📚 Year 11 CCEA Statistics Unit Test Mock Paper Walkthrough | CCEA 统计单元测试模拟卷解析
This article provides a complete walkthrough of a mock unit test for Year 11 CCEA Statistics. Each question is carefully designed to reflect the style and content of the CCEA syllabus, covering data collection, representation, averages and spread, probability, correlation and regression. Step‑by‑step solutions in both English and Chinese help you consolidate key skills and avoid common pitfalls.
本文为 Year 11 CCEA 统计单元测试模拟卷提供完整解析。每道题目紧扣 CCEA 课程要求，涵盖数据收集、数据表示、均值与离散程度、概率、相关与回归等核心内容。双语分步解析旨在帮助你巩固要点，规避常见失分点。
1. Census vs Sampling | 普查与抽样
A school wants to estimate the average daily screen time of its 1200 students. Explain why a census might be impractical and suggest a suitable sampling method. Justify your choice.
某校想估计其 1200 名学生每天的平均屏幕使用时间。请解释为什么普查可能不切实际，并提出一种合适的抽样方法，说明理由。
A census would require measuring screen time for every student, which is time‑consuming, expensive and logistically difficult. Sampling gathers data from a subset, saving resources. A stratified random sample is recommended. First, split the population into strata (e.g. Year 8, Year 9, Year 10, Year 11). Then, determine the number of students to sample from each stratum proportionally to its size. Finally, within each stratum, select students using a simple random technique (e.g. random number generator). This method ensures every subgroup is fairly represented and reduces selection bias.
普查需要测量每一名学生的屏幕时间，耗时、昂贵且组织困难。抽样只从一个子集收集数据，节省资源。推荐采用分层随机抽样。首先，将总体分层（例如按年级分为 8、9、10、11 年级）。然后，根据各层大小按比例确定每层应抽取的学生数。最后，在每层内使用简单随机方法（如随机数生成器）选取学生。此方法保证各子群均被公平代表，降低了选择偏差。
2. Stem‑and‑Leaf Diagram & Box Plot | 茎叶图与箱线图
The data below show the number of minutes 13 students spent on a quiz:
15, 22, 23, 25, 27, 31, 32, 34, 34, 36, 41, 42, 48.
Construct a stem‑and‑leaf diagram, find the median and quartiles, and draw a box plot.
以下数据是 13 名学生在一次小测中所用时间（分钟）：15, 22, 23, 25, 27, 31, 32, 34, 34, 36, 41, 42, 48。请画出茎叶图，求出中位数和四分位数，并绘制箱线图。
Stem‑and‑leaf diagram (key: 1|5 = 15):
| Stem | Leaf |
|---|---|
| 1 | 5 |
| 2 | 2 3 5 7 |
| 3 | 1 2 4 4 6 |
| 4 | 1 2 8 |
茎叶图（图例：1|5 = 15）： 如上表。
Ordered data: 15, 22, 23, 25, 27, 31, 32, 34, 34, 36, 41, 42, 48. n = 13. Median is the 7th value: 32. Q₁ is at position (13+1)/4 = 3.5, so Q₁ = (23+25)/2 = 24. Q₃ is at position 3(13+1)/4 = 10.5, so Q₃ = (36+41)/2 = 38.5. The five‑number summary: min = 15, Q₁ = 24, median = 32, Q₃ = 38.5, max = 48. IQR = 38.5 – 24 = 14.5. The box plot is drawn on a scale with a box from 24 to 38.5, a vertical line at 32, and whiskers extending to 15 and 48.
排序后数据：15, 22, 23, 25, 27, 31, 32, 34, 34, 36, 41, 42, 48。n = 13。中位数是第 7 个值：32。Q₁ 位置 = (13+1)/4 = 3.5，故 Q₁ = (23+25)/2 = 24。Q₃ 位置 = 3(13+1)/4 = 10.5，故 Q₃ = (36+41)/2 = 38.5。五数概括：最小值 15，Q₁ = 24，中位数 32，Q₃ = 38.5，最大值 48。IQR = 14.5。绘制箱线图时，箱子从 24 到 38.5，中间线在 32，触须延伸至 15 和 48。
3. Mean, Range and Standard Deviation | 均值、极差与标准差
For the data set: 12, 15, 18, 21, 24, calculate the mean, the range and the population standard deviation.
对于数据集：12, 15, 18, 21, 24，计算均值、极差和总体标准差。
Mean μ = (12+15+18+21+24) ÷ 5 = 90 ÷ 5 = 18. Range = 24 – 12 = 12. To find the standard deviation, compute the squared deviations from the mean: (12–18)² = 36, (15–18)² = 9, (18–18)² = 0, (21–18)² = 9, (24–18)² = 36. Sum = 36+9+0+9+36 = 90. Variance σ² = 90 ÷ 5 = 18. Standard deviation σ = √18 ≈ 4.24 (3 s.f.).
均值 μ = 18。极差 = 24 – 12 = 12。计算标准差需先求各值与均值的离差平方：(12–18)² = 36，(15–18)² = 9，(18–18)² = 0，(21–18)² = 9，(24–18)² = 36。总和 = 90。方差 σ² = 90 ÷ 5 = 18。标准差 σ = √18 ≈ 4.24（保留三位有效数字）。
4. Estimating the Mean from a Grouped Frequency Table | 根据分组频率表估计均值
A survey recorded the number of hours spent on homework per week by 32 students. The results are grouped:
| Hours | Frequency |
|---|---|
| 0 – <10 | 4 |
| 10 – <20 | 8 |
| 20 – <30 | 12 |
| 30 – <40 | 6 |
| 40 – <50 | 2 |
Estimate the mean number of hours spent on homework per week.
一项调查记录了 32 名学生每周用于家庭作业的时间（小时），上表为分组数据。请估计每周作业时间的平均值。
Use the midpoint of each class: 5, 15, 25, 35, 45. Multiply each midpoint by its frequency: 5×4 = 20, 15×8 = 120, 25×12 = 300, 35×6 = 210, 45×2 = 90. Total of these products = 20+120+300+210+90 = 740. Estimated mean = 740 ÷ 32 = 23.125 hours. When rounded to a sensible degree, approximately 23.1 hours.
使用每组的组中值：5, 15, 25, 35, 45。各组中点乘频数：5×4 = 20，15×8 = 120，25×12 = 300，35×6 = 210，45×2 = 90。乘积总和 = 740。估计均值 = 740 ÷ 32 = 23.125 小时。适当约整后约为 23.1 小时。
5. Probability Tree Diagram (Without Replacement) | 不放回概率树图
A bag contains 5 red and 3 blue balls. Two balls are drawn at random one after the other, without replacement. Draw a tree diagram and calculate the probability that (a) both balls are red, (b) at least one ball is blue.
袋中有 5 个红球和 3 个蓝球。连续随机抽取两球，不放回。画出树图并计算：(a) 两球都是红色的概率；(b) 至少一球是蓝色的概率。
Tree diagram structure:
First draw: P(Red) = 5/8, P(Blue) = 3/8.
Second draw after a Red: P(Red) = 4/7, P(Blue) = 3/7.
Second draw after a Blue: P(Red) = 5/7, P(Blue) = 2/7.
Probability (both red) = (5/8) × (4/7) = 20/56 = 5/14.
Probability (at least one blue) = 1 – P(both red) = 1 – 5/14 = 9/14.
Alternatively, add the paths RR? No, RR is both red, so the complement works neatly.
树图结构：第一次抽取：P(红) = 5/8，P(蓝) = 3/8；若第一次抽到红后，第二次：P(红) = 4/7，P(蓝) = 3/7；若第一次抽到蓝后，第二次：P(红) = 5/7，P(蓝) = 2/7。
概率（两球皆红）= (5/8) × (4/7) = 20/56 = 5/14。
概率（至少一蓝）= 1 – P(两球皆红) = 1 – 5/14 = 9/14。亦可直接累加含蓝路径的概率，得相同结果。
6. Mutually Exclusive and Independent Events | 互斥事件与独立事件
Given P(A) = 0.3, P(B) = 0.4 and P(A ∩ B) = 0.1. Determine whether events A and B are (i) mutually exclusive, (ii) independent. Give a reason in each case.
已知 P(A) = 0.3，P(B) = 0.4，P(A ∩ B) = 0.1。判断事件 A 和 B 是否 (i) 互斥；(ii) 独立，并分别说明理由。
(i) Mutually exclusive events require P(A ∩ B) = 0. Here P(A ∩ B) = 0.1 ≠ 0, so A and B are not mutually exclusive.
(ii) For independence, we test P(A ∩ B) = P(A) × P(B). P(A) × P(B) = 0.3 × 0.4 = 0.12. Since 0.1 ≠ 0.12, the events are not independent.
(i) 互斥事件要求 P(A ∩ B) = 0。此处 0.1 ≠ 0，故 A 与 B 不是互斥事件。
(ii) 检验独立性需满足 P(A ∩ B) = P(A)P(B)。P(A)P(B) = 0.3 × 0.4 = 0.12，而 0.1 ≠ 0.12，因此两事件不独立。
7. Spearman’s Rank Correlation Coefficient | 斯皮尔曼秩相关系数
Ten students are ranked on their Maths and Physics scores. Calculate Spearman’s rank correlation coefficient.
| Student | Maths rank | Physics rank |
|---|---|---|
| A | 1 | 2 |
| B | 2 | 1 |
| C | 3 | 4 |
| D | 4 | 3 |
| E | 5 | 5 |
| F | 6 | 7 |
| G | 7 | 6 |
| H | 8 | 10 |
| I | 9 | 9 |
| J | 10 | 8 |
十名学生的数学和物理排名如上表。计算斯皮尔曼秩相关系数。
Difference d = (Maths rank – Physics rank). Values: –1, 1, –1, 1, 0, –1, 1, –2, 0, 2.
d²: 1, 1, 1, 1, 0, 1, 1, 4, 0, 4. Σ d² = 14.
Spearman’s formula: rs = 1 – (6 Σ d²) / [n(n² – 1)]. Here n = 10, n(n² – 1) = 10(100 – 1) = 10 × 99 = 990.
rs = 1 – (6 × 14) / 990 = 1 – 84/990 = 1 – 0.084848… ≈ 0.915 (3 s.f.).
A value close to +1 indicates a strong positive correlation between the ranks.
差值 d =（数学排名 – 物理排名）。得：–1, 1, –1, 1, 0, –1, 1, –2, 0, 2。
d²：1, 1, 1, 1, 0, 1, 1, 4, 0, 4。Σ d² = 14。
斯皮尔曼公式：rs = 1 – (6 Σ d²) / [n(n² – 1)]。n = 10，n(n² – 1) = 10×99 = 990。
rs = 1 – (84/990) = 1 – 0.084848… ≈ 0.915。
数值接近 +1，说明排名之间有很强的正相关。
8. Equation of the Regression Line | 回归线方程
The table shows values of x and y for five observations.
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y |
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