 = 9人。 (c) 篮球角度120°，因此占比 = 120/360 = 1/3（最简形式）。
5. Question 5: Mean, Median, Mode | 第5题：平均数、中位数、众数
Question: The ages of a junior chess club members are: 11, 12, 10, 12, 13, 11, 10, 12, 14, 11. Calculate the mean, median, and mode.
题目：一个少年国际象棋俱乐部成员的年龄为：11, 12, 10, 12, 13, 11, 10, 12, 14, 11。请计算平均数、中位数和众数。
Mean: Add all values. Sum = 10+10+11+11+11+12+12+12+13+14 = 116. Number of values = 10. Mean = 116 ÷ 10 = 11.6.
平均数：将所有数值相加。总和 = 10+10+11+11+11+12+12+12+13+14 = 116。数据个数 = 10。平均数 = 116 ÷ 10 = 11.6。
Median: First order the data: 10, 10, 11, 11, 11, 12, 12, 12, 13, 14. With 10 numbers, the median is the average of the 5th and 6th values. 5th = 11, 6th = 12. Median = (11+12) ÷ 2 = 11.5.
中位数：先排序：10, 10, 11, 11, 11, 12, 12, 12, 13, 14。有10个数，中位数是第5和第6个数的平均数。第5个 = 11，第6个 = 12。中位数 = (11+12) ÷ 2 = 11.5。
Mode: The most frequent value is 11 and 12, both occurring 3 times. The data set is bimodal.
众数：出现次数最多的值是11和12，各出现3次。该数据集是双峰的。
6. Question 6: Range and Comparing Distributions | 第6题：极差与分布比较
Question: Two basketball teams record points in a tournament: Team P: 45, 50, 55, 60, 40; Team Q: 30, 70, 35, 65, 50. (a) Calculate the range for each team. (b) Which team is more consistent? Explain why.
题目：两支篮球队在锦标赛中的得分记录为：Team P: 45, 50, 55, 60, 40；Team Q: 30, 70, 35, 65, 50。 (a) 计算每支球队的极差。 (b) 哪支球队发挥更稳定？请说明理由。
(a) Range Team P = 60 − 40 = 20. Range Team Q = 70 − 30 = 40. (b) Team P has a smaller range, meaning their scores are more tightly clustered. Although the term ‘consistent’ is better judged by variation, the range gives a simple measure – lower range suggests more consistency.
(a) Team P 极差 = 60 − 40 = 20。Team Q 极差 = 70 − 30 = 40。 (b) Team P 的极差较小，意味着他们的得分更集中。虽然“稳定”最好用变异程度来判断，但极差提供了一个简单度量——极差较小通常表明更稳定。
7. Question 7: Introduction to Probability | 第7题：概率入门
Question: A bag contains 3 red, 5 blue, and 2 green counters. One counter is chosen at random. Find the probability that it is (a) red, (b) blue or green, (c) not red. Write your answer as a fraction in simplest form.
题目：一个袋子里装有3个红色、5个蓝色和2个绿色筹码。随机抽取一个筹码。求抽到下列筹码的概率：(a) 红色，(b) 蓝色或绿色，(c) 非红色。用最简分数表示。
Total counters = 3 + 5 + 2 = 10. (a) P(red) = 3/10. (b) Blue or green: there are 5 + 2 = 7 favourable outcomes, so P(blue or green) = 7/10. (c) ‘Not red’ is the complement of red: 1 − 3/10 = 7/10, or directly 7/10.
筹码总数 = 3 + 5 + 2 = 10。 (a) P(红色) = 3/10。 (b) 蓝色或绿色：有利结果有 5 + 2 = 7 个，P(蓝色或绿色) = 7/10。 (c) “非红色”是红色的补集：1 − 3/10 = 7/10，或直接 7/10。
8. Question 8: Venn Diagrams | 第8题：韦恩图
Question: In a class of 30 students, 18 like pizza, 15 like pasta, and 8 like both. (a) Draw a Venn diagram. (b) How many students like only pizza? (c) How many students like neither?
题目：在一个30名学生的班级中，18人喜欢披萨，15人喜欢意大利面，8人两者都喜欢。 (a) 画出韦恩图。 (b) 有多少名学生只喜欢披萨？ (c) 多少名学生两种都不喜欢？
Place 8 in the intersection. Only pizza = 18 − 8 = 10. Only pasta = 15 − 8 = 7. Total in the circles = 10 + 8 + 7 = 25. Neither = 30 − 25 = 5. The Venn diagram shows two overlapping circles labelled appropriately.
在交集区域填入8。只喜欢披萨 = 18 − 8 = 10。只喜欢意大利面 = 15 − 8 = 7。圆圈内总人数 = 10 + 8 + 7 = 25。两者都不喜欢 = 30 − 25 = 5。韦恩图显示两个重叠圆并正确标注。
9. Question 9: Interpreting Dual Bar Charts | 第9题：解读复式条形图
Question: A dual bar chart shows the number of boys and girls who chose swimming, football, and tennis. Swimming: boys 8, girls 12; Football: boys 15, girls 5; Tennis: boys 6, girls 10. (a) Which sport is most popular overall? (b) What is the ratio of boys to girls in football? (c) True or false: more girls than boys chose tennis.
题目：复式条形图显示了选择游泳、足球和网球的男生和女生人数。游泳：男生8人，女生12人；足球：男生15人，女生5人；网球：男生6人，女生10人。 (a) 总体上哪项运动最受欢迎？ (b) 足球项目中男生与女生的比例是多少？ (c) 判断对错：选择网球的女生比男生多。
(a) Total per sport: Swimming 20, Football 20, Tennis 16. Swimming and Football are equally most popular with 20 total. (b) Boys:girls in football = 15:5 = 3:1 in simplest form. (c) True, because 10 > 6.
(a) 每项运动总人数：游泳20人，足球20人，网球16人。游泳和足球并列最受欢迎，均为20人。 (b) 足球的男女比例 = 15:5 = 3:1（最简形式）。 (c) 正确，因为10 > 6。
10. Question 10: Mixed Problem-Solving | 第10题：综合问题解决
Question: A survey question asks, “How many hours of sport do you do each week?” with options: 0-2, 2-4, 4-6, more than 6. (a) Suggest one improvement to the response options. (b) A student surveys 40 peers and records: 5, 12, 15, 8 respectively for the four intervals. Calculate the mean using interval midpoints. (c) Explain why the mean is only an estimate.
题目：某调查问题为：“你每周进行多少小时体育运动？”选项为：0-2, 2-4, 4-6, 大于6。 (a) 对回答选项提出一个改进建议。 (b) 一名学生调查了40名同学，四个区间分别记录到：5, 12, 15, 8人。使用区间中点计算平均数。 (c) 解释为什么算出的平均数只是估计值。
(a) The intervals overlap (2 appears in both 0-2 and 2-4). Improved options: 0 ≤ h < 2, 2 ≤ h < 4, 4 ≤ h < 6, h ≥ 6. (b) Midpoints: 1, 3, 5, 7 (assume more than 6 has midpoint 7 for simplicity). Estimated total hours = 5×1 + 12×3 + 15×5 + 8×7 = 5 + 36 + 75 + 56 = 172. Mean estimate = 172 ÷ 40 = 4.3 hours. (c) It is an estimate because we do not know the exact hours of each student, only the group they fall into. The true mean may differ.
(a) 区间有重叠（2同时出现在0-2和2-4中）。改进选项：0 ≤ h < 2, 2 ≤ h < 4, 4 ≤ h < 6, h ≥ 6。 (b) 各区间中点：1, 3, 5, 7（假定大于6的中点为7以便计算）。总小时数估计 = 5×1 + 12×3 + 15×5 + 8×7 = 5 + 36 + 75 + 56 = 172。平均数估计 = 172 ÷ 40 = 4.3 小时。 (c) 这只是一个估计值，因为我们不知道每个学生的确切运动时间，只知道他们所在的区间。真实平均数可能不同。
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