📚 AS CIE Statistics: Unit Test Mock Paper Analysis | AS CIE 统计：单元测试模拟卷解析
Mock examinations are a powerful tool for diagnosing weaknesses and building confidence before the real AS Statistics paper. This article presents a carefully constructed unit test covering the core topics of the CIE AS syllabus, followed by detailed solutions and commentary. Each question targets specific skills, from data representation to the binomial distribution, and the analysis highlights common pitfalls, efficient methods, and examiner expectations.
模拟考试是诊断薄弱环节、建立临场信心的有效工具。本文提供一套精心设计的单元测试卷，覆盖 CIE AS 统计课程的核心主题，并附有详细解答与评注。每道题目针对特定技能，从数据表示到二项分布，解析将突出常见陷阱、高效解题方法以及考官的期望。
1. Overview of the Mock Paper | 模拟卷概览
The mock paper consists of 9 questions with a total of 50 marks, to be completed in 75 minutes. It mirrors the style of CIE Paper 5 (Probability & Statistics 1), covering representation of data, measures of central tendency and variation, probability, permutations and combinations, discrete random variables, and the binomial distribution. A correct final answer alone is not enough; clear working and proper notation are essential.
本模拟卷共 9 题，满分 50 分，建议用时 75 分钟。它仿照 CIE 试卷 5（概率与统计 1）的风格，涵盖数据表示、集中趋势与变异度量、概率、排列组合、离散随机变量和二项分布。仅给出正确答案是不够的；清晰的步骤和规范的符号至关重要。
2. Question 1: Histogram and Frequency Density | 第1题：直方图与频率密度
A histogram is drawn for a continuous variable where the class widths are unequal. The frequency density for the interval 10 ≤ x < 20 is 1.8, and the frequency is 18. For the interval 20 ≤ x < 35, the frequency is 24. Calculate the frequency density for this second interval and sketch the corresponding bar.
直方图用于类宽不等的连续变量。区间 10 ≤ x < 20 的频率密度为 1.8，频数为 18。对于区间 20 ≤ x < 35，频数为 24。请计算第二个区间的频率密度，并画出对应的条形。
Solution: Frequency density = frequency ÷ class width. First interval: class width = 10, so 1.8 = 18/10 confirms. Second interval: width = 35 − 20 = 15, so frequency density = 24 ÷ 15 = 1.6. The bar should have height 1.6 on the frequency density scale, spanning from 20 to 35.
解答：频率密度 = 频数 ÷ 类宽。第一个区间：类宽 = 10，故 1.8 = 18/10 验证无误。第二个区间：宽度 = 35 − 20 = 15，频率密度 = 24 ÷ 15 = 1.6。条形高度应为 1.6，横跨 20 到 35。
Many candidates mistakenly treat frequency density as frequency or use the class midpoint when drawing the bar. Always label the horizontal axis with the variable and the vertical axis with ‘Frequency density’.
许多考生误将频率密度当作频数，或者用组中值来画条形。务必在横轴上标注变量名称，在纵轴上标注“频率密度”。
3. Question 2: Mean and Standard Deviation from Grouped Data | 第2题：分组数据的均值与标准差
The table summarises the times, in minutes, taken by 40 students to complete a puzzle.
| Time (min) | Frequency |
|---|---|
| 0-4 | 6 |
| 5-9 | 14 |
| 10-14 | 12 |
| 15-19 | 5 |
| 20-30 | 3 |
Estimate the mean and standard deviation using appropriate midpoints.
下表汇总了 40 名学生完成拼图所需的时间（分钟）。使用合适的组中值估计均值和标准差。
Solution: Midpoints: 2, 7, 12, 17, 25 (for the last 20-30). Σf = 40. Σfm = 6×2 + 14×7 + 12×12 + 5×17 + 3×25 = 12 + 98 + 144 + 85 + 75 = 414. Mean = 414/40 = 10.35 min. For standard deviation, Σfm² = 6×4 + 14×49 + 12×144 + 5×289 + 3×625 = 24 + 686 + 1728 + 1445 + 1875 = 5758. Variance = (5758/40) − (10.35)² = 143.95 − 107.1225 = 36.8275. Standard deviation = √36.8275 ≈ 6.07 min.
解答：组中值：2, 7, 12, 17, 25（最后 20-30）。Σf = 40。Σfm = 414，均值 = 414/40 = 10.35 分钟。Σfm² = 5758，方差 = 143.95 − 107.1225 = 36.8275，标准差 ≈ √36.8275 ≈ 6.07 分钟。
Using the correct midpoints for open‑ended classes is critical. The last class is treated as having width 11 (20 to 30 inclusive), so midpoint = (20+30)/2 = 25. Always use the exact class boundaries.
正确使用开放类别的组中值十分关键。最后一类的宽度为 11（20 至 30 含），故中点为 25。始终使用确切的组界。
4. Question 3: Cumulative Frequency and Median | 第3题：累积频率与中位数
Using the data from Question 2, construct a cumulative frequency table and graph, then estimate the median time and the interquartile range. Also state the percentage of students who took less than 8 minutes.
利用第2题的数据，构建累积频率表并绘图，然后估计时间的中位数和四分位距。并说明少于 8 分钟的学生百分比。
Solution: Cumulative frequencies: 6, 20, 32, 37, 40. Upper class boundaries: 4.5, 9.5, 14.5, 19.5, 30.5. Plot (4.5,6), (9.5,20), (14.5,32), (19.5,37), (30.5,40). Median at 20.5th value: from graph ≈ 9.5 min. Q₁ at 10.25th value: ≈ 6.3 min. Q₃ at 30.75th value: ≈ 13.8 min. IQR = 13.8 − 6.3 = 7.5 min. For < 8 min, locate 8 on the upper boundary axis, read cumulative frequency ≈ 14.5, so percentage = (14.5/40)×100% = 36.25%.
解答：累积频数：6, 20, 32, 37, 40。上组界：4.5, 9.5, 14.5, 19.5, 30.5。描点连线。中位数对应 20.5 个值，由图 ≈ 9.5 分钟。Q₁（10.25th）≈ 6.3 分钟，Q₃（30.75th）≈ 13.8 分钟，IQR = 7.5 分钟。少于 8 分钟：在横轴 8 处读取累积频数 ≈ 14.5，百分比 = (14.5/40)×100% = 36.25%。
Candidates must use upper class boundaries when plotting cumulative frequency. A smooth curve should be drawn, and interpolation lines must be clearly shown to earn marks for reading the median and quartiles.
考生在绘制累积频率图时必须使用上组界。应绘制平滑曲线，并清楚标出插值线，以获取读取中位数和四分位数的分数。
5. Question 4: Probability and Venn Diagrams | 第4题：概率与韦恩图
In a group of 60 students, 35 study Mathematics, 28 study Physics, and 12 study both. A student is chosen at random. Draw a Venn diagram and find the probability that the student studies (a) neither subject, (b) Mathematics only, (c) Physics but not Mathematics, (d) at least one of the subjects.
在 60 名学生中，35 人学习数学，28 人学习物理，12 人同时学习两科。随机选择一名学生。画出韦恩图，并求该生（a）两科都不学的概率，（b）只学数学的概率，（c）学物理但不学数学的概率，（d）至少学一科的概率。
Solution: Let M = Math, P = Physics. Both = 12. Math only = 35 − 12 = 23. Physics only = 28 − 12 = 16. Total in union = 23 + 12 + 16 = 51. Neither = 60 − 51 = 9. (a) P(neither) = 9/60 = 0.15. (b) P(Math only) = 23/60 ≈ 0.3833. (c) P(Physics only) = 16/60 = 0.2667. (d) P(at least one) = 51/60 = 0.85. The Venn diagram should show two overlapping circles labelled appropriately.
解答：数学与物理的交集为 12。仅数学：35 − 12 = 23。仅物理：28 − 12 = 16。并集 = 23 + 12 + 16 = 51。两科都不学 = 60 − 51 = 9。（a）P(都不学) = 9/60 = 0.15。（b）P(仅数学) = 23/60 ≈ 0.3833。（c）P(仅物理) = 16/60 = 0.2667。（d）P(至少一科) = 51/60 = 0.85。韦恩图应显示两个相交的圆并正确标注。
Always define events clearly and present probabilities as exact fractions or decimals to three significant figures. The ‘neither’ region is often forgotten; remember it lies outside both circles but inside the rectangle.
务必清晰地定义事件，并将概率表达到三位有效数字的精确分数或小数。“两科都不学”的区域常被遗忘；请记住它位于两个圆之外、矩形之内的位置。
6. Question 5: Permutations and Combinations | 第5题：排列与组合
A committee of 4 people is to be chosen from 7 men and 5 women. Find the number of ways the committee can be formed if (a) there are no restrictions, (b) it must contain exactly 2 women, (c) it must contain more men than women. Additionally, in how many ways can the 4 chosen people be arranged in a row for a photograph if the two women must stand together?
从 7 名男性和 5 名女性中选出一个 4 人委员会。求以下情况的选法数：（a）无限制，（b）恰好包含 2 名女性，（c）男性多于女性。另外，如果选出的 4 人排成一排拍照，且两名女性必须相邻，有多少种排列方式？
Solution: (a) ¹²C₄ = 495. (b) Choose 2 women from 5: ⁵C₂ = 10; choose 2 men from 7: ⁷C₂ = 21; total = 10 × 21 = 210. (c) More men than women: cases 3M1W and 4M0W. 3M1W: ⁷C₃ × ⁵C₁ = 35 × 5 = 175. 4M0W: ⁷C₄ = 35. Total = 210. For arrangement: treat the two women as one block, so 3 items (block + 2 men). Arrange 3 items: 3! = 6. Within the block, women can be arranged in 2! = 2 ways. Total = 6 × 2 = 12 ways per chosen group. Note: this part assumes the exact committee composition from part (b) (2W,2M), so multiply by 210 if required, but the question asks for the arrangement for a given chosen group, so 12.
解答：（a）¹²C₄ = 495。（b）从 5 女中选 2：⁵C₂ = 10；从 7 男中选 2：⁷C₂ = 21；共 210。（c）男多于女：情形 3男1女 和 4男0女。3男1女：⁷C₃ × ⁵C₁ = 35 × 5 = 175；4男0女：⁷C₄ = 35；合计 210。排列：将两名女性视为一个整体，共 3 个对象（整体 + 2 男），排列数 3! = 6；整体内女性排列有 2! = 2 种；共 12 种。此问通常针对已选定的特定小组进行计算。
Combinations (order does not matter) are used for selection, while permutations matter for lining up. The multiplication principle is key: do not add when combining independent choices. In part (c), always list the mutually exclusive cases and sum them.
选择用组合（顺序无关），排队用排列。乘法原理至关重要：切勿在独立选择时将数目相加。对于（c）部分，应列出互斥的情形并求和。
7. Question 6: Discrete Random Variables | 第6题：离散随机变量
The discrete random variable X has the following probability distribution:
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X=x) | 0.2 | a | 0.3 | b |
Given that E(X) = 2.4, find the values of a and b. Then find Var(X) and P(1 < X ≤ 3).
离散随机变量 X 的概率分布如上表。已知 E(X) = 2.4，求 a 和 b 的值。然后求 Var(X) 和 P(1 < X ≤ 3)。
Solution: Sum of probabilities = 1: 0.2 + a + 0.3 + b = 1 → a + b = 0.5. E(X) = Σx·p = 1×0.2 + 2a + 3×0.3 + 4b = 0.2 + 2a + 0.9 + 4b = 1.1 + 2a + 4b = 2.4. So 2a + 4b = 1.3. Solve the system: from a = 0.5 − b, substitute: 2(0.5 − b) + 4b = 1.3 → 1 − 2b + 4b = 1.3 → 2b = 0.3 → b = 0.15, a = 0.35. E(X²) = 1²×0.2 + 4×0.35 + 9×0.3 + 16×0.15 = 0.2 + 1.4 + 2.7 + 2.4 = 6.7. Var(X) = E(X²) − [E(X)]² = 6.7 − 5.76 = 0.94. P(1 < X ≤ 3) = P(X=2) + P(X=3) = 0.35 + 0.3 = 0.65.
解答：概率和为 1：0.2 + a + 0.3 + b = 1 → a + b = 0.5。E(X) = 1.1 + 2a + 4b = 2.4，故 2a + 4b = 1.3。解方程组得 b = 0.15，a = 0.35。E(X²) = 0.2 + 1.4 + 2.7 + 2.4 = 6.7。Var(X) = 6.7 − (2.4)² = 0.94。P(1 < X ≤ 3) = 0.35 + 0.3 = 0.65。
Forming two equations using total probability and expectation is the standard method. Always check your variance is non‑negative. For probabilities of intervals, pay attention to strict vs. non‑strict inequalities.
利用总概率与期望值建立两个方程是标准方法。始终检查方差是否为非负值。对于区间概率，务必注意严格不等式与非严格不等式的区别。
8. Question 7: Expectation and Variance of Linear Functions | 第7题：线性函数的期望与方差
A game involves spinning a wheel where the score X has E(X) = 5.2 and Var(X) = 1.96. The prize money £Y is given by Y = 2X + 3. Find E(Y) and Var(Y). If the cost to play is £12, what is the expected profit per game?
某游戏需旋转一个转盘，得分 X 满足 E(X) = 5.2，Var(X) = 1.96。奖金 £Y 由 Y = 2X + 3 给出。求 E(Y) 和 Var(Y)。如果每次游戏成本为 £12，求每局的期望利润。
Solution: E(Y) = E(2X + 3) = 2E(X) + 3 = 2×5.2 + 3 = 13.4. Var(Y) = Var(2X + 3) = 2² Var(X) = 4 × 1.96 = 7.84. Profit P = Y − 12. E(P) = E(Y) − 12 = 13.4 − 12 = 1.4. So expected profit is £1.40 per game.
解答：E(Y) = 2E(X) + 3 = 13.4。Var(Y) = 4 × Var(X) = 7.84。利润 P = Y − 12，E(P) = 13.4 − 12 = 1.4，故每局期望利润为 £1.40。
Linear transformations of random variables are straightforward: E(aX + b) = aE(X) + b, Var(aX + b) = a² Var(X). Adding a constant does not affect variance. Remember to subtract the fixed cost to find profit.
随机变量的线性变换很直接：E(aX + b) = aE(X) + b，Var(aX + b) = a² Var(X)。加常数不影响方差。求利润时记得减去固定成本。
9. Question 8: Binomial Distribution | 第8题：二项分布
In a factory, 8% of light bulbs produced are defective. A random sample of 20 bulbs is selected. Find the probability that (a) exactly 2 are defective, (b) fewer than 3 are defective, (c) more than 1 is defective. State the assumptions needed for a binomial model.
某工厂生产的灯泡中有 8% 为次品。随机抽取 20 个灯泡作为样本。求以下概率：（a）恰好有 2 个次品，（b）少于 3 个次品，（c）多于 1 个次品。并说明使用二项分布所需的假设。
Solution: X ~ B(20, 0.08). (a) P(X=2) = ²⁰C₂ (0.08)² (0.92)¹⁸ = 190 × 0.0064 × 0.92¹⁸. Compute 0.92¹⁸ ≈ 0.215, so ≈ 190 × 0.0064 × 0.215 ≈ 0.261. (b) P(X < 3) = P(X=0) + P(X=1) + P(X=2). P(X=0) = 0.92²⁰ ≈ 0.1887. P(X=1) = 20 × 0.08 × 0.92¹⁹ ≈ 20 × 0.08 × 0.205 ≈ 0.328. Total ≈ 0.1887 + 0.328 + 0.261 = 0.778. (c) P(X > 1) = 1 − P(X ≤ 1) = 1 − (0.1887 + 0.328) = 0.4833. Assumptions: bulbs are independent, probability of defective is constant, each bulb is either defective or not, fixed number of trials.
解答：X ~ B(20, 0.08)。（a）P(X=2) = ²⁰C₂ (0.08)² (0.92)¹⁸ ≈ 0.261。（b）P(X<3) = P(0)+P(1)+P(2) ≈ 0.1887 + 0.328 + 0.261 = 0.778。（c）P(X>1) = 1 − P(X ≤ 1) = 1 − 0.5167 = 0.4833。假设：各灯泡相互独立，次品概率恒定，每个灯泡只有好坏两种结果，试验次数固定。
Use the binomial formula accurately. With small p, probabilities can be small; computing (0.92)¹⁸ requires a calculator. In an exam, you may be given binomial tables, but here you must show the formula. Always verify that all binomial conditions are satisfied when applying the model.
准确使用二项公式。由于 p 较小，概率值可能很小；计算 (0.92)¹⁸ 需要计算器。考试中或可查表，但此处需展示公式。应用模型前，务必确认所有二项条件均满足。
10. Question 9: Normal Approximation to Binomial | 第9题：二项分布的正态近似
A company claims that 70% of its customers are satisfied. In a survey of 150 customers, find the probability that more than 110 are satisfied using a normal approximation. Apply a continuity correction.
某公司声称其顾客满意率为 70%。在对 150 名顾客的调查中，使用正态近似（含连续性校正）求多于 110 人满意的概率。
Solution: X ~ B(150, 0.7). Mean = np = 150 × 0.7 = 105. Variance = npq = 150 × 0.7 × 0.3 = 31.5, standard deviation = √31.5 ≈ 5.612. Normal approximation: X ≈ N(105, 31.5). Need P(X > 110). With continuity correction, P(X > 110) ≈ P(X > 110.5) for the normal. Standardise: z = (110.5 − 105) / 5.612 ≈ 5.5 / 5.612 ≈ 0.980. Then P(Z > 0.980) = 1 − Φ(0.980) ≈ 1 − 0.8365 = 0.1635.
解答：X ~ B(150, 0.7)。均值 = 105，方差 = 31.5，标准差 ≈ 5.612。正态近似 X ≈ N(105, 31.5)。需求 P(X > 110)，经连续性校正后变为 P(X > 110.5)。标准化：z ≈ 0.980，P(Z > 0.980) ≈ 1 − 0.8365 = 0.1635。
Continuity correction is essential for binomial to normal approximation. Because we are looking for more than 110, the normal cut‑off is 110.5. Also check that np and nq are greater than 5 to justify using the normal approximation.
连续性校正在二项到正态近似中必不可少。由于求的是多于 110，正态截断点为 110.5。同时应检查 np 与 nq 均大于 5，以验证可近似使用正态分布。
11. Common Mistakes and Tips | 常见错误与提示
Many marks are lost through careless errors: misreading class boundaries, forgetting to use frequency density, mixing up combinations and permutations, ignoring continuity corrections, and misapplying variance rules. Always double‑check the conditions for each distribution. Ensure your calculator is in degree mode where necessary, and present probability values with appropriate rounding (3 significant figures unless otherwise stated).
许多分数因粗心而丢失：误读组界，忘记使用频率密度，混淆组合与排列，忽略连续性校正，以及错误应用方差公式。务必核查每个分布的条件。确保计算器设置合适，概率值按适当精度（通常三位有效数字）呈现。
- Histograms: use frequency density = frequency / class width.
- Grouped data: use class midpoints for mean and standard deviation.
- Cumulative frequency: plot against upper class boundary.
- Probability: show clear working and use correct set notation.
- Binomial & normal: state the distribution and parameters before calculating.
- 直方图：使用频率密度 = 频数 ÷ 类宽。
- 分组数据：使用组中值计算均值和标准差。
- 累积频率：对上组界描点。
- 概率：展示清晰步骤并使用正确的集合符号。
- 二项与正态：先写出分布及其参数再计算。
12. Final Advice | 最后建议
Practice under timed conditions and review every mistake. The AS Statistics paper rewards methodical working and precise communication. By analysing a mock paper in depth, you transform errors into learning opportunities. Remember, statistics is not just about numerical answers; it is about explaining variability and drawing meaningful conclusions from data.
在限时条件下练习并反思每个错误。AS 统计试卷看重条理清晰的解答和准确的表达。通过深入分析模拟卷，你可以将错误转化为学习契机。请记住，统计学不只看数字答案，更在于解释变异性并从数据中得出有意义的结论。
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