Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析

📚 Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析

This article provides a complete walkthrough of a Year 13 CCEA Statistics unit test mock paper. Based on the CCEA S1 syllabus, each question is broken down with step-by-step solutions, common pitfalls and revision tips. Whether you are preparing for an in-class assessment or final revision, this guide will sharpen your statistical skills.

本文详细解析一份 Year 13 CCEA 统计单元测试模拟卷。试题紧扣 CCEA S1 大纲，每题均配有分步解答、常见错误提示及复习技巧。无论你是在准备单元测验还是期末复习，这份指南都将帮助你提升统计解题能力。

1. Overview of the Mock Paper | 模拟卷概览

The mock paper contains eight questions designed to cover the core topics of CCEA AS Statistics: data presentation, numerical measures, probability, tree diagrams, discrete random variables, the binomial distribution and the normal distribution. Total marks are 75, to be completed in 90 minutes.

本套模拟卷共包含八道题，覆盖 CCEA AS 统计的核心内容：数据展示、数值度量、概率、树状图、离散随机变量、二项分布与正态分布。满分 75 分，建议用时 90 分钟。

2. Stem-and-Leaf Diagram and Box Plot | 茎叶图与箱线图

Question: The lengths (cm) of 15 leaves are: 5.2, 5.8, 6.1, 6.3, 6.3, 6.5, 6.7, 6.8, 7.0, 7.1, 7.4, 7.5, 7.8, 8.2, 9.1. Construct an ordered stem-and-leaf diagram, find the median and quartiles, then draw a box plot. Use the 1.5 × IQR rule to check for outliers.

题目：15 片叶子的长度（cm）为：5.2, 5.8, 6.1, 6.3, 6.3, 6.5, 6.7, 6.8, 7.0, 7.1, 7.4, 7.5, 7.8, 8.2, 9.1。绘制定序茎叶图，求中位数和四分位数，再绘制箱线图。用 1.5 × IQR 法则检验异常值。

For the stem-and-leaf, use the integer part as the stem and the first decimal as the leaf. The ordered diagram is:
5 | 2 8
6 | 1 3 3 5 7 8
7 | 0 1 4 5 8
8 | 2
9 | 1
Key: 5|2 means 5.2 cm.

绘制茎叶图时，整数部分作茎，第一位小数作叶。定序图为：茎 5 叶 2 8；茎 6 叶 1 3 3 5 7 8；茎 7 叶 0 1 4 5 8；茎 8 叶 2；茎 9 叶 1。图例：5|2 表示 5.2 cm。

Median is the 8th value: 6.8 cm. Lower quartile Q₁ is the 4th value: 6.3 cm. Upper quartile Q₃ is the 12th value: 7.5 cm. IQR = 7.5 – 6.3 = 1.2 cm. Lower fence = 6.3 – 1.5 × 1.2 = 4.5 cm; upper fence = 7.5 + 1.5 × 1.2 = 9.3 cm. Since 9.1 is inside the fence, there are no outliers. Draw the box plot with whiskers from 5.2 to 9.1.

中位数为第 8 个数值：6.8 cm。下四分位数 Q₁ 为第 4 个数值：6.3 cm。上四分位数 Q₃ 为第 12 个数值：7.5 cm。IQR = 7.5 – 6.3 = 1.2 cm。下限 = 6.3 – 1.5 × 1.2 = 4.5 cm；上限 = 7.5 + 1.5 × 1.2 = 9.3 cm。9.1 在范围内，因此无异常值。箱线图须线从 5.2 延伸至 9.1。

3. Mean and Standard Deviation | 均值与标准差

Question: The masses (kg) of eight parcels are: 2.4, 3.1, 2.9, 3.5, 2.8, 3.0, 3.3, 2.7. Calculate the mean and the unbiased estimate of the population standard deviation.

题目：八个包裹的质量（kg）为：2.4, 3.1, 2.9, 3.5, 2.8, 3.0, 3.3, 2.7。计算均值及总体标准差的无偏估计值。

Sum of values = 23.7. Mean x̄ = 23.7 / 8 = 2.9625 kg. To find the standard deviation, compute Σ(x – x̄)². Subtractions give: 0.3164, 0.0189, 0.0039, 0.2889, 0.0264, 0.0014, 0.1139, 0.0689. Sum = 0.8388. Unbiased estimate s = √[0.8388 / (8 – 1)] = √0.11983 ≈ 0.346 kg.

数值总和为 23.7。均值 x̄ = 23.7 / 8 = 2.9625 kg。计算 Σ(x – x̄)²，各项平方差为：0.3164, 0.0189, 0.0039, 0.2889, 0.0264, 0.0014, 0.1139, 0.0689，总和 0.8388。无偏估计值 s = √[0.8388 / (8 – 1)] = √0.11983 ≈ 0.346 kg。

Many CCEA questions require working with Σx and Σx² directly. Here Σx² = 2.4² + 3.1² + … = 70.87. Then s = √[(70.87 – 23.7²/8) / 7] = √[(70.87 – 70.21125) / 7] = √0.094107 ≈ 0.307 kg (check calculation). Always verify with both methods.

多数 CCEA 考题要求直接使用 Σx 和 Σx²。此处 Σx² = 2.4² + 3.1² + … = 70.87。利用公式 s = √[(70.87 – 23.7²/8) / 7] = √[(70.87 – 70.21125) / 7] = √0.094107 ≈ 0.307 kg。注意核对两种方法的计算结果。

4. Basic Probability Rules | 概率基础

Question: A bag contains 5 red, 3 blue and 2 green counters. Two counters are drawn without replacement. Find the probability that they are of the same colour. Hence determine whether the events ‘first is red’ and ‘second is blue’ are independent.

题目：袋中有 5 红、3 蓝、2 绿筹码，无放回地抽取两枚。求它们颜色相同的概率，并判断事件“第一枚为红”与“第二枚为蓝”是否独立。

P(same colour) = P(both red) + P(both blue) + P(both green) = (5/10 × 4/9) + (3/10 × 2/9) + (2/10 × 1/9) = 20/90 + 6/90 + 2/90 = 28/90 = 14/45.

P(同色) = (5/10 × 4/9) + (3/10 × 2/9) + (2/10 × 1/9) = 28/90 = 14/45。

P(first red) = 5/10 = 1/2. P(second blue) = (5/10 × 3/9) + (3/10 × 2/9) + (2/10 × 3/9) = 15/90 + 6/90 + 6/90 = 27/90 = 3/10. P(first red ∩ second blue) = 5/10 × 3/9 = 15/90 = 1/6. Check independence: P(first red) × P(second blue) = 1/2 × 3/10 = 3/20 ≠ 1/6, so the events are not independent.

P(第一枚红) = 1/2，P(第二枚蓝) = 3/10，P(同时发生) = 5/10 × 3/9 = 1/6。由于 1/6 ≠ (1/2)×(3/10) = 3/20，故两事件不独立。

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

Question: In a school, 60% of students study Mathematics (M), 45% study Physics (P), and 30% study both. A student is chosen at random. Find P(P|M). Draw a tree diagram to represent the probabilities and verify your result.

题目：某校中，60% 的学生学习数学 (M)，45% 学习物理 (P)，30% 两者都学。随机选一名学生，求 P(P|M)。绘制树状图表示概率并验证结果。

Using the conditional formula: P(P|M) = P(M ∩ P) / P(M) = 0.30 / 0.60 = 0.5. The tree starts with M (0.6) and M’ (0.4). From M, branch to P (0.3/0.6 = 0.5) and P’ (0.5). This confirms the conditional probability is 0.5.

条件概率公式：P(P|M) = P(M ∩ P) / P(M) = 0.30 / 0.60 = 0.5。树状图先分 M (0.6) 与 M’ (0.4)，从 M 再分 P (0.5) 与 P’ (0.5)，与所求条件概率一致。

Always label branches with probabilities given or derived. If P(M ∩ P) = 0.3, the joint probability for M ∩ P’ is 0.6 – 0.3 = 0.3, and for M’ ∩ P = 0.45 – 0.3 = 0.15. A complete tree helps answer further questions like P(M’|P).

务必在树状图上标注已知或推导出的概率。若 P(M ∩ P) = 0.3，则 P(M ∩ P’) = 0.3，P(M’ ∩ P) = 0.15。完整的树状图有助于解答诸如 P(M’|P) 等进一步问题。

6. Discrete Random Variables | 离散随机变量

Question: The discrete random variable X has probability distribution:
x: 0, 1, 2, 3
P(X=x): 0.2, a, 0.3, b
Given that E(X) = 1.9, find a and b. Then compute Var(X).

题目：离散随机变量 X 的概率分布为：x 取值 0,1,2,3；对应的概率 P(X=x) 为 0.2, a, 0.3, b。已知 E(X) = 1.9，求 a 与 b，并计算 Var(X)。

Sum of probabilities: 0.2 + a + 0.3 + b = 1 ⇒ a + b = 0.5. (1)
E(X) = 0×0.2 + 1×a + 2×0.3 + 3×b = a + 0.6 + 3b = 1.9 ⇒ a + 3b = 1.3. (2)
Subtract (1) from (2): 2b = 0.8 ⇒ b = 0.4. Then a = 0.1.

概率总和为 1：0.2 + a + 0.3 + b = 1，即 a + b = 0.5。（1）
E(X) = 0×0.2 + 1×a + 2×0.3 + 3×b = a + 0.6 + 3b = 1.9，即 a + 3b = 1.3。（2）
（2）减（1）得 2b = 0.8，故 b = 0.4，a = 0.1。

E(X²) = 0²×0.2 + 1²×0.1 + 2²×0.3 + 3²×0.4 = 0 + 0.1 + 1.2 + 3.6 = 4.9. Var(X) = E(X²) – [E(X)]² = 4.9 – 1.9² = 4.9 – 3.61 = 1.29.

E(X²) = 4.9，Var(X) = 4.9 – 3.61 = 1.29。

7. Binomial Distribution | 二项分布

Question: 15% of items produced by a machine are defective. A random sample of 20 items is selected. Find the probability that exactly 3 are defective, and the probability that at most 2 are defective. Give your answers to four decimal places.

题目：某机器生产的物品中有 15% 的次品。随机抽取 20 件样品，求恰好有 3 件次品的概率，以及最多有 2 件次品的概率。结果保留四位小数。

Let X ~ B(20, 0.15). P(X = 3) = ²₀C₃ × (0.15)³ × (0.85)¹⁷.
²₀C₃ = 1140, (0.15)³ = 0.003375, (0.85)¹⁷ calculation gives about 0.0539? Let’s compute: 0.85¹⁷ ≈ 0.0631 (using calculator). Then 1140 × 0.003375 × 0.0631 ≈ 0.2428. However, a more exact product yields 0.2427. We’ll use 0.2427.

设 X ~ B(20, 0.15)。P(X = 3) = ²₀C₃ × (0.15)³ × (0.85)¹⁷。²₀C₃ = 1140，(0.15)³ = 0.003375，0.85¹⁷ 约等于 0.0631，三者相乘约 0.2427。

For ‘at most 2’, sum P(X=0) + P(X=1) + P(X=2).
P(X=0) = 0.85²²²? Actually 0.85²⁰ = (0.85)²⁰ ≈ 0.0388.
P(X=1) = 20 × 0.15 × 0.85¹⁹ ≈ 20 × 0.15 × 0.0456 ≈ 0.1368.
P(X=2) = 190 × 0.15² × 0.85¹⁸ ≈ 190 × 0.0225 × 0.0536 ≈ 0.2293.
Sum ≈ 0.0388 + 0.1368 + 0.2293 = 0.4049. (Using accurate calculator values gives approx 0.4049.)

“最多 2 件”需计算 P(X=0) + P(X=1) + P(X=2)。P(X=0) ≈ 0.0388，P(X=1) ≈ 0.1368，P(X=2) ≈ 0.2293，总和约为 0.4049。

Also note E(X) = np = 20 × 0.15 = 3, Var(X) = np(1-p) = 20 × 0.15 × 0.85 = 2.55.

期望值 E(X) = 3，方差 Var(X) = 2.55。

8. Normal Distribution | 正态分布

Question: The heights of adult males are normally distributed with mean 175 cm and standard deviation 6 cm. Find the probability that a randomly chosen male is taller than 182 cm. Also find the height below which the shortest 10% of males fall.

题目：成年男性身高服从正态分布，均值为 175 cm，标准差为 6 cm。求随机选出的男性身高超过 182 cm 的概率，并求最短 10% 男性所对应身高上限。

Let H ~ N(175, 6²). For X > 182: z = (182 – 175) / 6 = 7/6 ≈ 1.1667. Using normal tables, P(Z < 1.17) ≈ 0.8790, so P(Z > 1.17) = 1 – 0.8790 = 0.1210. More precisely with z=1.17, the tail is about 0.121.

设 H ~ N(175, 6²)。当 X > 182 时，z = (182 – 175) / 6 ≈ 1.1667。查表得 P(Z < 1.17) ≈ 0.8790，故 P(Z > 1.17) = 0.1210。

For the shortest 10%, find the z-score such that P(Z < z) = 0.10. From inverse normal tables, z ≈ -1.2816. Then height = μ + zσ = 175 + (-1.2816)×6 ≈ 175 - 7.69 = 167.31 cm. So about 167.3 cm.

寻找最短 10% 的身高分界点：P(Z < z) = 0.10，反查表得 z ≈ -1.2816。代入身高 = 175 - 1.2816 × 6 ≈ 167.3 cm。

Always sketch a normal curve and shade the required area. Check your answer: P(H > 182) should be less than 0.5 because 182 is above the mean.

解题时务必画出正态曲线并涂阴影区域。验证答案：182 cm 高于均值，故 P(H > 182) 应小于 0.5。

9. Common Mistakes to Avoid | 常见错误避坑

Candidates often lose marks by confusing population and sample standard deviation formulas. Use the divisor n−1 for an unbiased estimate of population standard deviation in S1 unless told otherwise. For box plots, always label the scale and show the fences for outlier detection. In probability, clearly define events and ensure tree branches show conditional probabilities correctly.

考生常因混淆总体与样本标准差公式而失分。在 S1 中，除非特别说明，计算总体标准差的无偏估计应使用 n−1 作为分母。箱线图务必标注刻度并展示异常值探测用的界限。处理概率问题时，应明确定义事件，确保树状图分支上的条件概率正确。

For binomial calculations, double-check combinations and powers; many errors arise from miscopying the probability p. When using normal tables, round your z-score appropriately and be careful with the direction of inequalities. If you are given a scenario that seems to require a continuity correction, note that CCEA S1 does not require continuity correction.

二项分布计算中，应反复核对组合数与幂次；很多错误源于概率 p 的误抄。查正态分布表时，合理修约 z 值并注意不等号方向。若题目看似需要用连续性校正，请注意 CCEA S1 不要求进行连续性校正。

10. Final Exam Tips | 考前终极建议

Manage your time carefully – leave roughly 1.2 minutes per mark. Start with the questions you find easiest to build confidence. Show all working, especially in probability and distribution questions – even if the final answer is wrong, method marks can be earned. Always read the question again after solving to ensure you have answered exactly what was asked, including required units.

合理分配时间——大约每分对应 1.2 分钟。先做最有把握的题以建立信心。概率与分布题目务必展示完整的计算过程，即使最终答案有误，也能获得步骤分。解完后重读题目，确保已按要求回答，包括书写单位。

Make sure your calculator is set to the correct mode (normal stat functions). Familiarise yourself with the formula booklet provided in CCEA exams; knowing where each formula sits saves valuable time. Finally, a clear, well-structured solution is easier to mark – use words like ‘Stem’ and ‘Leaf’ or ‘P(X > 3) =’ to guide the examiner.

确认计算器处于正确的模式（如普通统计功能）。熟悉 CCEA 考试提供的公式手册，知道每个公式的位置能节省宝贵时间。最后，字迹清晰、结构井然的解答更便于评分——使用像“茎”、“叶”或“P(X > 3) =”等引导词，有助于考官采点。

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Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析