📚 OCR Year 7 Statistics Mock Paper Walkthrough | OCR 7年级统计单元测试模拟卷解析
This article provides a detailed, question-by-question walkthrough of a mock unit test designed for the OCR Year 7 Statistics syllabus. The paper covers key topics including averages, frequency tables, bar charts, pie charts, scatter graphs, basic probability, data comparison, misleading graphs, and questionnaire design. Each section presents the question in English and Chinese, followed by a step-by-step solution with bilingual explanations. Use this guide to consolidate your understanding and prepare effectively for your end-of-unit assessment.
本文为OCR考试局7年级统计单元测试模拟卷提供逐题解析。试卷涵盖平均数、频率表、条形图、饼图、散点图、基础概率、数据比较、误导性图表和问卷设计等核心主题。每个小节均给出中英文原题，并配以双语分步解答。通过本指南巩固知识点，为单元测验做好充分准备。
1. Mean, Median, Mode & Range | 平均数、中位数、众数和范围
Question 1: Ten students scored the following marks in a test: 12, 15, 14, 10, 18, 16, 13, 14, 15, 12. (a) Calculate the mean. (b) Find the median. (c) Identify the mode. (d) Work out the range.
题目一：十名学生在测验中的分数如下：12, 15, 14, 10, 18, 16, 13, 14, 15, 12。(a) 计算平均数。 (b) 找出中位数。 (c) 找出众数。 (d) 求全距。
Solution: First, add all the scores together: 12 + 15 + 14 + 10 + 18 + 16 + 13 + 14 + 15 + 12 = 139. There are 10 values, so the mean is 139 ÷ 10 = 13.9. Use the formula:
解答：首先将所有分数相加：12 + 15 + 14 + 10 + 18 + 16 + 13 + 14 + 15 + 12 = 139。一共有10个数值，因此平均数为 139 ÷ 10 = 13.9。所用公式：
Mean = Σx ÷ n
For the median, write the marks in order from smallest to largest: 10, 12, 12, 13, 14, 14, 15, 15, 16, 18. With an even number of data points, the median is the mean of the 5th and 6th values: (14 + 14) ÷ 2 = 14.
求中位数时，先将分数从小到大排序：10, 12, 12, 13, 14, 14, 15, 15, 16, 18。数据个数为偶数，中位数是第5和第6个值的平均数：(14 + 14) ÷ 2 = 14。
The mode is the value that appears most often. Here, 12, 14 and 15 each appear twice, so the data set has three modes: 12, 14 and 15.
众数是出现次数最多的数值。此处12、14和15各出现两次，因此该数据集有三个众数：12、14和15。
The range is the difference between the highest and lowest scores: 18 − 10 = 8.
全距为最高分与最低分之差：18 − 10 = 8。
2. Frequency Tables | 频率表
Question 2: A survey asked 24 people about their favourite colour. The results are shown in the frequency table below. Colour | Frequency: Red = 6, Blue = 9, Green = 4, Yellow = 3, Other = 2. (a) How many people took part? (b) What is the modal colour? (c) Find the median colour.
题目二：一项调查询问了24人最喜欢哪种颜色。结果如频率表所示：红色6人，蓝色9人，绿色4人，黄色3人，其他2人。(a) 共有多少人参与？ (b) 众数颜色是什么？ (c) 找出中位数颜色。
Solution: The total number of people is the sum of all frequencies: 6 + 9 + 4 + 3 + 2 = 24. The mode is the colour with the highest frequency, which is Blue (9 votes). To find the median colour, calculate cumulative frequencies: Red 6, Blue 15 (6+9), Green 19, Yellow 22, Other 24. The median position is (24 + 1) ÷ 2 = 12.5, so we need the 12th and 13th data points. Both lie in the Blue category (positions 7 to 15), therefore the median colour is Blue.
解答：总人数为所有频率之和：6 + 9 + 4 + 3 + 2 = 24。众数是频率最高的颜色，即蓝色（9票）。要找出中位数颜色，需计算累积频率：红6，蓝15（6+9），绿19，黄22，其他24。中位数所在位置为 (24 + 1) ÷ 2 = 12.5，即第12和第13个数据点。这两者都在蓝色类别中（第7至15位），因此中位数颜色为蓝色。
3. Bar Charts | 条形图
Question 3: A bar chart displays the number of students in four Year 7 classes: Class 7A has 28 students, 7B has 30, 7C has 25, and 7D has 32. (a) Which class has the most students? (b) How many students are there in total? (c) Describe two things that must be included when drawing this bar chart.
题目三：某条形图显示了四个7年级班级的学生人数：7A班28人，7B班30人，7C班25人，7D班32人。(a) 哪个班级人数最多？ (b) 总共有多少名学生？ (c) 绘制此条形图时必须包含哪两点要素？
Solution: The class with the most students is 7D (32 students). The total number of students is 28 + 30 + 25 + 32 = 115. When drawing the bar chart, you must label both axes — the horizontal axis for classes, the vertical axis for frequency (number of students) with a suitable scale — and give the chart a title. Bars should be of equal width with even spacing between them.
解答：人数最多的班级是7D班（32人）。学生总数为 28 + 30 + 25 + 32 = 115。绘制条形图时，必须标注两条坐标轴——横轴为班级，纵轴为频数（学生人数）并选择合适的刻度——同时为图表添加标题。条形宽度应保持一致，间距均匀。
4. Pie Charts | 饼图
Question 4: 40 students recorded how they travel to school: Walk = 15, Cycle = 9, Bus = 6, Car = 10. (a) What fraction of students walk? (b) Calculate the angle for each sector of a pie chart. (c) Explain how to construct the pie chart.
题目四：40名学生记录了上学交通方式：步行15人，骑自行车9人，公交车6人，私家车10人。(a) 步行的学生占几分之几？ (b) 计算饼图中每个扇形的角度。 (c) 解释如何绘制该饼图。
Solution: The fraction who walk is 15/40, which simplifies to 3/8. To find each angle, multiply each fraction by 360°. Walk: 15/40 × 360° = 135°. Cycle: 9/40 × 360° = 81°. Bus: 6/40 × 360° = 54°. Car: 10/40 × 360° = 90°. Always check the angles sum to 360°: 135 + 81 + 54 + 90 = 360. To construct the chart, draw a circle and a radius. Use a protractor to measure and draw each central angle in turn, then label each sector with the category name and percentage or angle.
解答：步行学生所占比例为15/40，化简为3/8。计算每个角度时，用相应分数乘以360°。步行：15/40 × 360° = 135°。自行车：9/40 × 360° = 81°。公交车：6/40 × 360° = 54°。私家车：10/40 × 360° = 90°。务必检查角度之和为360°：135 + 81 + 54 + 90 = 360。绘制饼图时，先画一个圆及一条半径。使用量角器依次测量并画出每个圆心角，然后为每个扇形标注类别名称和百分比或角度。
5. Scatter Graphs & Correlation | 散点图与相关性
Question 5: The table shows the number of hours spent revising and the test score achieved. Hours (x): 1, 2, 3, 4, 5, 6. Score (y): 40, 50, 55, 65, 70, 80. (a) Plot the points on a scatter graph. (b) Describe the correlation. (c) Use the trend to estimate the score for a student who revised 4.5 hours.
题目五：表格显示了复习小时数与测验成绩的关系。小时数 (x)：1, 2, 3, 4, 5, 6。成绩 (y)：40, 50, 55, 65, 70, 80。(a) 在散点图上描出这些点。 (b) 描述相关性。 (c) 利用趋势估计复习4.5小时的学生可能取得的成绩。
Solution: After plotting, the points rise steadily from bottom left to top right, indicating a positive correlation: as revision time increases, the test score tends to increase. The relationship looks roughly linear. To estimate the score for 4.5 hours, draw a line of best fit and read the y-value at x = 4.5. Based on the pattern, a reasonable estimate is around 67 or 68 marks.
解答：描点后可见，这些点从左下方向右上方稳步上升，表明呈正相关：随着复习时间增加，测验成绩也倾向于提高。该关系大致呈线性。要估计复习4.5小时的成绩，需画出最佳拟合线，并在 x = 4.5 处读取 y 值。根据趋势，合理的估计值约为67或68分。
6. Basic Probability | 基础概率
Question 6: A bag contains 3 red balls, 2 blue balls and 5 yellow balls. One ball is taken at random. (a) Find P(red). (b) Find P(blue). (c) Find P(not yellow). (d) What is P(green)?
题目六：一个袋子里有3个红球、2个蓝球和5个黄球。随机取出一个球。(a) 求 P(红球)。 (b) 求 P(蓝球)。 (c) 求 P(非黄球)。 (d) P(绿球) 是多少？
Solution: Total number of balls = 3 + 2 + 5 = 10. P(red) = number of red balls / total = 3/10. P(blue) = 2/10 = 1/5. P(not yellow) means the ball is red or blue, so probability = (3+2)/10 = 5/10 = 1/2. Since there are no green balls in the bag, P(green) = 0.
解答：球的总数 = 3 + 2 + 5 = 10。P(红球) = 红球个数 / 总球数 = 3/10。P(蓝球) = 2/10 = 1/5。P(非黄球) 表示取出的是红球或蓝球，概率为 (3+2)/10 = 5/10 = 1/2。由于袋中没有绿球，P(绿球) = 0。
7. Mutually Exclusive and Exhaustive Events | 互斥且穷尽的事件
Question 7: A probability spinner has three colours: red, blue, and green. The probability of landing on red is 0.5, and on blue is 0.3. (a) What is the probability of landing on green? (b) Explain why these three outcomes are mutually exclusive and exhaustive.
题目七：一个概率转盘有三种颜色：红、蓝、绿。转到红色的概率是0.5，转到蓝色的概率是0.3。(a) 转到绿色的概率是多少？ (b) 解释为什么这三种结果是互斥且穷尽的。
Solution: Since the only possible outcomes are red, blue, and green, their probabilities must add to 1. Hence, P(green) = 1 − (0.5 + 0.3) = 0.2. The outcomes are mutually exclusive because the spinner cannot land on two colours at the same time. They are exhaustive because together they cover all possible results of a single spin.
解答：由于仅有可能的结果是红、蓝、绿，它们的概率之和必须为1。因此，P(绿色) = 1 − (0.5 + 0.3) = 0.2。这些结果是互斥的，因为转盘无法同时停留于两种颜色。它们也是穷尽的，因为这些结果共同涵盖了一次转动中所有可能出现的情况。
8. Comparing Data Sets | 比较数据集Published by TutorHao | Year 7 统计 Revision Series | aleveler.com
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