📚 Year 7 OCR Statistics: Mock Test Walkthrough | Year 7 OCR 统计学：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock test designed for the Year 7 OCR Statistics unit. Each question is broken down with step-by-step solutions and helpful tips to strengthen your understanding of data handling, averages, charts, and probability.
本文为 Year 7 OCR 统计学单元测试模拟卷的详细解析。每道题都配有逐步解答和学习技巧，帮助你巩固数据处理、平均数、图表和概率等知识点。
1. Overview of the Mock Test | 模拟卷概述
This mock test consists of eight questions that mirror the style of OCR assessments. They cover primary and secondary data, bar charts, pictograms, averages and range, pie charts, line graphs, and basic probability. You should aim to complete the test in about 40 minutes, showing all working clearly.
这份模拟卷包含八道题，模拟了OCR的考试风格。题目涵盖初级数据与次级数据、条形图、象形图、平均数与范围、饼图、折线图以及基础概率。你应该在40分钟内完成测试，并清晰地展示所有解题步骤。
The questions progress from straightforward data identification to interpreting more complex representations. The mark allocation is shown for each part, just like in a real test.
题目从简单的数据辨认逐渐过渡到解读更复杂的统计图表。每部分的分数分配都与真实考试一致。
2. Question 1: Data Types and Collection | 第1题：数据类型与收集
Question: A group of Year 7 students recorded the number of hours they spent on homework each day for one week. The results for Monday to Friday were: 1.5, 2, 0.5, 1, 2.5. (a) Is this data primary or secondary? Explain. (2 marks) (b) Suggest one way to improve the data collection. (1 mark)
问题：一组 Year 7 学生记录了一周中每天做作业的小时数。周一至周五的结果为：1.5, 2, 0.5, 1, 2.5。(a) 这些数据是初级数据还是次级数据？请解释。(2分) (b) 提出一条改进数据收集方法的建议。(1分)
Solution (a): The data is primary data because the students gathered it themselves through direct observation or measurement. They did not obtain it from an existing source such as a website or a book.
解答 (a)：这些数据是初级数据，因为学生们通过直接观察或测量亲自收集了信息。他们并非从网站或书本等已有来源获得数据。
(b): A good improvement would be to record the time to the nearest quarter of an hour or to include weekend days to get a fuller picture of the weekly homework pattern.
(b)：一个合理的改进是将时间记录精确到一刻钟（15分钟），或者将周末也纳入记录，以便更完整地反映一周的作业模式。
3. Question 2: Bar Charts | 第2题：条形图
Question: 30 students were asked their favourite colour. The results are shown in the table. (a) Draw a fully labelled bar chart. (3 marks) (b) Write down the mode. (1 mark) (c) What fraction of the students chose Blue? Give your answer in its simplest form. (2 marks)
问题：30名学生接受了“最喜欢的颜色”调查。结果如下表。(a) 绘制一个标注完整的条形图。(3分) (b) 写出众数。(1分) (c) 选择蓝色的学生占几分之几？请化为最简分数。(2分)
| Colour | 颜色 | Frequency | 频率 |
|---|---|
| Red | 红色 | 8 |
| Blue | 蓝色 | 12 |
| Green | 绿色 | 5 |
| Yellow | 黄色 | 3 |
| Other | 其他 | 2 |
For the bar chart, draw two axes. Label the horizontal axis 'Colour' and the vertical axis 'Frequency' vertical axis. Draw bars of equal width, without gaps, rising to the correct frequencies.
绘制条形图时，需要画出两条坐标轴。横轴标注为“颜色”，纵轴标注为“频率”。纵轴可采用1厘米代表2名学生的比例尺。各条形宽度相等、之间不留间隙，高度与对应频数一致。
The mode is the colour with the highest frequency, which is Blue (12 students).
众数是出现频率最高的颜色，即蓝色（12名学生）。
ng Blue is 12 out of 30. Simplify by dividing numerator and denominator by 6: 12/30 = 2/5.
选择蓝色的分数为30之12。分子分母同时除以6可化简：12/30 = 2/5。
4. Question 3: Averages and Range | 第3题：平均数与范围
Question: Here are the scores from a spelling test: 8, 9, 7, 6, 9, 8, 7, 10, 8. (a) Find the mode. (1 mark) (b) Find the median. (2 marks) (c) Calculate the mean. (2 marks) (d) Work out the range. (1 mark)
问题：以下是一次拼写测试的分数：8, 9, 7, 6, 9, 8, 7, 10, 8。(a) 找出众数。(1分) (b) 找出中位数。(2分) (c) 计算平均数。(2分) (d) 求出范围。(1分)
First, write the scores in ascending order: 6, 7, 7, 8, 8, 8, 9, 9, 10. The mode is the value that appears most often. Here 8 appears three times, so the mode is 8.
首先，将分数从小到大排列：6, 7, 7, 8, 8, 8, 9, 9, 10。众数是出现次数最多的数值。此处8出现了三次，所以众数为8。
The median is the middle value. With 9 numbers, the 5th value is the median. The 5th number in the ordered list is 8, so the median is 8.
中位数是位于中间的数。共有9个数据，第5个为中位数。有序列表中第5个数字是8，因此中位数为8。
To find the mean, add all the scores together and divide by the number of scores.
计算平均数时，将所有分数相加，再除以分数的个数。
Sum = 8 + 9 + 7 + 6 + 9 + 8 + 7 + 10 + 8 = 72
Mean = 72 ÷ 9 = 8
求和：8+9+7+6+9+8+7+10+8 = 72。
平均数 = 72 ÷ 9 = 8
ce between the highest and lowest values. Highest = 10, lowest = 6, so range = 10 - 6 = 4.
范围（极差）是最大值与最小值之差。最大值=10，最小值=6，所以范围 = 10 - 6 = 4。
5. Question 4: Pie Charts | 第4题：饼图
Question: A survey asked 40 people to choose their favourite fruit. The results were: Apple 10, Banana 15, Orange 8, Grape 7. (a) Calculate the angle for each sector of a pie chart. (3 marks) (b) Draw the pie chart, labelling each sector. (2 marks) (c) What percentage of the people chose Banana? (1 mark)
问题：一项调查询问了40人最喜欢的水果。结果为：苹果10人，香蕉15人，橙子8人，葡萄7人。(a) 计算饼图中各扇形的角度。(3分) (b) 绘制饼图并标注各扇形。(2分) (c) 选择香蕉的人数占百分比多少？(1分)
ull circle has 360°. The angle for each fruit is calculated as (frequency ÷ total) × 360°.
总人数为40。整圆为360°。每种水果的扇形角度计算方法为：(频数 ÷ 总数) × 360°。
Apple: (10 ÷ 40) × 360° = 0.25 × 360° = 90°
苹果：(10 ÷ 40) × 360° = 0.25 × 360° = 90°
Banana: (15 ÷ 40) × 360° = 0.375 × 360° = 135°
香蕉：(15 ÷ 40) × 360° = 0.375 × 360° = 135°
Orange: (8 ÷ 40) × 360° = 0.2 × 360° = 72°
橙子：(8 ÷ 40) × 360° = 0.2 × 360° = 72°
Grape: (7 ÷ 40) × 360° = 0.175 × 360° = 63°
葡萄：(7 ÷ 40) × 360° = 0.175 × 360° = 63°
When drawing, use a protractor to measure each angle from the same starting radius. Label each sector clearly.
绘图时，使用量角器从同一起始半径出发量取各个角度。每个扇形都要清晰标注。
The percentage who chose Banana is (15/40) × 100% = 37.5%.
选择香蕉的百分比为 (15/40) × 100% = 37.5%。
6. Question 5: Pictograms | 第5题：象形图
Question: A pictogram uses a book symbol to represent 4 books read. The table below shows the number of books read on three days. Complete the pictogram and answer the questions. (a) How many books were read on Tuesday? (1 mark) (b) How many books were read in total over the three days? (2 marks)
问题：某象形图用一个书本符号代表4本已读的书。下表显示了三天内阅读的书籍数量。完成象形图并回答问题。(a) 周二阅读了多少本书？(1分) (b) 三天总共阅读了多少本书？(2分)
| Day | 星期 | Number of symbols | 符号数量 |
|---|---|
| Monday | 星期一 | 3 |
| Tuesday | 星期二 | 2 ½ |
| Wednesday | 星期三 | 4 |
Each full symbol represents 4 books. Half a symbol represents 2 books.
每个完整符号代表4本书。半个符号代表2本书。
ls give 8 books, plus half a symbol (2 books) gives 10 books. Wednesday: 4 × 4 = 16 books.
星期一：3 × 4 = 12本。星期二：2个完整符号为8本，加上半个符号（2本）共10本。星期三：4 × 4 = 16本。
Total books = 12 + 10 + 16 = 38 books.
总本数 = 12 + 10 + 16 = 38本。
7. Question 6: Line Graphs | 第6题：折线图
Question: The temperature (°C) was recorded every 2 hours from 8am to 6pm: 8am 12°C, 10am 15°C, 12pm 18°C, 2pm 20°C, 4pm 19°C, 6pm 16°C. (a) Plot a line graph of temperature against time. (3 marks) (b) Describe the trend in temperature during this period. (1 mark) (c) Estimate the temperature at 1pm. (1 mark)
问题：从上午8点到下午6点，每两小时记录一次气温(°C)：8am 12°C，10am 15°C，12pm 18°C，2pm 20°C，4pm 19°C，6pm 16°C。(a) 绘制温度随时间变化的折线图。(3分) (b) 描述该时段内温度的变化趋势。(1分) (c) 估算下午1点的气温。(1分)
ale, such as 1 cm = 2°C. Plot points accurately and join them with straight line segments.
将时间标在横轴上，温度标在纵轴上。选取合适的比例尺，例如1厘米代表2°C。准确描点并用直线段连接各点。
The temperature increased from 8am to 2pm, reaching a maximum of 20°C at 2pm. After that it decreased gradually until 6pm. The overall trend shows a rise and then a fall.
温度从上午8点至下午2点持续上升，在下午2点达到最高值20°C，之后逐渐下降至下午6点。整体趋势表现为先升后降。
1pm is exactly halfway between 12pm (18°C) and 2pm (20°C). A good estimate is 19°C, assuming a steady rise.
下午1点正好在12pm（18°C）和2pm（20°C）的中点。假设温度匀速上升，合理的估算值为19°C。
8. Question 7: Probability | 第7题：概率
Question: A bag contains 3 red, 2 blue and 5 green counters. A counter is taken at random. (a) What is the probability it is red? (1 mark) (b) What is the probability it is blue or green? (2 marks) (c) What is the probability it is not red? (1 mark)
问题：一个袋子里装有3个红色、2个蓝色和5个绿色筹码。随机取出一个。(a) 取出红色筹码的概率是多少？(1分) (b) 取出蓝色或绿色筹码的概率是多少？(2分) (c) 取出非红色筹码的概率是多少？(1分)
Probability is calculated as (number of favourable outcomes) ÷ (total number of outcomes).
筹码总数 = 3 + 2 + 5 = 10。概率的计算方法为（有利结果的数量）÷（所有可能结果的总数）。
P(red) = 3/10
P(红色) = 3/10
r green, add the number of blue and green counters: 2 + 5 = 7. So P(blue or green) = 7/10.
对于蓝色或绿色，将蓝色与绿色筹码的数量相加：2 + 5 = 7。所以 P(蓝色或绿色) = 7/10。
The event 'not red' includes blue and green counters, which is the same as the previous part. P(not red) = 7/10. Alternatively, 1 - P(red) = 1 - 3/10 = 7/10.
“非红色”这个事件包括蓝色和绿色筹码，与上一问的结果相同。P(非红色) = 7/10。或者，用 1 - P(红色) = 1 - 3/10 = 7/10 计算亦可。
9. Question 8: Comparing Data Sets | 第8题：比较数据集
Question: an = 68, range = 35. Compare the performance and consistency of the two classes. (2 marks)
问题：A班和B班参加了同一场科学测验（满分100）。A班：平均分 = 65，极差 = 20。B班：平均分 = 68，极差 = 35。比较两个班级的表现和成绩的稳定性。(2分)
On average, Class B performed slightly better because their mean score (68) is higher than Class A's mean (65). However, Class A's scores are more consistent, as shown by the smaller range of 20 compared to 35 in Class B. A larger range indicates that Class B had a wider spread of scores, with some students scoring much lower or much higher than the mean.
平均来看，B班的表现略好，因为他们的平均分（68分）高于A班的平均分（65分）。然而，A班的成绩更稳定，其极差为20，比B班的35要小。较大的极差表明B班的分数分布更分散，部分学生的成绩远低于或远高于平均分。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
Mistake 1: Forgetting to order the data before finding the median. Always sort from smallest to largest first.
错误1：在寻找中位数前忘记将数据排序。务必先将数据从小到大排列。
Mistake 2: Confusing the mean with the mode or median. Remember the mean uses all values and involves division, the mode is the most frequent, and the median is the central value.
错误2：混淆平均数、众数和中位数。记住平均数用到所有数值并涉及除法，众数是出现次数最多的值，中位数则是最中间的值。
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