📚 Year 9 CAIE Statistics: Unit Test Mock Paper Walkthrough | 九年级 CAIE 统计：单元测试模拟卷解析
This article provides a step-by-step walkthrough of a Year 9 CAIE Statistics unit test mock paper. It covers key topics such as data types, frequency tables, charts, averages, measures of spread, box plots, probability, and scatter graphs. Each question is explained in detail, with solutions and exam tips, to help you build confidence and accuracy for your real assessment.
本文将对一套九年级 CAIE 统计单元测试模拟卷进行逐步解析。内容涵盖数据类型、频数表、图表、平均数、散布度量、箱线图、概率和散点图等核心主题。每道题均配有详细的解答过程与考试技巧，旨在帮助你增强信心、提升答题准确性，为正式考试做好准备。
1. Data Types and Frequency Tables | 数据类型与频数表
Question 1: A class of 25 students was asked about the number of pets they own. The responses were: 0, 1, 2, 1, 0, 3, 1, 2, 0, 1, 1, 2, 0, 1, 3, 0, 2, 1, 0, 1, 2, 0, 1, 2, 0. (a) State whether this data is qualitative or quantitative, and discrete or continuous. (b) Construct a frequency distribution table. (c) How many students have exactly one pet?
问题1：一个班级的25名学生被问及他们拥有的宠物数量。回答如下：0, 1, 2, 1, 0, 3, 1, 2, 0, 1, 1, 2, 0, 1, 3, 0, 2, 1, 0, 1, 2, 0, 1, 2, 0。（a）说明这组数据是定性数据还是定量数据，是离散数据还是连续数据。（b）构建频数分布表。（c）有多少名学生恰好有一只宠物？
Step 1 – Identify the data type: The data are counts (numbers) so they are quantitative. Since you cannot have a fraction of a pet, the counts are discrete (whole numbers). Therefore, the data are quantitative discrete.
步骤1——识别数据类型：这些数据是计数（数字），因此属于定量数据。由于宠物的数量不能是分数，这些计数是离散的（整数）。因此，数据为定量离散数据。
Step 2 – Tally and construct the frequency table. List each distinct number of pets (0, 1, 2, 3). Count how many times each appears: 0 appears 8 times, 1 appears 9 times, 2 appears 6 times, 3 appears 2 times. Check total: 8+9+6+2=25, which matches the number of students.
步骤2——计数并构建频数表。列出每个不同的宠物数量（0, 1, 2, 3）。统计每个数字出现的次数：0出现8次，1出现9次，2出现6次，3出现2次。检验合计：8+9+6+2=25，与学生总数一致。
| Number of pets (x) | Tally | Frequency (f) |
|---|---|---|
| 0 | |||| ||| | 8 |
| 1 | |||| |||| | 9 |
| 2 | |||| | | 6 |
| 3 | || | 2 |
(c) From the frequency table, the number of students with exactly one pet is 9.
（c）根据频数表，恰好有一只宠物的学生人数为9。
2. Bar Charts and Pie Chart Angles | 条形图与饼图角度
Question 2: The frequency table below shows the favourite sports of 40 students: Football 12, Basketball 8, Tennis 5, Swimming 10, Other 5. (a) Draw a bar chart to represent this data. (b) Calculate the angle for the ‘Football’ sector in a pie chart. (c) What fraction of students chose Swimming? Give your answer in its simplest form.
问题2：下面的频数表显示了40名学生最喜爱的运动：足球12人，篮球8人，网球5人，游泳10人，其他5人。（a）绘制条形图来展示这些数据。（b）计算饼图中“足球”扇区的角度。（c）选择游泳的学生占几分之几？请以最简分数形式给出答案。
Bar chart tips: Use sport categories on the horizontal axis and frequency on the vertical axis. Draw bars of equal width, with heights equal to the frequencies. Always label axes and give the chart a title, e.g. ‘Favourite Sports of 40 Students’.
条形图提示：以运动类别为横轴，频数为纵轴。绘制等宽的条形，高度等于对应的频数。务必给坐标轴加标签并为图表设置标题，如“40名学生最喜爱的运动”。
Pie chart angle calculation: Total frequency = 12+8+5+10+5 = 40. The whole circle represents 360°. For Football, frequency = 12. Angle = (12/40) × 360° = (3/10) × 360° = 108°. So the Football sector angle is 108°.
饼图角度计算：总频数 = 12+8+5+10+5 = 40。整个圆代表360°。足球频数为12。角度 = (12/40) × 360° = (3/10) × 360° = 108°。因此足球扇区角度为108°。
Fraction for Swimming: Frequency for Swimming is 10, total is 40. Fraction = 10/40 = 1/4. So students who chose Swimming represent one quarter of the total.
游泳的比例：游泳频数为10，总数为40。分数 = 10/40 = 1/4。因此选择游泳的学生占总人数的四分之一。
3. Mean, Median and Mode | 平均数、中位数与众数
Question 3: The numbers of books read by 7 students in a month are: 3, 4, 2, 5, 2, 6, 3. (a) Calculate the mean number of books. (b) Find the median. (c) State the mode. (d) Explain why the median might be a better measure of central tendency than the mean if a value of 20 is added to the set.
问题3：7名学生在一个月内阅读的书籍数量如下：3, 4, 2, 5, 2, 6, 3。（a）计算平均阅读量。（b）求中位数。（c）写出众数。（d）若在数据集中加入一个数值20，解释为什么中位数可能比平均数更能代表集中趋势。
Mean: Sum = 3+4+2+5+2+6+3 = 25. Number of values n = 7. Mean = 25/7 ≈ 3.57 (to 2 decimal places).
平均数：求和 = 3+4+2+5+2+6+3 = 25。数值个数 n = 7。平均数 = 25/7 ≈ 3.57（保留两位小数）。
Median: First arrange in order: 2, 2, 3, 3, 4, 5, 6. The middle value (4th) is 3. So median = 3.
中位数：首先排序：2, 2, 3, 3, 4, 5, 6。中间值（第4个）是3，因此中位数 = 3。
Mode: The number that appears most often is 2 and 3 (both appear twice). This data set is bimodal with modes 2 and 3.
众数：出现次数最多的数字是2和3（各出现两次）。该数据集为双峰，众数为2和3。
Effect of an outlier: Adding 20 makes the new set: 2,2,3,3,4,5,6,20. The mean becomes (25+20)/8 = 45/8 = 5.625, which is pulled upwards away from most values. The median is now the average of the 4th and 5th values (3 and 4) = 3.5, which remains closer to the typical number of books. The median is more resistant to extreme values.
异常值的影响：加入20后新数据集为：2,2,3,3,4,5,6,20。平均数变为 (25+20)/8 = 45/8 = 5.625，被拉高偏离了大多数数值。中位数现在是第4和第5个值（3和4）的平均数 = 3.5，仍然更接近典型阅读量。中位数对极端值的抵抗力更强。
4. Range and Quartiles | 全距与四分位数
Question 4: The heights (in cm) of 11 plants are: 12, 15, 14, 11, 18, 16, 13, 14, 17, 14, 19. (a) Find the range. (b) Determine the lower quartile (Q1), median (Q2) and upper quartile (Q3). (c) Calculate the interquartile range (IQR).
问题4：11株植物的高度（单位 cm）为：12, 15, 14, 11, 18, 16, 13, 14, 17, 14, 19。（a）求全距。（b）确定下四分位数（Q1）、中位数（Q2）和上四分位数（Q3）。（c）计算四分位距（IQR）。
Range: Maximum = 19, minimum = 11. Range = 19 – 11 = 8 cm.
全距：最大值 = 19，最小值 = 11。全距 = 19 – 11 = 8 cm。
Order the data: 11, 12, 13, 14, 14, 14, 15, 16, 17, 18, 19. n=11. Median (Q2) is the 6th value: 14.
数据排序：11, 12, 13, 14, 14, 14, 15, 16, 17, 18, 19。n=11。中位数（Q2）为第6个值：14。
Lower half (before median): 11, 12, 13, 14, 14. Q1 is the median of this half, the 3rd value: 13.
下（前）半部分：11, 12, 13, 14, 14。Q1是这一半的中位数，第3个值：13。
Upper half (after median): 15, 16, 17, 18, 19. Q3 is the median of this half, the 3rd value: 17.
上（后）半部分：15, 16, 17, 18, 19。Q3是这一半的中位数，第3个值：17。
IQR = Q3 – Q1 = 17 – 13 = 4 cm.
IQR = Q3 – Q1 = 17 – 13 = 4 cm。
5. Box-and-Whisker Plots | 箱线图
Question 5: Using the five-number summary from Question 4 (min=11, Q1=13, median=14, Q3=17, max=19), construct a box plot. Then describe what the box plot tells you about the distribution of plant heights.
问题5：利用第4题中的五数概括（最小=11，Q1=13，中位=14，Q3=17，最大=19），绘制一个箱线图。然后描述该箱线图揭示了植物高度分布的哪些信息。
Drawing the box plot: Draw a horizontal scale from 10 to 20. Mark the five key points. Draw a box from Q1 (13) to Q3 (17). Inside the box, draw a vertical line at the median (14). Extend whiskers from the box ends to the minimum (11) and maximum (19). The box plot is complete.
绘制箱线图：绘制从10到20的水平刻度。标出五个关键点。从Q1（13）到Q3（17）画一个矩形箱。在箱内中位数（14）处画一条竖线。从箱两端向最小值（11）和最大值（19）延伸须线。箱线图完成。
Interpretation: The median is closer to Q1 than to Q3, and the left whisker is slightly longer than the right whisker, suggesting a mild positive skew. However, with a small data set, the distribution appears fairly symmetrical overall. The IQR of 4 cm shows that the middle 50% of heights are concentrated within a narrow range.
解读：中位数离Q1比离Q3更近，左须稍长于右须，表明有轻微的正偏态。但鉴于数据量少，整体分布看起来相当对称。四分位距为4 cm，显示中间50%的高度集中在一个较窄的范围内。
6. Basic Probability | 基础概率
Question 6: A bag contains 5 red marbles, 3 blue marbles and 2 green marbles. One marble is selected at random. (a) What is the probability that the marble is blue? (b) What is the probability that it is not red? (c) Two marbles are selected one after the other without replacement. What is the probability that both are green? (d) Explain why the events ‘selecting a red marble’ and ‘selecting a blue marble’ in a single draw are mutually exclusive.
问题6：一个袋子里装有5颗红色弹珠、3颗蓝色弹珠和2颗绿色弹珠。随机选取一颗弹珠。（a）抽到蓝色弹珠的概率是多少？（b）抽到的不是红色的概率是多少？（c）每次抽取一颗且不放回，连续抽取两颗。两颗都是绿色的概率是多少？（d）解释为什么在单次抽取中，“抽到红色弹珠”和“抽到蓝色弹珠”这两个事件是互斥的。
Total marbles = 5 + 3 + 2 = 10. (a) P(blue) = number of blue / total = 3/10.
弹珠总数 = 5 + 3 + 2 = 10。（a）P(蓝色) = 蓝色数量 / 总数 = 3/10。
(b) P(not red) = 1 – P(red) = 1 – 5/10 = 1 – 1/2 = 1/2. Alternatively, favourable outcomes are blue + green = 3+2=5, so probability = 5/10 = 1/2.
（b）P(非红色) = 1 – P(红色) = 1 – 5/10 = 1 – 1/2 = 1/2。也可从有利结果看，蓝色+绿色 = 3+2=5，概率 = 5/10 = 1/2。
(c) Without replacement: P(1st green) = 2/10 = 1/5. After one green is taken, 1 green remains out of 9 marbles. P(2nd green | 1st green) = 1/9. So P(both green) = (2/10) × (1/9) = 2/90 = 1/45.
（c）不放回：P(第一颗绿色) = 2/10 = 1/5。取出一颗绿色后，剩下9颗中还有1颗绿色。P(第二颗绿色 | 第一颗绿色) = 1/9。因此 P(两颗都绿) = (2/10) × (1/9) = 2/90 = 1/45。
(d) Mutually exclusive events: In a single draw, you cannot pick both a red and a blue marble at the same time. The two outcomes cannot happen together, so they are mutually exclusive.
（d）互斥事件：在单次抽取中，你不可能同时抽到一颗红色和一颗蓝色弹珠。这两个结果不可能同时发生，因此它们是互斥的。
7. Scatter Graphs and Correlation | 散点图与相关性
Question 7: The table shows the hours spent revising and the marks obtained in a test for 6 students: (2, 45), (3, 52), (5, 60), (1, 30), (4, 58), (6, 72). (a) Plot a scatter graph of marks versus revision hours. (b) Describe the type and strength of correlation shown. (c) Draw a line of best fit by eye. Estimate the mark for a student who revised for 4.5 hours. (d) Explain why it may not be appropriate to use this line to predict marks for 10 hours of revision.
问题7：表格显示了6名学生的复习时间和考试分数：(2, 45), (3, 52), (5, 60), (1, 30), (4, 58), (6, 72)。（a）绘制分数相对于复习时间的散点图。（b）描述其中显示的相关类型和强度。（c）凭目测画一条最佳拟合线。估算一名复习了4.5小时学生的分数。（d）解释为什么用这条线来预测复习10小时后的分数可能不合适。
Plotting: On graph paper, let horizontal axis = hours (0 to 7) and vertical axis = marks (20 to 80). Plot points (2,45), (3,52), (5,60), (1,30), (4,58), (6,72). The pattern shows an upward slope.
绘图：在坐标纸上，令横轴为时间（0到7），纵轴为分数（20到80）。描点：(2,45), (3,52), (5,60), (1,30), (4,58), (6,72)。图形呈现上升趋势。
Correlation: As revision hours increase, marks tend to increase. The points lie fairly close to a straight line, indicating a strong positive correlation.
相关性：随着复习时间增加，分数也倾向于增加。散点比较靠近直线，说明存在强正相关。
Line of best fit: Draw a straight line that passes near most points, balancing points above and below the line. Using the line, at 4.5 hours the estimated mark is approximately 64 (this may vary slightly depending on the line drawn).
最佳拟合线：画一条经过大多数点附近的直线，使线上方和下方的点大致平衡。使用该直线，当 x=4.5 小时，估算分数约为64分（画线不同略有差异）。
Extrapolation caution: 10 hours is far beyond the range of the given data (1–6 hours). The relationship may not remain linear beyond this range; other factors could influence marks. Predictions outside the data range are unreliable.
外推注意事项：10小时远超出给定数据的范围（1–6小时）。在这个范围之外，线性关系可能不再保持；其他因素可能影响分数。超出数据范围的预测是不可靠的。
8. Mixed Problem: Averages, Probability and Chart Interpretation | 综合题：平均数、概率与图表解读
Question 8: The pie chart represents the favourite school subjects of 180 Year 9 students. The angles are: Maths 120°, Science 90°, English 60°, History 50°, Geography 40°. (a) How many students chose Science as their favourite subject? (b) Calculate the mean number of choices if each student picks one subject. (c) If one student is selected at random, what is the probability that their favourite subject is either Maths or Science? (d) Explain why drawing a bar chart might be better than a pie chart for comparing the exact numbers of students choosing History and Geography.
问题8：饼图展示了180名九年级学生最喜爱的学校科目。各扇区角度为：数学120°，科学90°，英语60°，历史50°，地理40°。（a）有多少名学生选择科学作为最爱科目？（b）如果每名学生选择一科，计算选择数的平均数。（c）如果随机选出一名学生，其最爱科目是数学或科学的概率是多少？（d）解释为何在比较选择历史与地理的确切学生人数时，条形图可能优于饼图。
(a) Science: angle 90° out of 360° total. Fraction = 90/360 = 1/4. Number of students = (1/4) × 180 = 45.
（a）科学：扇区角度90°，总角度360°。比例 = 90/360 = 1/4。学生数 = (1/4) × 180 = 45。
(b) ‘Mean number of choices’ is a bit tricky: Since each student makes one choice, the total number of choices is 180. The mean number of choices per category is total/number of categories = 180/5 = 36. Alternatively, from the frequencies (Maths 60, Science 45, English 30, History 25, Geography 20), the sum is 180 and mean frequency per subject = 180/5 = 36.
（b）平均选择数：因每名学生只选一科，选择总数为180。每科平均选择数 = 总人数 / 科目数 = 180/5 = 36。也可以从频数（数学60，科学45，英语30，历史25，地理20）算出总和180，每科平均频数 = 180/5 = 36。
(c) P(Maths or Science) = P(Maths) + P(Science) = (60/180) + (45/180) = 105/180 = 7/12 (simplified). Alternatively using angles: (120°+90°)/360° = 210/360 = 7/12.
（c）P(数学或科学) = P(数学) + P(科学) = (60/180) + (45/180) = 105/180 = 7/12。或利用角度：(120°+90°)/360° = 210/360 = 7/12。
(d) Bar chart vs pie chart: In a pie chart it can be difficult to compare the sizes of sectors that are close in angle, like History (50°) and Geography (40°). A bar chart shows the exact frequencies as heights, making direct comparison of these two categories much clearer.
（d）条形图与饼图对比：在饼图中，当扇区角度比较接近时（如历史的50°和地理的40°），难以准确比较大小。条形图以高度明确显示频数，使这两个类别的直接对比清晰得多。
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