Year 11 CAIE Statistics: Unit Test Mock Paper Walkthrough | CAIE 统计 单元测试模拟卷解析

📚 Year 11 CAIE Statistics: Unit Test Mock Paper Walkthrough | CAIE 统计 单元测试模拟卷解析

This walkthrough breaks down a full Unit Test mock paper designed for Year 11 CAIE Statistics. It covers descriptive statistics, probability, binomial and normal distributions, correlation, regression, time series and sampling. Each question is followed by step‑by‑step solutions in both English and Chinese to help you master the techniques and avoid common pitfalls.

本文详细解析一份专为 Year 11 CAIE 统计设计的单元测试模拟卷，内容涵盖描述统计、概率、二项分布与正态分布、相关与回归、时间序列以及抽样方法。每道题均配有中英双语的分步解答，助你掌握解题技巧并避开常见错误。

1. Test Overview and Structure | 模拟卷概览与结构

This mock paper mimics a one‑hour unit test with 8 questions totalling 60 marks. The topics are distributed as shown in the table below. Working through this paper systematically will strengthen your ability to interpret questions and apply the correct statistical methods under timed conditions.

本模拟卷仿照一小时的单元测试设计，共 8 题，满分 60 分。各主题分布如下表所示。系统地完成这套试卷，能提升你在限时条件下理解题意、选用正确统计方法的能力。

2. Question 1: Descriptive Statistics – Mean, Median, IQR and Box Plot | 问题 1：描述统计——均值、中位数、四分位距与箱线图

Question: The data set shows the scores of 9 students in a test: 12, 15, 18, 21, 22, 25, 30, 35, 40. Calculate the mean, median, mode, range, interquartile range (IQR) and describe how to draw a box‑and‑whisker plot. [8 marks]

题目： 某次测验中 9 名学生的成绩为：12, 15, 18, 21, 22, 25, 30, 35, 40。请计算均值、中位数、众数、极差、四分位距 (IQR)，并说明如何绘制箱线图。[8 分]

Step 1 – Order and count: The data are already in ascending order. There are n = 9 observations.

步骤 1——排序与计数： 数据已按升序排列，观测值个数 n = 9。

Step 2 – Median (Q₂): The middle value of 9 ordered items is the 5th value. Median = 22.

步骤 2——中位数 (Q₂)： 9 个有序数据中，中间的第 5 个值为 22。中位数 = 22。

Step 3 – Lower and upper quartiles: The lower half (first 4 values) is 12, 15, 18, 21. Q₁ = (15+18)/2 = 16.5. The upper half (last 4 values) is 25, 30, 35, 40. Q₃ = (30+35)/2 = 32.5.

步骤 3——下四分位数与上四分位数： 下半组（前 4 个数）为 12, 15, 18, 21，Q₁ = (15+18)/2 = 16.5。上半组（后 4 个数）为 25, 30, 35, 40，Q₃ = (30+35)/2 = 32.5。

Step 4 – Interquartile range: IQR = Q₃ − Q₁ = 32.5 − 16.5 = 16.

步骤 4——四分位距： IQR = Q₃ − Q₁ = 32.5 − 16.5 = 16。

Step 5 – Mean: Sum = 12+15+18+21+22+25+30+35+40 = 218. Mean = 218 ÷ 9 ≈ 24.22.

步骤 5——均值： 总和 = 12+15+18+21+22+25+30+35+40 = 218。均值 = 218 ÷ 9 ≈ 24.22。

Step 6 – Mode and range: There is no repeating value, so no mode. Range = 40 − 12 = 28.

步骤 6——众数和极差： 没有重复数值，因此无众数。极差 = 40 − 12 = 28。

Step 7 – Box plot description: A box is drawn from Q₁ to Q₃ (16.5 to 32.5) with a vertical line at the median (22). Whiskers extend to the minimum (12) and maximum (40), since no data point lies beyond 1.5×IQR from the quartiles.

步骤 7——箱线图描述： 从 Q₁ 到 Q₃（16.5 至 32.5）绘制一个矩形箱体，在中位数 22 处画一条竖线。须线延伸至最小值 12 和最大值 40，因为所有数据点与四分位数的距离均未超过 1.5×IQR。

3. Question 2: Probability – Tree Diagrams and Conditional Probability | 问题 2：概率——树状图与条件概率

Question: A bag contains 4 red (R) and 6 blue (B) balls. Two balls are drawn at random without replacement. (a) Draw a tree diagram to represent all outcomes. (b) Find the probability that both balls are the same colour. (c) Find the probability that at least one ball is blue. [7 marks]

题目： 一个袋子里有 4 个红球 (R) 和 6 个蓝球 (B)。随机抽取两个球，不放回。(a) 画出树状图表示所有可能结果。(b) 求两个球颜色相同的概率。(c) 求至少有一个蓝球的概率。[7 分]

Tree diagram description: First draw: P(R) = 4/10 = 2/5, P(B) = 6/10 = 3/5. If R is taken first, second draw: P(R) = 3/9, P(B) = 6/9. If B is taken first, second draw: P(R) = 4/9, P(B) = 5/9. The four paths are RR, RB, BR, BB.

树状图描述： 第一次抽取：P(R) = 4/10 = 2/5，P(B) = 6/10 = 3/5。若第一次为 R，第二次：P(R) = 3/9，P(B) = 6/9。若第一次为 B，第二次：P(R) = 4/9，P(B) = 5/9。四条路径分别为 RR、RB、BR、BB。

Part (b) same colour: P(RR) = (4/10) × (3/9) = 12/90 = 2/15. P(BB) = (6/10) × (5/9) = 30/90 = 5/15. P(same colour) = 2/15 + 5/15 = 7/15.

小题 (b) 同色概率： P(RR) = (4/10) × (3/9) = 12/90 = 2/15. P(BB) = (6/10) × (5/9) = 30/90 = 5/15. P(同色) = 2/15 + 5/15 = 7/15。

Part (c) at least one blue: The complement is ‘both red’. P(at least one B) = 1 − P(RR) = 1 − 2/15 = 13/15. Alternatively, add P(RB) + P(BR) + P(BB) = (4/10)×(6/9) + (6/10)×(4/9) + 5/15 = 24/90 + 24/90 + 30/90 = 78/90 = 13/15, confirming the result.

小题 (c) 至少一个蓝球： 对立事件为“两个都是红球”。P(至少一个 B) = 1 − P(RR) = 1 − 2/15 = 13/15。也可用 P(RB) + P(BR) + P(BB) = (4/10)×(6/9) + (6/10)×(4/9) + 5/15 = 24/90 + 24/90 + 30/90 = 78/90 = 13/15 进行验证。

4. Question 3: Binomial Distribution – Calculating Probabilities and Expectation | 问题 3：二项分布——概率计算与期望

Question: A fair coin is tossed 8 times. Let X be the number of heads. (a) State the distribution of X. (b) Find P(X = 5). (c) Find P(X ≥ 7). (d) Calculate E(X) and Var(X). [8 marks]

题目： 抛掷一枚均匀硬币 8 次，令 X 为正面朝上的次数。(a) 说明 X 服从何种分布。(b) 计算 P(X = 5)。(c)

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