📚 Year 9 CAIE Statistics: Unit Test Mock Exam Analysis | Year 9 CAIE 统计：单元测试模拟卷解析
This article provides a full walkthrough of a Year 9 CAIE Statistics unit test mock exam. Each section breaks down a typical exam question, explains the key concepts, and highlights common pitfalls. Use this analysis to sharpen your understanding before the real test.
本文为一份 Year 9 CAIE 统计单元测试模拟卷提供完整解析。每个部分拆解一道典型考题，讲解核心概念并指出常见错误。请利用这份解析在真正考试前打磨你的理解。
1. Classifying Data Types | 数据分类
A question in the mock asks: "Classify each of the following as qualitative or quantitative. If quantitative, state whether it is discrete or continuous: (a) the number of pets owned, (b) favourite pizza topping, (c) time taken to run 100 metres, (d) shoe size."
模拟卷中有这样一道题：“将下列各项分为定性或定量数据。如果是定量数据，指出是离散还是连续：(a) 拥有的宠物数量，(b) 最喜欢的披萨配料，(c) 跑 100 米所用时间，(d) 鞋码。”
The number of pets is a count; it is quantitative discrete. Favourite pizza topping is a category without numerical meaning, so it is qualitative. The time taken to run 100 m can take any value on a continuous scale, so it is quantitative continuous. Shoe size, although often given as whole or half numbers, can only take specific values and is therefore quantitative discrete.
宠物数量是一个计数；它是定量离散数据。最喜欢的披萨配料是一个没有数值意义的类别，所以是定性数据。跑 100 米所用时间可以在连续尺度上取任意值，因此是定量连续数据。鞋码虽然常以整数或半码表示，但只能取特定值，因此是定量离散数据。
2. Organising Data in Frequency Tables | 用频数表整理数据
The exam presents this data set of the number of books read by 20 students in a month: 2, 3, 1, 2, 4, 3, 2, 5, 1, 2, 3, 4, 2, 3, 1, 2, 4, 5, 3, 2. The task is to complete a frequency table and draw a bar chart.
考试给出 20 名学生一个月内阅读书籍数量的数据集：2, 3, 1, 2, 4, 3, 2, 5, 1, 2, 3, 4, 2, 3, 1, 2, 4, 5, 3, 2。要求完成频数表并绘制条形图。
| Number of books | Tally | Frequency |
| 1 | ||| | 3 |
| 2 | |||| || | 7 |
| 3 | |||| | 5 |
| 4 | ||| | 3 |
| 5 | || | 2 |
Count the occurrences of each value to fill the frequency column. The bar chart should have the number of books on the horizontal axis, labelled from 1 to 5, and the frequency on the vertical axis. Each bar must be of equal width, and the height corresponds to the frequency. Remember to label both axes and give the chart a title, such as 'Books Read in a Month'.
统计每个数值出现的次数以填写频数列。条形图的横轴应为书籍数量，标注 1 至 5，纵轴为频数。每个条形宽度应相等，高度对应频数。记得标注两轴，并给图表加上标题，如“一个月内阅读的书籍数量”。
3. Calculating Mean, Median, and Mode | 计算平均数、中位数和众数
Using the same data set, the next question asks: "Find the mean, median, and mode of the number of books read."
使用相同的数据集，下一道题要求：“求出所读书籍数量的平均数、中位数和众数。”
le value. With 20 data points, the median is the average of the 10th and 11th values. The 10th value is 2 and the 11th is 3, so median = (2+3)/2 = 2.5. To find the mean, sum all values: (3×1) + (7×2) + (5×3) + (3×4) + (2×5) = 3+14+15+12+10 = 54. Mean = 54 ÷ 20 = 2.7.
首先，把所有值按顺序排列：1,1,1,2,2,2,2,2,2,2,3,3,3,3,3,4,4,4,5,5。众数是出现最频繁的值，即 2。中位数是中间值。有 20 个数据点，中位数是第 10 和第 11 个值的平均数。第 10 个值是 2，第 11 个是 3，因此中位数 = (2+3)/2 = 2.5。计算平均数，求所有值之和：(3×1) + (7×2) + (5×3) + (3×4) + (2×5) = 3+14+15+12+10 = 54。平均数 = 54 ÷ 20 = 2.7。
Mean = Σx ÷ n = 54 ÷ 20 = 2.7
4. Calculating Range and Interquartile Range | 计算极差和四分位距
A question on spread states: "For the books data, calculate the range and the interquartile range (IQR)."
一道关于离散程度的题目说：“根据书籍数据，计算极差和四分位距 (IQR)。”
Range = highest value – lowest value = 5 – The median of this half is the average of the 5th and 6th values, both 2, so Q₁ = 2. Q₃ is the median of the upper half (last 10 values: 3,3,3,3,3,4,4,4,5,5). The 5th and 6th values are 3 and 4, so Q₃ = (3+4)/2 = 3.5. IQR = Q₃ – Q₁ = 3.5 – 2 = 1.5.
极差 = 最大值 – 最小值 = 5 – 1 = 4。对于四分位距，找出下四分位数 (Q₁) 和上四分位数 (Q₃)。Q₁ 是下半部分数据（前 10 个值：1,1,1,2,2,2,2,2,2,2）的中位数。该半部分的中位数是第 5 和 第 6 个值的平均，两者均为 2，因此 Q₁ = 2。Q₃ 是上半部分数据（后 10 个值：3,3,3,3,3,4,4,4,5,5）的中位数。第 5 和第 6 个值是 3 和 4，因此 Q₃ = (3+4)/2 = 3.5。IQR = Q₃ – Q₁ = 3.5 – 2 = 1.5。
IQR = Q₃ – Q₁ = 3.5 – 2 = 1.5
5. Probability from Simple Events | 简单事件的概率
The mock exam includes a probability bag question: "A bag contains 5 red, 3 blue, and 2 green counters. One counter is taken at random. Find the probability that it is (a) blue, (b) not red, (c) yellow."
模拟卷包含一道抽签概率题：“一个袋子里有 5 个红色、3 个蓝色和 2 个绿色筹码。随机取出一个。求它是 (a) 蓝色，(b) 不是红色，(c) 黄色的概率。”
Total counters = 5 + 3 + 2 = 10. (a) P(blue) = number of blue / total = 3/10. (b) 'Not red' iven as fractions, decimals, or percentages, but fractions in simplest form are preferred.
总筹码数 = 5 + 3 + 2 = 10。(a) P(蓝色) = 蓝色数量 / 总数 = 3/10。(b) “不是红色”包括蓝色和绿色，共 3+2 = 5 个筹码。P(不是红色) = 5/10 = 1/2。(c) 没有黄色筹码，因此 P(黄色) = 0。概率可以用分数、小数或百分数表示，但推荐使用最简分数。
6. Interpreting Pie Charts | 解读饼图
A pie chart shows how 36 students travel to school: Car 120°, Bus 90°, Walk 100°, Cycle 50°. The question asks: "How many students walk to school? What fraction goes by bus?"
一个饼图显示 36 名学生的上学交通方式：私家车 120°，公交车 90°，步行 100°，自行车 50°。题目问：“有多少名学生步行上学？乘公交车的比例是多少？”
nt = 10 students. The bus angle is 90°, fraction = 90/360 = 1/4. Alternatively, number of bus students = 90° ÷ 10 = 9, fraction = 9/36 = 1/4. Always check your angle sums to 360°.
饼图的总角度为 360°。每名学生对应 360° ÷ 36 = 10°。步行的角度是 100°，因此步行人数 = 100° ÷ 10°/人 = 10 名学生。公交车的角度是 90°，比例 = 90/360 = 1/4。或者，坐公交车的学生数 = 90° ÷ 10 = 9，比例 = 9/36 = 1/4。务必检查角度总和是否为 360°。
7. Reading Line Graphs | 读取折线图
You are shown a line graph of temperature recorded every 2 hours from 08:00 to 20:00. The question asks: "At what time was the highest temperature? What was the temperature increase between 10:00 and 14:00?"
试卷给出一个从 08:00 到 20:00 每两小时记录一次温度的折线图。题目问：“最高温度出现在什么时间？在 10:00 至 14:00 之间温度上升了多少？”
urs at 16:00 with a temperature of 24°C. At 10:00 the temperature was 16°C, and at 14:00 it was 22°C. The increase is 22 – xes carefully; use a ruler to align points with the scales. Check whether the vertical axis starts at zero or a broken scale, as this can affect visual impression.
从图中看，最高点出现在 16:00，温度为 24°C。在 10:00，温度为 16°C，14:00 时为 22°C。上升值为 22 – 16 = 6°C。始终仔细从坐标轴读取数值；用直尺将数据点与刻度对齐。检查纵轴是从零开始还是截断刻度，这会影响视觉印象。
8. Estimating the Mean from Grouped Data | 从分组数据估算平均数
The next question gives a grouped frequency table of test marks: 0-10 (frequency 4), 10-20 (7), 20-30 (9), 30-40 (5). "Calculate an estimate of the mean mark."
下一题给出测验分数的分组频数表：0-10（频数 4），10-20（7），20-30（9），30-40（5）。“计算平均分数的估算值。”
| Marks | Midpoint (x) | Frequency (f) | f × x |
| 0-10 | 5 | 4 | 20 |
| 10-20 | 15 | 7 | 105 |
| 20-30 | 25 | 9 | 225 |
| 30-40 | 35 | 5 | 175 |
| Total | 25 | 525 |
75 = 525. Total frequency is 25. Estimated mean = Σfx ÷ Σf = 525 ÷ 25 = 21. The estimate assumes data points are evenly spread within each interval, so it is an approximation.
找出每组的组中值：(0+10)/2 = 5，(10+20)/2 = 15，等等。将每个组中值乘以其频数，并求和：Σfx = 20 + 105 + 225 + 175 = 525。总频数为 25。估算平均数 = Σfx ÷ Σf = 525 ÷ 25 = 21。此估算假设数据点在每个区间内均匀分布，因此是一个近似值。
Estimated mean = Σfx ÷ Σf = 525 ÷ 25 = 21
9. Scatter Graphs and Correlation | 散点图与相关性
The mock exam plots ice cream sales against temperature (in °C). Data points: (15, 120), (18, 140), (20, 155), (22, 170), (25, 190), (28, 210). "Draw a scatter graph. Describe the correlation. Draw a line of best fit and estimate sales at 24°C."
模拟卷绘制了冰淇淋销量（个）与温度（°C）的散点图。数据点：(15, 120), (18, 140), (20, 155), (22, 170), (25, 190), (28, 210)。“绘制散点图。描述相关性。画出最佳拟合线，并估算 24°C 时的销量。”
Plot points on axes with temperature on the x-axis and sales on the y-axis. The points show a clear upward trend, so there is positive correlation. Draw a straight line of best fit passing through as many points as possible with roughly equal numbers of points above and below. Extend the line to 24°C and read the corresponding sales value (around 182). Avoid forcing the line through the origin unless data suggests it.
在坐标轴上标出各点，x 轴为温度，y 轴为销量。这些点呈现明显的上升趋势，因此存在正相关。画一条最佳拟合直线，尽可能穿过更多的点，并使线上方和下方的点数大致相等。将线延长至 24°C，读取对应的销量值（约 182）。除非数据表明，不要强制让直线通过原点。
10. Combined Events and Tree Diagrams | 组合事件与树状图
A final higher-challenge question: "The probability of rain on a Saturday is 0.3. If it rains, the probability that a match is cancelled is 0.8. If it does not rain, the cancellation probability is 0.1. Draw a tree diagram and find the probability that the match is cancelled."
最后一道高难度题：“周六下雨的概率是 0.3。如果下雨，比赛取消的概率是 0.8。如果不下雨，取消的概率是 0.1。画出树状图，并求比赛取消的概率。”
nd Not Cancelled (0.9). The match is cancelled in two ways: Rain & Cancelled (0.3 × 0.8 = 0.24), Not Rain & Cancelled (0.7 × 0.1 = 0.07). Total probability = 0.24 + 0.07 = 0.31. Always multiply along branches and add relevant joint probabilities.
树状图：第一层分支为下雨 (0.3) 和不下雨 (0.7)。从下雨出发，有两个分支：取消 (0.8) 和未取消 (0.2)。从不下雨出发，取消 (0.1) 和未取消 (0.9)。比赛取消有两种途径：下雨且取消 (0.3 × 0.8 = 0.24)，不下雨且取消 (0.7 × 0.1 = 0.07)。总概率 = 0.24 + 0.07 = 0.31。始终沿分支相乘，并将相关的联合概率相加。
P(Cancelled) = (0.3 × 0.8) + (0.7 × 0.1) = 0.24 + 0.07 = 0.31
Published by TutorHao | Statistics Revision Series | aleveler.com
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