Year 12 CCEA Statistics: Unit Test Mock Paper Walkthrough | Year 12 CCEA 统计：单元测试模拟卷解析

📚 Year 12 CCEA Statistics: Unit Test Mock Paper Walkthrough | Year 12 CCEA 统计：单元测试模拟卷解析

This walkthrough provides a detailed, step-by-step solution to a typical CCEA Year 12 AS Statistics unit test. The paper focuses on AS Unit 1: Descriptive Statistics and Probability, covering data handling, probability rules, discrete random variables, bivariate data and the normal distribution. Each solution is paired with revision notes to reinforce key concepts.

本文详细解析了一份典型的 CCEA Year 12 AS 统计单元测试模拟卷。试卷聚焦 AS 单元 1：描述性统计与概率，涵盖数据处理、概率规则、离散随机变量、双变量数据与正态分布。每道题的解答均配有复习要点，巩固关键概念。

1. Raw Data Summary and Outliers | 原始数据汇总与异常值

A teacher recorded the test scores of 11 students: 33, 38, 41, 44, 45, 48, 51, 53, 55, 60, 92. Find the median, lower quartile (Q1), upper quartile (Q3) and interquartile range (IQR). Identify any outliers using the 1.5 × IQR rule and construct a box plot.

某教师记录了 11 名学生的测验分数：33, 38, 41, 44, 45, 48, 51, 53, 55, 60, 92。求中位数、下四分位数 (Q1)、上四分位数 (Q3) 与四分位距 (IQR)。用 1.5 × IQR 规则识别异常值并绘制箱线图。

Step 1: Order the data (already sorted). The median is the 6th value because (11+1)/2 = 6, so median = 48.

步骤 1：数据已排序。中位数是第 6 个值，因为 (11+1)/2 = 6，所以中位数 = 48。

Step 2: For Q1, consider the lower half before the median: 33, 38, 41, 44, 45. Median of these five is the 3rd: Q1 = 41.

步骤 2：Q1 取中位数之前的下半部分：33, 38, 41, 44, 45。这五个数的中位数是第 3 个：Q1 = 41。

Step 3: For Q3, take the upper half after the median: 51, 53, 55, 60, 92. Median is the 3rd: Q3 = 55.

步骤 3：Q3 取中位数之后的上半部分：51, 53, 55, 60, 92。中位数是第 3 个：Q3 = 55。

Step 4: IQR = Q3 – Q1 = 55 – 41 = 14.

步骤 4：IQR = Q3 – Q1 = 55 – 41 = 14。

Step 5: Outlier boundaries: lower fence = Q1 – 1.5 × IQR = 41 – 21 = 20; upper fence = Q3 + 1.5 × IQR = 55 + 21 = 76. Any value below 20 or above 76 is an outlier. The score 92 exceeds 76, so it is an outlier.

步骤 5：异常值界限：下界 = Q1 – 1.5 × IQR = 41 – 21 = 20；上界 = Q3 + 1.5 × IQR = 55 + 21 = 76。任何低于 20 或高于 76 的值为异常值。分数 92 超过 76，因此是异常值。

Step 6: Box plot construction: draw a scale. The box stretches from Q1 (41) to Q3 (55) with a line at the median (48). Whiskers extend to the smallest non-outlier (33) and the largest non-outlier (60). Plot 92 as an individual point.

步骤 6：绘制箱线图：设定刻度。箱体从 Q1 (41) 延伸到 Q3 (55)，中位数 (48) 处画线。触须延伸至最小非异常值 (33) 和最大非异常值 (60)。将 92 作为独立点标出。

This question tests your ability to summarise data and detect unusual values using the IQR method, a staple in CCEA AS unit 1.

此题考查运用 IQR 方法汇总数据并检测异常值的能力，是 CCEA AS 单元 1 的基本题型。

2. Grouped Frequency and Histogram | 分组数据与直方图

The masses of 50 apples are summarised in the table: (0–50 g): 8, (50–100): 15, (100–150): 18, (150–200): 6, (200–300): 3. Estimate the mean mass and draw a histogram using frequency density.

50 个苹果的重量汇总如下：(0–50 g): 8, (50–100): 15, (100–150): 18, (150–200): 6, (200–300): 3。估计平均重量并用频率密度绘制直方图。

Step 1: Find class midpoints: 25, 75, 125, 175, 250. Multiply each midpoint by frequency: 25×8=200, 75×15=1125, 125×18=2250, 175×6=1050, 250×3=750. Sum of fx = 200+1125+2250+1050+750 = 5375. Total frequency = 50. Estimated mean = 5375 / 50 = 107.5 g.

步骤 1：求组中值：25, 75, 125, 175, 250。将每个组中值乘以频数：25×8=200, 75×15=1125, 125×18=2250, 175×6=1050, 250×3=750。∑fx = 200+1125+2250+1050+750 = 5375。总频数 = 50。估计均值 = 5375 / 50 = 107.5 g。

Step 2: To construct a histogram, calculate frequency density = frequency / class width. Class widths: 50, 50, 50, 50, 100. Densities: 8/50=0.16, 15/50=0.3, 18/50=0.36, 6/50=0.12, 3/100=0.03.

步骤 2：绘制直方图需计算频率密度 = 频数 / 组距。组距：50, 50, 50, 50, 100。密度：8/50=0.16, 15/50=0.3, 18/50=0.36, 6/50=0.12, 3/100=0.03。

Step 3: Draw axes: x-axis with mass boundaries, y-axis as frequency density. Plot bars with heights corresponding to density. The final class (200–300) will be wider but shorter because of the lower density.

步骤 3：绘制坐标轴：x 轴为重量界限，y 轴为频率密度。根据密度高度绘制矩形条。最后一组 (200–300) 因密度较低而较宽但更矮。

Always remember that in a histogram, area represents frequency. CCEA exam questions frequently ask you to estimate the mean and justify why it is an estimate – because raw data are not available.

务必记住，直方图中面积表示频数。CCEA 考试常要求估计均值并说明为何是估计值——因为无法获得原始数据。

3. Basic Probability Rules | 基本概率规则

Events A and B are independent with P(A) = 0.4 and P(B) = 0.3. Find P(A ∩ B), P(A ∪ B) and P(A’ ∩ B’). Explain why the addition rule for non-mutually exclusive events differs from that for mutually exclusive events.

事件 A 与 B 独立，P(A) = 0.4，P(B) = 0.3。求 P(A ∩ B)、P(A ∪ B) 和 P(A’ ∩ B’)。解释非互斥事件的加法规则为何不同于互斥事件。

Step 1: For independent events, P(A ∩ B) = P(A) × P(B) = 0.4 × 0.3 = 0.12.

步骤 1：独立事件，P(A ∩ B) = P(A) × P(B) = 0.4 × 0.3 = 0.12。

Step 2: P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.4 + 0.3 – 0.12 = 0.58. The subtraction avoids double-counting the overlap.

步骤 2：P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.4 + 0.3 – 0.12 = 0.58。减法避免了重复计算重叠部分。

Step 3: P(A’ ∩ B’) = P((A ∪ B)’) = 1 – P(A ∪ B) = 1 – 0.58 = 0.42. This represents neither A nor B occurring.

步骤 3：P(A’ ∩ B’) = P((A ∪ B)’) = 1 – P(A ∪ B) = 1 – 0.58 = 0.42。表示 A 和 B 都不发生。

Step 4: Mutually exclusive events cannot occur together, so P(A ∩ B)=0, giving P(A ∪ B) = P(A) + P(B). With non‑mutually exclusive events, the intersection must be subtracted to correctly compute the union probability.

步骤 4：互斥事件不能同时发生，故 P(A ∩ B)=0，得 P(A ∪ B) = P(A) + P(B)。非互斥事件必须减去交集才能正确计算并集概率。

This question reinforces the core probability formulas that appear in nearly every CCEA statistics paper.

此题强化了几乎每份 CCEA 统计试卷都会出现的核心概率公式。

4. Conditional Probability and Tree Diagrams | 条件概率与树状图

In a school, 60% of students study Maths (M) and 45% study Physics (P). 30% study both. A student is chosen at random. Find the probability the student studies Physics given they study Maths, and illustrate with a two-way table and tree diagram.

某学校 60% 学生学习数学 (M)，45% 学习物理 (P)，30% 两门都学。随机选一名学生。求该生学习物理的条件下也学习数学的概率，并用双向表和树状图说明。

Step 1: Use the conditional probability formula: P(P|M) = P(P ∩ M) / P(M) = 0.30 / 0.60 = 0.5.

步骤 1：使用条件概率公式：P(P|M) = P(P ∩ M) / P(M) = 0.30 / 0.60 = 0.5。

Step 2: Construct a two-way table: rows for Maths (Yes/No), columns for Physics (Yes/No). Fill in the intersections using given percentages.

步骤 2：构建双向表：行表示数学（是/否），列表示物理（是/否）。用给定百分比填表。

	Physics Yes	Physics No	Total
Maths Yes	30	30	60
Maths No	15	25	40
Total	45	55	100

Step 3: From the table, P(Physics Yes | Maths Yes) = 30/60 = 0.5, matching the formula.

步骤 3：由表知，P(物理是 | 数学是) = 30/60 = 0.5，与公式一致。

Step 4: Tree diagram: first branch Maths (0.6) and not Maths (0.4). For each, branch Physics with conditional probabilities: P(Physics | Maths) = 0.30/0.60 = 0.5, so P(not Physics | Maths) = 0.5. P(Physics | not Maths) = 0.15/0.40 = 0.375.

步骤 4：树状图：第一层分支为数学（0.6）与非数学（0.4）。每个分支下再画物理条件概率：P(物理|数学) = 0.5，故 P(非物理|数学)=0.5。P(物理|非数学)=0.15/0.40=0.375。

Being able to switch between tables, diagrams and formulae is a vital skill for CCEA exam success.

能在表格、图表和公式之间灵活转换是 CCEA 考试成功的关键技能。

5. Discrete Random Variables and Expectation | 离散随机变量与期望

The probability distribution of a discrete random variable X is: x: 1, 2, 3, 4 with P(X=x): 0.2, 0.3, 0.1, 0.4. Find E(X), Var(X) and E(2X + 5).

离散随机变量 X 的概率分布为：x: 1, 2, 3, 4，P(X=x): 0.2, 0.3, 0.1, 0.4。求 E(X)、Var(X) 以及 E(2X + 5)。

Step 1: E(X) = Σ x·P(X=x) = 1×0.2 + 2×0.3 + 3×0.1 + 4×0.4 = 0.2 + 0.6 + 0.3 + 1.6 = 2.7.

步骤 1：E(X) = Σ x·P(X=x) = 1×0.2 + 2×0.3 + 3×0.1 + 4×0.4 = 0.2 + 0.6 + 0.3 + 1.6 = 2.7。

Step 2: E(X²) = Σ x²·P(X=x) = 1×0.2 + 4×0.3 + 9×0.1 + 16×0.4 = 0.2 + 1.2 + 0.9 + 6.4 = 8.7.

步骤 2：E(X²) = Σ x²·P(X=x) = 1×0.2 + 4×0.3 + 9×0.1 + 16×0.4 = 0.2 + 1.2 + 0.9 + 6.4 = 8.7。

Step 3: Var(X) = E(X²) – [E(X)]² = 8.7 – (2.7)² = 8.7 – 7.29 = 1.41.

步骤 3：Var(X) = E(X²) – [E(X)]² = 8.7 – (2.7)² = 8.7 – 7.29 = 1.41。

Step 4: Using linearity of expectation, E(2X + 5) = 2E(X) + 5 = 2×2.7 + 5 = 5.4 + 5 = 10.4.

步骤 4：利用期望的线性性质，E(2X + 5) = 2E(X) + 5 = 2×2.7 + 5 = 5.4 + 5 = 10.4。

Expectation and variance of discrete distributions appear frequently in both Unit 1 and Unit 2 papers. Remember that E(aX+b) = aE(X)+b, but Var(aX+b) = a²Var(X) — a shift does not affect variance.

离散分布的期望和方差经常出现在单元 1 和单元 2 的试卷中。记住 E(aX+b) = aE(X)+b，但 Var(aX+b) = a²Var(X)——平移不影响方差。

6. Bivariate Data: Correlation and Regression Line | 双变量数据：相关与回归线

The number of hours of revision (x) and test scores (y) for five students are: (2, 45), (4, 55), (6, 65), (8, 75), (10, 85). Calculate the product moment correlation coefficient (PMCC) and the equation of the regression line of y on x. Interpret the slope.

五名学生的复习小时数 (x) 与测验分数 (y) 如下：(2, 45), (4, 55), (6, 65), (8, 75), (10, 85)。计算积矩相关系数 (PMCC) 与 y 对 x 的回归线方程。解释斜率的意义。

Step 1: Summaries: n=5, Σx=30, Σy=325, Σx²=4+16+36+64+100=220, Σy²=2025+3025+4225+5625+7225=22125, Σxy=90+220+390+600+850=2150.

步骤 1：汇总：n=5, Σx=30, Σy=325, Σx²=4+16+36+64+100=220, Σy²=2025+3025+4225+5625+7225=22125, Σxy=90+220+390+600+850=2150。

Step 2: PMCC formula: r = [nΣxy – (Σx)(Σy)] / √{[nΣx² – (Σx)²][nΣy² – (Σy)²]}.
Numerator = 5×2150 – 30×325 = 10750 – 9750 = 1000.
Denominator = √[(5×220 – 900)(5×22125 – 105625)] = √[(1100–900)(110625–105625)] = √[200×5000] = √1,000,000 = 1000.

步骤 2：PMCC 公式：r = [nΣxy – (Σx)(Σy)] / √{[nΣx² – (Σx)²][nΣy² – (Σy)²]}.
分子 = 5×2150 – 30×325 = 10750 – 9750 = 1000。
分母 = √[(5×220 – 900)(5×22125 – 105625)] = √[(1100–900)(110625–105625)] = √[200×5000] = √1,000,000 = 1000。

Step 3: PMCC r = 1000/1000 = 1. A perfect positive linear correlation indicates that all points lie exactly on a straight line.

步骤 3：PMCC r = 1000/1000 = 1。完全正向线性相关说明所有点恰好在一条直线上。

Step 4: Regression line y = a + bx. b = [nΣxy – ΣxΣy] / [nΣx² – (Σx)²] = 1000 / 200 = 5. a = (Σy/n) – b(Σx/n) = (325/5) – 5×(30/5) = 65 – 30 = 35. Equation: y = 35 + 5x.

步骤 4：回归线 y = a + bx。b = [nΣxy – ΣxΣy] / [nΣx² – (Σx)²] = 1000 / 200 = 5。a = (Σy/n) – b(Σx/n) = (325/5) – 5×(30/5) = 65 – 30 = 35。方程：y = 35 + 5x。

Step 5: The slope of 5 means that for every extra hour of revision, the test score increases by 5 marks on average. This interpretation must mention “on average” in the context of regression.

步骤 5：斜率 5 表示每增加 1 小时复习，分数平均增加 5 分。解释回归时务必使用“平均”一词。

CCEA often asks for interpretation of slope and intercept within the specific context of the question.

CCEA 常要求结合题目背景解释斜率与截距。

7. Normal Distribution Calculations | 正态分布计算

The lengths of bolts produced by a machine are normally distributed with mean μ = 50 mm and standard deviation σ = 2 mm. Find the probability that a randomly chosen bolt has a length between 47 mm and 53 mm, and determine the length below which the shortest 5% of bolts fall.

某机器生产的螺栓长度服从正态分布，均值 μ = 50 mm，标准差 σ = 2 mm。求随机选出的螺栓长度介于 47 mm 与 53 mm 之间的概率，并确定最短的 5% 螺栓长度上限。

Step 1: Convert to z-scores. For x = 47, z = (47 – 50)/2 = –1.5. For x = 53, z = (53 – 50)/2 = 1.5.

步骤 1：转换为 z 值。对于 x = 47，z = (47 – 50)/2 = –1.5；对于 x = 53，z = (53 – 50)/2 = 1.5。

Step 2: Using standard normal tables: P(Z < 1.5) = 0.9332, P(Z < –1.5) = 1 – 0.9332 = 0.0668. Therefore P(–1.5 < Z < 1.5) = 0.9332 – 0.0668 = 0.8664.

步骤 2：查标准正态分布表：P(Z < 1.5) = 0.9332，P(Z < –1.5) = 1 – 0.9332 = 0.0668。因此 P(–1.5 < Z < 1.5) = 0.9332 – 0.0668 = 0.8664。

Step 3: For the shortest 5%, find the z-value such that P(Z < z) = 0.05. From tables, z ≈ –1.6449. The corresponding x is: μ + zσ = 50 – 1.6449×2 = 50 – 3.2898 = 46.71 mm (to 2 d.p.). So 5% of bolts are shorter than 46.71 mm.

步骤 3：最短的 5%，求满足 P(Z < z) = 0.05 的 z 值。查表得 z ≈ –1.6449。对应的 x 为：μ + zσ = 50 – 1.6449×2 = 50 – 3.2898 = 46.71 mm（保留两位小数）。因此 5% 的螺栓长度短于 46.71 mm。

Remember to sketch the normal curve and shade the area of interest; this prevents sign errors. CCEA provides normal tables in the exam formula booklet.

记得绘制正态曲线并标出关注区域；这样可避免符号错误。CCEA 考试提供公式手册中的正态分布表。

8. Sampling Methods and Bias | 抽样方法与偏差

A school wants to survey students’ opinions on a new canteen menu. It plans to interview the first 50 students who enter the canteen on Monday. Name this sampling method, explain a potential bias and suggest a better stratified sample design.

某学校想调查学生对新食堂菜单的意见，计划在周一采访最先进入食堂的 50 名学生。说出该抽样方法的名称，解释潜在的偏差，并推荐更好的分层抽样设计。

Step 1: The described method is convenience sampling (or opportunity sampling), because it selects readily available participants rather than randomly.

步骤 1：所述方法为便利抽样（或机会抽样），因为它选择最容易接触的参与者而非随机选取。

Step 2: Potential bias: students who arrive early may have particular characteristics (e.g., younger years, specific dietary habits) and exclude those arriving later or bringing packed lunches. The sample may not represent the whole student population.

步骤 2：潜在偏差：早到的学生可能具有特定特征（如低年级、特殊饮食习惯），而排除晚到或自带午餐的学生。样本可能无法代表全体学生。

Step 3: A stratified sample would divide students into meaningful strata, such as Year Groups (Year 12, Year 13, Year 14), and randomly select a proportional number from each group. For a school of 600 with 200 in each year, selecting 17, 17 and 16 respectively would give a total of 50, ensuring all years are fairly represented.

步骤 3：分层抽样可将学生分为有意义的层，如年级（Year 12, Year 13, Year 14），然后从每层随机抽取比例适当的人数。若全校 600 人，各年级 200 人，则抽取 17、17、16 人，总计 50，确保各年级公平体现。

Stratified sampling reduces bias and improves representativeness when population subgroups are known.

当总体子群已知时，分层抽样可减少偏差并提高代表性。

Being able to evaluate sampling methods is essential for the statistical enquiry cycle component in CCEA assessments.

能够评价抽样方法对 CCEA 评估中统计探究循环部分至关重要。

Published by TutorHao | Statistics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Year 12 CCEA Statistics: Unit Test Mock Paper Walkthrough | Year 12 CCEA 统计：单元测试模拟卷解析