📚 Year 10 CAIE Statistics Unit Test Paper Walkthrough | Year 10 CAIE 统计：单元测试模拟卷解析
This walkthrough will guide you through a typical Year 10 CAIE Statistics unit test, covering measures of central tendency, data representation, probability, and sampling. Each question is solved step by step, with explanations in both English and Chinese to reinforce key concepts and exam skills.
本文将逐题解析一份典型的 Year 10 CAIE 统计单元测试卷，涵盖集中趋势度量、数据表示、概率和抽样等内容。每题都配有详细步骤与中英双语解释，帮助你巩固关键概念并提升应试技巧。
1. Question 1: Mean, Mode, and Median | 第1题：平均数、众数与中位数
The first question gives the dataset 5, 7, 8, 6, 9, 7, 5, 8, 6, 10 and asks for the mean, mode, and median. To compute the mean, sum all values: 5 + 7 + 8 + 6 + 9 + 7 + 5 + 8 + 6 + 10 = 71, then divide by the count of numbers, which is 10, giving a mean of 7.1. For the mode, we look for the most frequent value. Here 5, 6, 7, and 8 each appear exactly twice, so the dataset is multimodal with modes 5, 6, 7, and 8. The median is found by ordering the data from smallest to largest: 5, 5, 6, 6, 7, 7, 8, 8, 9, 10. With an even number of data points (10), the median is the average of the two middle values (5th and 6th): (7 + 7) / 2 = 7.
第一题给出了数据集 5, 7, 8, 6, 9, 7, 5, 8, 6, 10，要求计算平均数、众数和中位数。计算平均数时，先求总和：5 + 7 + 8 + 6 + 9 + 7 + 5 + 8 + 6 + 10 = 71，再除以数据个数 10，得出平均数为 7.1。众数是出现次数最多的数值。在这里，5、6、7、8 各出现了两次，因此该数据集是多众数的，众数为 5、6、7 和 8。求中位数时，先将数据从小到大排序：5, 5, 6, 6, 7, 7, 8, 8, 9, 10。由于数据个数为偶数（10 个），中位数取中间两个数（第 5 和第 6 个）的平均值：(7 + 7) / 2 = 7。
2. Question 2: Estimated Mean from Grouped Data | 第2题：从分组数据估算平均数
A frequency table gives the time in minutes for students to solve a puzzle. The class intervals and frequencies are: 0 ≤ t < 10 (4 students), 10 ≤ t < 20 (7), 20 ≤ t < 30 (12), 30 ≤ t < 40 (15), 40 ≤ t < 50 (8), 50 ≤ t < 60 (4). To estimate the mean, we use the midpoint of each interval. The midpoints are 5, 15, 25, 35, 45, and 55. Multiply each midpoint by its frequency: 5×4 = 20, 15×7 = 105, 25×12 = 300, 35×15 = 525, 45×8 = 360, 55×4 = 220. The sum of these products is 20 + 105 + 300 + 525 + 360 + 220 = 1530. The total frequency is 4 + 7 + 12 + 15 + 8 + 4 = 50. Therefore, the estimated mean = 1530 ÷ 50 = 30.6 minutes.
一个频数表列出了学生解谜所需的时间（分钟）。组距和频数分别为：0 ≤ t < 10（4 人），10 ≤ t < 20（7 人），20 ≤ t < 30（12 人），30 ≤ t < 40（15 人），40 ≤ t < 50（8 人），50 ≤ t < 60（4 人）。估算平均数时，使用各组中值。中值分别为 5, 15, 25, 35, 45, 55。将各中值乘以对应频数：5×4 = 20, 15×7 = 105, 25×12 = 300, 35×15 = 525, 45×8 = 360, 55×4 = 220。这些乘积之和为 20 + 105 + 300 + 525 + 360 + 220 = 1530。总频数为 50。因此，估计平均时间为 1530 ÷ 50 = 30.6 分钟。
3. Question 3: Cumulative Frequency Graph and Quartiles | 第3题：累积频数图与四分位数
A cumulative frequency table is first constructed from the grouped data. Using the same class intervals and frequencies as in Question 2, the cumulative frequencies are: 4 for t < 10, 4+7 = 11 for t < 20, 11+12 = 23 for t < 30, 23+15 = 38 for t < 40, 38+8 = 46 for t < 50, and 46+4 = 50 for t < 60. Plot the upper boundary of each interval against its cumulative frequency, then join points with a smooth curve. To estimate the median, find the value corresponding to the 25th student (50/2). From the graph, the median is approximately 31 minutes. The lower quartile Q1 is at the 12.5th student, giving about 22 minutes; the upper quartile Q3 at the 37.5th student gives about 41 minutes. The interquartile range IQR = Q3 – Q1 ≈ 41 – 22 = 19 minutes.
首先从分组数据构建累积频数表。仍然使用第2题的组距和频数，累积频数为：t < 10 为 4，t < 20 为 11，t < 30 为 23，t < 40 为 38，t < 50 为 46，t < 60 为 50。以每个区间的上界对累积频数描点，然后用平滑曲线连接。估计中位数时，找到第 25 个学生（50/2）对应的值，从图中读出中位数约为 31 分钟。下四分位数 Q₁ 对应第 12.5 个学生，约为 22 分钟；上四分位数 Q₃ 对应第 37.5 个学生，约为 41 分钟。四分位距 IQR = Q₃ – Q₁ ≈ 41 – 22 = 19 分钟。
4. Question 4: Basic Probability from a Bag of Coloured Balls | 第4题：袋中取球的基本概率
A bag contains 3 red balls, 2 blue balls, and 5 green balls, making a total of 10 balls. (a) The probability of selecting a red ball = number of red balls / total number of balls = 3/10. (b) The event “blue or green” means selecting a ball that is either blue or green. The favourable outcomes are 2 + 5 = 7, so the probability = 7/10. (c) “Not red” is the complement of “red”, so its probability = 1 – P(red) = 1 – 3/10 = 7/10. Notice that (b) and (c) give the same result, which makes sense because “blue or green” is exactly the complement of “red” in this scenario.
一个袋子里有 3 个红球、2 个蓝球和 5 个绿球，总共有 10 个球。(a) 抽到红球的概率 = 红球数量 / 总球数 = 3/10。(b) “蓝球或绿球”意味着抽到的球是蓝色或绿色，有利结果数为 2 + 5 = 7，所以概率 = 7/10。(c) “不是红球”是“红球”的互补事件，其概率 = 1 – P(红球) = 1 – 3/10 = 7/10。可以看到 (b) 和 (c) 的结果相同，这是因为在这个情境中“蓝或绿”恰好是“红”的补集。
5. Question 5: Probability of a Sum of 8 with Two Dice | 第5题：两个骰子点数之和为 8 的概率
When two fair dice are rolled, there are 6 × 6 = 36 equally likely outcomes. The combinations that give a sum of 8 are (2,6), (3,5), (4,4), (5,3), and (6,2). Count these carefully: there are 5 favourable outcomes. Therefore, P(sum = 8) = 5/36. It is common to represent this using a sample space diagram; each cell in a 6×6 grid represents a pair, and shading the cells where row + column = 8 helps confirm the count.
掷两个公平的骰子，共有 6 × 6 = 36 个等可能结果。点数之和为 8 的组合有：(2,6), (3,5), (4,4), (5,3), (6,2)，一共 5 个有利结果。因此，P(和为 8) = 5/36。通常可以用样本空间图表示，在 6×6 的表格中，每一个格子代表一个数对，将行数加列数等于 8 的格子涂上阴影，很容易验证数量为 5。
6. Question 6: Drawing a Histogram from Frequency Density | 第6题：根据频数密度绘制直方图
The heights of 50 plants are recorded in the following class intervals: 0–10 cm, 10–20 cm, 20–30 cm, 30–40 cm, 40–60 cm. The frequencies are 6, 14, 18, 8, 4 respectively. Note that the last interval has a width of 20 cm. Frequency density = frequency ÷ class width. For 0–10: 6/10 = 0.6; 10–20: 14/10 = 1.4; 20–30: 18/10 = 1.8; 30–40: 8/10 = 0.8; 40–60: 4/20 = 0.2. The histogram is drawn with height equal to the frequency density on the vertical axis and class boundaries on the horizontal axis. The area of each bar represents the frequency. This corrects for unequal class widths, ensuring the graph is not misleading.
记录了 50 株植物的高度，分组如下：0–10 cm, 10–20 cm, 20–30 cm, 30–40 cm, 40–60 cm，对应的频数分别为 6, 14, 18, 8, 4。注意最后一组的组距是 20 cm。频数密度 = 频数 ÷ 组距。计算得：0–10: 6/10 = 0.6；10–20: 14/10 = 1.4；20–30: 18/10 = 1.8；30–40: 8/10 = 0.8；40–60: 4/20 = 0.2。绘制直方图时，纵轴高度为频数密度，横轴为组界。每个长方形的面积代表该组的频数。当组距不相等时，这种方法可以避免图形产生误导。
7. Question 7: Scatter Graphs, Correlation, and Line of Best Fit | 第7题：散点图、相关性与最佳拟合线
A scatter graph plots revision hours (x) against test scores (y). The points show a general upward trend, indicating positive correlation. A line of best fit is drawn through the centre of the points, balancing the number above and below the line. To estimate the test score for 5 hours of revision, locate x = 5, draw a vertical line to the best-fit line, then read horizontally to the y-axis. The estimated score might be around 68. It is important to note that this predicted value is only valid within the range of the original data (interpolation); extending the line beyond the data (extrapolation) would be unreliable. The correlation coefficient r is not asked, but you can state that the relationship appears moderately strong.
散点图以复习时数（x 轴）对测验成绩（y 轴）描点。这些点整体呈现出向上的趋势，表明存在正相关。最佳拟合线画在点的中心位置，使得在线上方和下方的点数大致平衡。若要估计复习 5 小时对应的成绩，先找到 x = 5，向上作垂线与拟合线相交，再水平读出 y 轴的值，估计成绩约为 68。需要注意的是，此预测值仅在原始数据范围内有效（内插）；如果将线延长到数据范围之外（外推），结果就不可靠。虽然没有要求计算相关系数 r，但可以判断这种相关关系强度为中等偏强。
8. Question 8: Random Sampling and Stratified Sampling | 第8题：随机抽样与分层抽样
A random sample is one where every member of the population has an equal chance of being selected. This can be achieved using a random number generator or drawing names from a hat. It helps to avoid bias. A stratified sample is obtained by dividing the population into distinct subgroups (strata) that share a common characteristic (e.g., year groups, gender), then taking a random sample from each stratum in proportion to its size. For example, if 30% of a school are Year 10 students, then 30% of the stratified sample should be Year 10. Stratified sampling ensures that all relevant groups are represented, which improves the accuracy of estimates for the whole population.
随机抽样是指总体中每个成员被抽中的概率都相等的抽样方法，可以通过随机数生成器或抽签来实现，有助于避免偏差。分层抽样则是先将总体按某一共同特征（如年级、性别）分成不同的子群（层），然后从每一层中按该层所占比例随机抽取样本。例如，如果某所学校中 Year 10 学生占 30%，那么在分层样本中也应有 30% 的 Year 10 学生。分层抽样能确保所有相关群体都被代表，从而提高对整体估计的准确性。
9. Question 9: Transforming the Mean and Standard Deviation | 第9题：平均数与标准差的变换
A dataset has mean μ = 20 and standard deviation σ = 3. If every value x is transformed to y = 2x + 5, the new mean becomes 2 × 20 + 5 = 45. Multiplying by 2 scales both the mean and the standard deviation by 2, while adding 5 shifts the mean but does not affect the spread. Therefore, the new standard deviation is 2 × 3 = 6. In general, if y = a x + b, then mean(y) = a·mean(x) + b, and standard deviation(y) = |a|·standard deviation(x). This property is frequently tested in CAIE Statistics and helps when standardising data or comparing distributions.
某数据集具有平均数 μ = 20，标准差 σ = 3。若将每个数值 x 变换为 y = 2x + 5，则新平均数为 2 × 20 + 5 = 45。乘以 2 使得平均数和标准差都扩大 2 倍，而加上常数 5 仅平移平均数，对离散程度无影响，因此新标准差为 2 × 3 = 6。一般地，若 y = a x + b，则 y 的平均数 = a·x 的平均数 + b，y 的标准差 = |a|·x 的标准差。这一性质在 CAIE 统计考试中经常出现，有助于数据标准化和分布比较。
10. Question 10: Box Plots and Comparing Distributions | 第10题：箱线图与分布比较
Two sets of exam results give five-number summaries. For Class A: minimum = 38, Q₁ = 55, median = 68, Q₃ = 82, maximum = 97. For Class B: minimum = 45, Q₁ = 60, median = 72, Q₃ = 80, maximum = 95. Box plots are drawn using a scale. The box spans from Q₁ to Q₃, with a line at the median. Whiskers extend to the minimum and maximum. Comparing the two, Class A has a lower minimum and a slightly higher maximum, showing a wider range (97 – 38 = 59) compared with Class B (95 – 45 = 50). However, Class B’s median is higher, suggesting a better central performance. The interquartile range for Class A is 82 – 55 = 27, while for Class B it is 20, indicating that the middle 50% of Class B’s scores are less spread out and more consistent.
两组考试成绩的五数概括如下：A 班：最小值 38，Q₁ = 55，中位数 68，Q₃ = 82，最大值 97；B 班：最小值 45，Q₁ = 60，中位数 72，Q₃ = 80，最大值 95。根据这些数据在同一比例尺上绘制箱线图，箱体从 Q₁ 到 Q₃，中间用横线标出中位数，触须延伸到最小值和最大值。比较两个班级，A 班的最低分更低、最高分略高，全距为 97 – 38 = 59，大于 B 班的 50，说明 A 班分数更分散。但 B 班的中位数更高，表明中心表现更优。A 班的四分位距为 82 – 55 = 27，B 班为 20，说明 B 班中间 50% 的成绩更为集中、一致性更好。
Published by TutorHao | Statistics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply