📚 Year 9 Edexcel Statistics: Unit Test Mock Paper Walkthrough | Year 9 Edexcel 统计：单元测试模拟卷解析
This article presents a complete walkthrough of a mock unit test for Year 9 Edexcel Statistics. We will work through ten carefully chosen questions that mirror the style of Edexcel assessments, covering data types, averages, charts, probability, sampling, and more. Each solution is explained step by step, with paired English and Chinese commentary to help bilingual learners master the essential skills.
本文完整解析一份 Year 9 Edexcel 统计单元测试模拟卷。我们精选了十道贴合 Edexcel 考试风格的题目，涵盖数据类型、平均数、图表、概率、抽样等核心主题。每道题都配有逐步解析，中英文一一对照，帮助双语学习者扎实掌握必备技能。
1. Question 1: Classifying Data | 问题1：数据分类
A student records the following variables from her classmates: (a) shoe size, (b) height in cm, (c) favourite colour, (d) number of pets. Classify each as categorical, discrete, or continuous data. Shoe size is numerical but only takes certain values (e.g. 4, 4.5, 5), so it is discrete. Height can take any value within a range, so it is continuous. Favourite colour is a word describing a category, so it is categorical. Number of pets is a count, which only gives whole numbers, so it is discrete.
一位同学记录了几个来自班级的变量：(a) 鞋码，(b) 身高（cm），(c) 最喜欢的颜色，(d) 宠物数量。请将每个变量归类为分类数据、离散数据或连续数据。鞋码是数值型但只能取特定值（如 4、4.5、5），因此是离散数据。身高可以在一个区间内取任意值，因此是连续数据。最喜欢的颜色是描述类别的词语，因此是分类数据。宠物数量是计数值，只给出整数，因此是离散数据。
2. Question 2: Mean from a Frequency Table | 问题2：从频数表求平均数
The table shows scores from a quiz: Score 1 (frequency 3), Score 2 (freq 5), Score 3 (freq 2), Score 4 (freq 2). To find the mean, add an ‘fx’ column: 1×3=3, 2×5=10, 3×2=6, 4×2=8. Sum of fx = 27. Sum of f = 12. Mean = 27 ÷ 12 = 2.25. Always check that your sum of frequencies matches the total number of data points. The mean tells us the average score per student was 2.25 marks.
表格显示了一次小测验的得分：1 分（频数 3）、2 分（频数 5）、3 分（频数 2）、4 分（频数 2）。要计算平均数，增加一列 ‘fx’：1×3=3，2×5=10，3×2=6，4×2=8。fx 总和 = 27，频数总和 = 12。平均数 = 27 ÷ 12 = 2.25。请务必检查频数总和是否与数据点总个数一致。平均数告诉我们每位学生的平均得分是 2.25 分。
3. Question 3: Interpreting a Bar Chart | 问题3：解读条形图
A bar chart shows the number of students choosing different sports: Football 14, Netball 10, Tennis 6, Swimming 12. Identify the most popular sport (Football) and find the total number of students asked. Total = 14+10+6+12 = 42. When drawing a bar chart, remember that bars should be equal in width, gaps between bars must be uniform, and the frequency axis must start from zero. Label both axes clearly and give the chart a title.
某条形图显示了选择不同运动的学生人数：足球 14、无板篮球 10、网球 6、游泳 12。指出最受欢迎的运动（足球）并求出被调查的学生总人数。总数 = 14+10+6+12 = 42。绘制条形图时，记住条形宽度应相等，条形之间的间距必须一致，频数轴必须从零开始。请清晰地标注两条坐标轴并为图表写上标题。
4. Question 4: Stem-and-Leaf Diagram and Quartiles | 问题4：茎叶图与四分位数
Data set: 15, 22, 23, 25, 31, 34, 36, 41. Draw an ordered stem-and-leaf diagram. Stems are 1,2,3,4. Leaf for 15 is ‘5’ on stem 1; stem 2 gets leaves 2,3,5; stem 3 gets 1,4,6; stem 4 gets 1. Always include a key (e.g. 2|3 means 23). For 8 values, the median lies between the 4th and 5th values: (25+31)/2 = 28. The lower quartile is the median of the first four numbers: (22+23)/2 = 22.5. Upper quartile is the median of the last four: (34+36)/2 = 35. Interquartile range = 35 – 22.5 = 12.5.
数据集：15, 22, 23, 25, 31, 34, 36, 41。画一个有序茎叶图。茎为 1、2、3、4。15 的叶子是 ‘5’ 写在茎 1 上；茎 2 的叶子是 2、3、5；茎 3 是 1、4、6；茎 4 是 1。始终要给出图例（例如 2|3 表示 23）。对于 8 个数据值，中位数位于第 4 和第 5 个值的中间：(25+31)/2 = 28。下四分位数是前四个数中的中位数：(22+23)/2 = 22.5。上四分位数是后四个数中的中位数：(34+36)/2 = 35。四分位距 = 35 – 22.5 = 12.5。
5. Question 5: Probability from a Two-Way Table | 问题5：利用双向表求概率
A two-way table shows students’ favourite subjects: 10 boys like Maths, 8 girls like Maths; 12 boys like Science, 6 girls like Science; 3 boys like English, 11 girls like English. Find the probability that a randomly chosen student is a girl who likes Science. Total students = 10+8+12+6+3+11 = 50. Girls liking Science = 6. Probability = 6/50 = 3/25 (or 0.12). Alternatively, find P(girl) = (8+6+11)/50 = 25/50 = 0.5. Conditional probabilities could also be asked: P(Science | girl) = 6/25 = 0.24.
一张双向表显示学生最喜欢的科目：喜欢数学的有 10 名男生、8 名女生；喜欢科学的有 12 名男生、6 名女生；喜欢英语的有 3 名男生、11 名女生。求随机选到一名喜欢科学的女生的概率。总人数 = 10+8+12+6+3+11 = 50。喜欢科学的女生 = 6。概率 = 6/50 = 3/25（或 0.12）。也可计算 P（女生）= (8+6+11)/50 = 0.5。还可能考条件概率：P（科学 | 女生）= 6/25 = 0.24。
6. Question 6: Scatter Graphs and Correlation | 问题6：散点图与相关性
Plot a scatter graph for hours of revision (x) and test marks (y): (2, 45), (3, 50), (5, 65), (1, 35), (4, 60). The points show a general upward trend, indicating positive correlation. Draw a line of best fit that passes through the middle of the points. Use the line to estimate the mark for 3.5 hours of revision: read across from x=3.5 to the line, then down to the y-axis, giving roughly 55 marks. Caution: extrapolating beyond the data range can be unreliable.
绘制复习小时数（x）与测验分数（y）的散点图：(2,45), (3,50), (5,65), (1,35), (4,60)。这些点整体呈上升趋势，表明存在正相关。画一条最适线使其从点群中间穿过。利用该线估计复习 3.5 小时的分数：从 x=3.5 向上对应到最适线，再水平对应到 y 轴，得到大约 55 分。注意：在数据范围之外进行外推可能不可靠。
7. Question 7: Sampling Methods | 问题7：抽样方法
A school wants to survey Year 9 students about lunch choices. Suggest a suitable sampling method and explain why. A simple random sample could be obtained by assigning each student a number and using a random number generator to pick 50. This avoids bias and gives every student an equal chance. Stratified sampling could ensure proportional representation from each tutor group. Avoid convenience sampling, such as asking only your friends, as it leads to biased results. Cluster sampling might be used if classes are selected randomly.
一学校想调查九年级学生对午餐的选择。请建议一种合适的抽样方法并解释理由。简单随机抽样可以通过给每位学生编号，再用随机数生成器选出 50 人来实现。这种方法避免了偏差，使每位学生入选机会均等。分层抽样可以保证每个导师组按比例被抽中。应避免便利抽样（例如只问自己的朋友），因为它会导致结果有偏差。如果随机整班抽选，则可使用整群抽样。
8. Question 8: Pie Chart Calculations | 问题8：饼图计算
A pie chart shows favourite fruit. The sector for apples has an angle of 120° and represents 30 students. Find the total number of students surveyed. Since 120° out of 360° corresponds to 30 students, the fraction is 120/360 = 1/3. Therefore, 1/3 of the total is 30, so total = 30 × 3 = 90 students. To find the number of students for a sector of 80°, use proportion: (80/360) × 90 = (2/9) × 90 = 20 students. Always check the sector angles add to 360°.
一个饼图展示了最喜欢的水果。苹果的扇区角度为 120°，代表 30 名学生。求被调查学生的总人数。120° 对应 360° 中的 120/360 = 1/3，所以总人数的 1/3 就是 30，总人数 = 30 × 3 = 90 名学生。要计算一个 80° 扇区所代表的学生数，用比例计算：(80/360) × 90 = (2/9) × 90 = 20 名学生。务必验证扇区角度总和为 360°。
9. Question 9: Range and Interquartile Range | 问题9：极差与四分位距
Given the data set: 12, 15, 18, 22, 24, 30, 35. The range is the difference between the largest and smallest values: 35 – 12 = 23. To find the IQR, first locate the median (Q2): the 4th value is 22. The lower half (12,15,18) has median 15 (Q1). The upper half (24,30,35) has median 30 (Q3). Thus IQR = Q3 – Q1 = 30 – 15 = 15. The IQR shows the spread of the middle 50% of the data, which is less affected by extreme values than the range.
给定数据集：12, 15, 18, 22, 24, 30, 35。极差是最大值与最小值之差：35 – 12 = 23。求四分位距 (IQR) 时，先找中位数 (Q2)：第 4 个数 22。较小的一半 (12,15,18) 的中位数是 15（Q1）；较大的一半 (24,30,35) 的中位数是 30（Q3）。因此 IQR = 30 – 15 = 15。IQR 体现了中间 50% 数据的离散程度，与极差相比，它受极端值的影响更小。
10. Question 10: Time Series and Trend | 问题10：时间序列与趋势
Quarterly sales figures (£000s) are: Q1: 24, Q2: 18, Q3: 30, Q4: 40, Q5: 28, Q6: 22, Q7: 34, Q8: 44. Plot these as a line graph. The overall trend is increasing, with regular seasonal fluctuations: sales peak in Q4 of each year and dip in Q2. To smooth out seasonality, calculate a 4-point moving average: first average for Q1-Q4: (24+18+30+40)/4 = 28; next for Q2-Q5: (18+30+40+28)/4 = 29; and so on. Plot the moving averages to see the trend more clearly.
季度销售额（千英镑）为：Q1: 24, Q2: 18, Q3: 30, Q4: 40, Q5: 28, Q6: 22, Q7: 34, Q8: 44。将这些数据绘制成折线图。整体趋势是上升的，并伴有规则的季节性波动：每年 Q4 销售额达到高峰，Q2 有所下降。为消除季节性影响，计算四项移动平均：首先是 Q1-Q4 的平均：(24+18+30+40)/4 = 28；然后是 Q2-Q5 的平均：(18+30+40+28)/4 = 29；以此类推。绘制移动平均值可以更清晰地看出趋势。
Published by TutorHao | Statistics Revision Series | aleveler.com
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