📚 Year 7 WJEC Statistics: Unit Test Mock Paper Analysis | 七年级 WJEC 统计：单元测试模拟卷解析
Welcome to this detailed walkthrough of a Year 7 WJEC Statistics unit test mock paper. This resource will guide you through typical exam questions, explain key concepts, and show you how to achieve full marks. By working through these solutions, you will strengthen your data handling skills and build confidence for the real test.
欢迎阅读这份七年级 WJEC 统计单元测试模拟卷的详细解析。本资源将带你完成典型的考试题目，解释关键概念，并展示如何获得满分。通过解答这些题目，你将加强数据处理技能，为真正的考试树立信心。
1. Question 1: Identifying Data Types | 题型一：识别数据类型
The question states: ‘A student records the favourite colour of each classmate. Is this data qualitative or quantitative? Explain your answer.’
题目内容：“一名学生记录了每位同学最喜欢的颜色。这组数据是定性数据还是定量数据？请解释你的答案。”
To classify data correctly, recall the definitions. Qualitative data describes qualities or categories and is non-numerical. Quantitative data is numerical and can be measured or counted.
要正确分类数据，请回忆定义。定性数据描述性质或类别，是非数值的。定量数据是数值型的，可以测量或计数。
Favourite colours such as ‘red’, ‘blue’ or ‘green’ are categories, not numbers. Therefore, the data collected is qualitative.
像“红色”、“蓝色”或“绿色”这样最喜欢的颜色是类别，而不是数字。因此，收集的数据是定性的。
A common misunderstanding is to think that because the student can count how many classmates chose each colour, the data becomes quantitative. However, the original responses (the colours themselves) remain qualitative; only the frequencies are quantitative.
一个常见的误解是，认为因为学生可以统计每种颜色有多少人选择，数据就变成了定量数据。然而，原始回答（颜色本身）仍然是定性的；只有频率是定量的。
2. Question 2: Interpreting a Bar Chart | 题型二：解读条形图
The question presents a bar chart showing the favourite fruits of 30 students. Fruits: Apple (12 votes), Banana (7), Orange (5), Pear (3), Grapes (3). Part (a) asks: ‘Which fruit is the most popular?’
题目展示了一张条形图，显示了 30 名学生最喜欢的水果。水果：苹果（12 票）、香蕉（7）、橙子（5）、梨（3）、葡萄（3）。第 (a) 小题问：“哪种水果最受欢迎？”
By reading the tallest bar, we see Apple has the highest frequency of 12. So the most popular fruit is Apple.
通过识别最高的条形，我们看到苹果的频率最高，为 12。因此，最受欢迎的水果是苹果。
Part (b) asks: ‘How many more students chose Apple than Pear?’ We subtract: 12 – 3 = 9 students.
第 (b) 小题问：“选择苹果的学生比选择梨的学生多多少人？” 我们做减法：12 – 3 = 9 名学生。
Part (c) asks: ‘What fraction of the students chose Banana? Give your answer in simplest form.’ There are 7 Banana votes out of 30 total. The fraction is 7/30, which is already in simplest form.
第 (c) 小题问：“选择香蕉的学生占学生总数的几分之几？请用最简分数表示。” 香蕉得票 7 票，总人数 30。分数是 7/30，已经是最简形式。
Always check the total on the chart equals the stated sum: 12+7+5+3+3=30, which matches.
请务必检查图表上的总和是否等于给定总数：12+7+5+3+3=30，结果吻合。
3. Question 3: Finding the Mode and Range | 题型三：求众数和极差
The question gives a list of numbers: 5, 7, 7, 8, 10, 12, 12, 12, 15. Part (a) asks for the mode.
题目给出一个数字列表：5, 7, 7, 8, 10, 12, 12, 12, 15。第 (a) 小题要求找出众数。
The mode is the value that appears most frequently. In this list, 12 appears three times, more than any other number. Therefore, the mode is 12.
众数是出现最频繁的数值。在这个列表中，12 出现了三次，比任何其他数字都多。因此，众数是 12。
Part (b) asks for the range. The range is the difference between the largest and smallest values: 15 – 5 = 10.
第 (b) 小题要求计算极差。极差是最大值与最小值之差：15 – 5 = 10。
It is important to write down your subtraction step clearly to avoid careless mistakes.
清楚地写出减法步骤非常重要，以避免粗心错误。
4. Question 4: Calculating the Mean from a Frequency Table | 题型四：根据频率表计算平均数
The question provides a frequency table showing the number of pets owned by students in a class:
题目提供了一个频率表，显示一个班级学生拥有的宠物数量：
| Number of pets (x) | Frequency (f) |
|---|---|
| 0 | 3 |
| 1 | 7 |
| 2 | 5 |
| 3 | 3 |
| 4 | 2 |
To find the mean, first calculate the total number of pets: multiply each x by its f, then sum: (0 × 3) + (1 × 7) + (2 × 5) + (3 × 3) + (4 × 2) = 0 + 7 + 10 + 9 + 8 = 34.
要计算平均数，首先计算宠物总数：将每个 x 与它的 f 相乘，然后求和：(0 × 3) + (1 × 7) + (2 × 5) + (3 × 3) + (4 × 2) = 0 + 7 + 10 + 9 + 8 = 34。
Next, find the total number of students: sum the frequencies: 3 + 7 + 5 + 3 + 2 = 20.
接下来，计算学生总数：将频率求和：3 + 7 + 5 + 3 + 2 = 20。
The mean is total pets ÷ total students = 34 ÷ 20 = 1.7. So the mean number of pets is 1.7.
平均数等于宠物总数 ÷ 学生总数 = 34 ÷ 20 = 1.7。因此，宠物的平均数量是 1.7。
Always check that your multiplication and addition are correct. Using a table to organise the ‘x × f’ column can be very helpful.
务必检查你的乘法和加法是否正确。使用表格来整理“x × f”列会非常有帮助。
5. Question 5: Drawing a Pie Chart | 题型五：绘制饼图
The question gives data on how students travel to school: Walk (12), Cycle (8), Bus (15), Car (5). Part (a) asks to calculate the angle for each sector.
题目给出了学生上学交通方式的数据：步行（12），骑车（8），公交（15），小汽车（5）。第 (a) 小题要求计算每个扇区的角度。
First, find the total number of students: 12 + 8 + 15 + 5 = 40.
首先，计算学生总数：12 + 8 + 15 + 5 = 40。
For each category, the angle is (frequency ÷ total) × 360°. Walk: (12 ÷ 40) × 360° = 0.3 × 360° = 108°. Cycle: (8 ÷ 40) × 360° = 72°. Bus: (15 ÷ 40) × 360° = 135°. Car: (5 ÷ 40) × 360° = 45°.
对于每个类别，角度 = (频率 ÷ 总数) × 360°。步行：(12 ÷ 40) × 360° = 0.3 × 360° = 108°。骑车：(8 ÷ 40) × 360° = 72°。公交：(15 ÷ 40) × 360° = 135°。小汽车：(5 ÷ 40) × 360° = 45°。
Check that the angles sum to 360°: 108 + 72 + 135 + 45 = 360, which is correct.
检查角度之和是否为 360°：108 + 72 + 135 + 45 = 360，结果正确。
Part (b) asks to draw the pie chart. Use a protractor to measure each angle accurately, label each sector clearly, and give the chart a title such as ‘Method of Travel to School’.
第 (b) 小题要求绘制饼图。用量角器准确测量每个角度，清楚地标注每个扇区，并给图表加上标题，例如“上学交通方式”。
6. Question 6: Finding the Median | 题型六：求中位数
The question gives an ordered set of data: 11, 14, 15, 18, 21. Find the median.
题目给出一组有序数据：11, 14, 15, 18, 21。要求找出中位数。
The median is the middle value when the numbers are arranged in order. Since there are 5 numbers (odd count), the median is the 3rd number: 15.
中位数是数字按顺序排列时的中间值。因为有 5 个数字（奇数个），中位数就是第 3 个数字：15。
Explain what the median tells us: it is a measure of central tendency that is not affected by extremely high or low values, unlike the mean.
请解释中位数所代表的含义：它是一组数据集中趋势的量度，与平均数不同，它不受极端高值或低值的影响。
For an even number of values, you would need to find the mean of the two middle numbers. In this case, it is straightforward.
对于偶数个值，你需要找出中间两个数的平均数。而这道题直截了当。
7. Question 7: Comparing Data with a Dual Bar Chart | 题型七：使用双条形图比较数据
The question shows a dual bar chart comparing boys’ and girls’ favourite lunch choices: Sandwiches, Pizza, Salad, Pasta. Boys: Sandwich (6), Pizza (10), Salad (4), Pasta (5). Girls: Sandwich (8), Pizza (7), Salad (9), Pasta (6).
题目展示了一张双条形图，比较男孩和女孩最喜欢的午餐选择：三明治、披萨、沙拉、意面。男孩：三明治（6），披萨（10），沙拉（4），意面（5）。女孩：三明治（8），披萨（7），沙拉（9），意面（6）。
Part (a) asks: ‘Which lunch option is preferred by more girls than boys?’ Compare each pair: Sandwiches girls (8) > boys (6); Salad girls (9) > boys (4); Pasta girls (6) > boys (5). So Sandwiches, Salad and Pasta have higher girls’ counts. However, the question might ask for the option with the greatest difference: Salad (difference 5).
第 (a) 小题问：“哪一种午餐选项女孩子比男孩子更喜欢？” 逐项比较：三明治女孩（8）> 男孩（6）；沙拉女孩（9）> 男孩（4）；意面女孩（6）> 男孩（5）。因此，三明治、沙拉和意面都是女孩计数更高。但题目可能询问差异最大的选项：沙拉（相差 5）。
Part (b) asks: ‘How many students chose Pizza in total?’ Add boys and girls: 10 + 7 = 17 students.
第 (b) 小题问：“共有多少名学生选择了披萨？” 将男孩和女孩相加：10 + 7 = 17 名学生。
Always read the question carefully to know whether you are comparing within one category or across categories.
务必仔细审题，以明确是在一个类别内还是比较跨类别数据。
8. Question 8: Interpreting a Line Graph | 题型八：解读折线图
The question provides a line graph showing the maximum temperature in °C for each day of a week: Mon 12°C, Tue 14°C, Wed 15°C, Thu 13°C, Fri 11°C, Sat 10°C, Sun 9°C.
题目提供了一张折线图，显示一周中每天的最高温度（°C）：周一 12°C，周二 14°C，周三 15°C，周四 13°C，周五 11°C，周六 10°C，周日 9°C。
Part (a) asks: ‘On which day was the temperature highest?’ The peak is Wednesday at 15°C.
第 (a) 小题问：“哪一天温度最高？” 最高点是周三，15°C。
Part (b) asks: ‘Describe the overall trend from Monday to Sunday.’ The temperature increased from Monday to Wednesday, then decreased steadily from Wednesday to Sunday. Overall, there is a downward trend after midweek.
第 (b) 小题问：“描述从周一到周日的整体趋势。” 温度从周一到周三上升，然后从周三到周日稳步下降。总体而言，周中之后呈现下降趋势。
Part (c) asks: ‘What is the difference between the highest and lowest temperatures?’ Subtract: 15 – 9 = 6°C.
第 (c) 小题问：“最高温度和最低温度相差多少？” 相减：15 – 9 = 6°C。
When describing trends, use words like increase, decrease, peak, steady, sharp drop. Avoid vague language.
描述趋势时，使用诸如上升、下降、峰值、平稳、急剧下降等词语。避免模糊的语言。
9. Question 9: Evaluating a Questionnaire Question | 题型九：评价调查问卷设计
The question shows a draft survey question: ‘How often do you exercise? □ Often □ Sometimes □ Never’. It asks to identify one problem and write a better question.
题目展示了一个调查问卷草案问题：“你多久锻炼一次？ □ 经常 □ 有时 □ 从不”。要求指出一个问题并写出更好的问题。
Problem: The response options are vague and subjective. What ‘often’ means to one person may differ from another’s interpretation. Also, the question does not specify a time frame.
问题：回答选项模糊且主观。对一个人而言，“经常” 的含义可能与另一个人的理解不同。此外，问题没有指定时间范围。
Improved question: ‘In a typical week, how many times do you exercise for at least 30 minutes? □ 0 □ 1-2 □ 3-4 □ 5 or more’. This makes the data quantitative, easier to compare, and removes ambiguity.
改进后的问题：“在一周内，你通常进行几次至少 30 分钟的锻炼？ □ 0 次 □ 1-2 次 □ 3-4 次 □ 5 次或以上”。这使数据定量化，更容易比较，并消除了歧义。
Other possible problems: overlapping options, leading questions, or missing an ‘Other’ option when necessary. Always check that the question is clear and unbiased.
其他可能的问题：选项重叠、诱导性问题，或在必要时缺少“其他”选项。务必检查问题是否清晰且不带偏见。
10. Question 10: Basic Probability from a Frequency Table | 题型十：从频率表计算基本概率
The question gives a frequency table of blood types in a sample: A (20), B (15), O (30), AB (10). Part (a) asks: ‘If a person is chosen at random, what is the probability they have blood type O?’
题目提供了某个样本中血型的频率表：A（20）、B（15）、O（30）、AB（10）。第 (a) 小题问：“如果随机选择一个人，此人是 O 型血的概率是多少？”
First, find the total number of people: 20 + 15 + 30 + 10 = 75.
首先，求出总人数：20 + 15 + 30 + 10 = 75。
Probability = (number of people with type O) ÷ (total number of people) = 30 ÷ 75. Simplify the fraction by dividing numerator and denominator by 15 to get 2/5.
概率 = (具有 O 型血的人数) ÷ (总人数) = 30 ÷ 75。分子分母同时除以 15 化简分数，得到 2/5。
Part (b) asks: ‘What is the probability that a randomly chosen person does NOT have type AB?’ People not AB = 75 – 10 = 65. Probability = 65/75 = 13/15.
第 (b) 小题问：“随机选择一个人，此人不具有 AB 型血的概率是多少？” 非 AB 型的人数 = 75 – 10 = 65。概率 = 65/75 = 13/15。
Always simplify fractions unless told otherwise. Write the probability as a fraction in simplest form.
除非另有说明，务必化简分数。将概率写成最简分数形式。
Published by TutorHao | Statistics Revision Series | aleveler.com
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