A-Level CAIE Statistics: Unit Test Mock Exam Analysis | A-Level CAIE 统计：单元测试模拟卷解析

📚 A-Level CAIE Statistics: Unit Test Mock Exam Analysis | A-Level CAIE 统计：单元测试模拟卷解析

This article provides a detailed, step-by-step walkthrough of a unit test mock exam designed for the CAIE A-Level Statistics syllabus. By examining selected questions, we consolidate key concepts, from data representation to hypothesis testing, and highlight common mistakes that students must avoid.

本文通过一套精心编制的CAIE A-Level统计单元测试模拟卷，详细拆解典型题目的解题过程。我们将巩固从数据表示到假设检验的核心知识点，并指出学生必须避免的常见错误。

1. Stem-and-Leaf Plot and Box Plot Analysis | 茎叶图与盒形图分析

Mock Question 1: The scores (out of 50) of 20 students are: 32, 45, 41, 37, 29, 48, 36, 44, 33, 40, 39, 42, 35, 31, 46, 38, 28, 34, 47, 43. Construct an ordered stem-and-leaf plot. Find the median, quartiles and identify any outliers. Draw a box-and-whisker plot. Calculate the sample mean and standard deviation.

模拟题1：20名学生的成绩（满分50分）如下：32, 45, 41, 37, 29, 48, 36, 44, 33, 40, 39, 42, 35, 31, 46, 38, 28, 34, 47, 43。绘制有序茎叶图；求中位数、四分位数并识别异常值；绘制盒形图；计算样本均值和样本标准差。

Step 1: Sort the data in ascending order: 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48.

步骤1：将数据按升序排列：28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48。

Step 2: Construct the stem-and-leaf plot using tens as stems and units as leaves. For stem 2, leaves are 8,9; for stem 3, leaves 1,2,3,4,5,6,7,8,9; for stem 4, leaves 0,1,2,3,4,5,6,7,8. Include a key, e.g. 2|8 means 28.

步骤2：以十位数为茎，个位数为叶绘制茎叶图。茎2的叶为8,9；茎3的叶为1,2,3,4,5,6,7,8,9；茎4的叶为0,1,2,3,4,5,6,7,8。附上说明，如2|8代表28。

Step 3: Determine the median position using (n+1)/2 = (20+1)/2 = 10.5, so take the average of the 10th and 11th values. The 10th value is 38, the 11th is 39, median Q₂ = (38+39)/2 = 38.5.

步骤3：用公式(n+1)/2 = (20+1)/2 = 10.5确定中位数位置，即第10和第11个数的平均值。第10个为38，第11个为39，中位数 Q₂ = (38+39)/2 = 38.5。

Step 4: For the lower quartile Q₁, consider the lower half of the data (first 10 values): 28 to 38. Its median is (5th+6th)/2 = (33+34)/2 = 33.5. Hence Q₁ = 33.5.

步骤4：下四分位数Q₁由数据的前半部分（前10个：28到38）计算，其中位数为第5和第6个的平均值：(33+34)/2 = 33.5。因此Q₁ = 33.5。

Step 5: For the upper quartile Q₃, consider the upper half (last 10 values): 39 to 48. Its median is (15th+16th)/2 = (43+44)/2 = 43.5. So Q₃ = 43.5.

步骤5：上四分位数Q₃由数据的后半部分（后10个：39到48）计算，其中位数是第15和第16个的平均值：(43+44)/2 = 43.5。因此Q₃ = 43.5。

Step 6: Compute the interquartile range: IQR = Q₃ – Q₁ = 43.5 – 33.5 = 10. Outlier boundaries are Q₁ – 1.5×IQR = 33.5 – 15 = 18.5, and Q₃ + 1.5×IQR = 43.5 + 15 = 58.5. All values lie between 18.5 and 58.5, so there are no outliers.

步骤6：计算四分位距 IQR = Q₃ – Q₁ = 43.5 – 33.5 = 10。异常值界限为 Q₁ – 1.5×IQR = 33.5 – 15 = 18.5 以及 Q₃ + 1.5×IQR = 43.5 + 15 = 58.5。所有数据均在18.5到58.5之间，因此没有异常值。

Step 7: Draw the box-and-whisker plot using these five-number summary values. The whiskers extend to the minimum (28) and maximum (48).

步骤7：用五数概括法绘制盒形图。胡须延伸至最小值28和最大值48。

Step 8: Compute the sample mean x̄ = Σx / n. Sum = 28+29+31+…+48 = 770. Hence x̄ = 770/20 = 38.5.

步骤8：计算样本均值 x̄ = Σx / n。总和 = 770，故 x̄ = 770/20 = 38.5。

Step 9: Compute the sample variance s² = Σ(x – x̄)² / (n-1). The squared deviations sum to (28-38.5)² + … + (48-38.5)² = 530. So s² = 530/19 ≈ 27.89, and sample standard deviation s = √27.89 ≈ 5.28.

步骤9：计算样本方差 s² = Σ(x – x̄)² / (n-1)。离差平方和为530，因此 s² = 530/19 ≈ 27.89，样本标准差 s = √27.89 ≈ 5.28。

A very common mistake is to divide by n instead of (n-1) when computing the sample variance: always use (n-1) for an unbiased estimate of the population variance.

一个非常常见的错误是在计算样本方差时除以n而不是(n-1)：务必使用(n-1)来得到总体方差的无偏估计。

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2. Permutations, Combinations and Conditional Probability | 排列、组合与条件概率

Mock Question 2: A committee of 4 people is to be chosen from 6 men and 5 women. Find the number of ways the committee can be formed if (a) there are no restrictions, (b) it must include exactly 2 women, (c) it must include at least 1 woman. Then, if the committee is chosen randomly, find the probability that exactly 2 women are selected, given that at least 1 woman is selected.

模拟题2：从6名男士和5名女士中选出4人组成委员会。求以下情况的选法数：(a) 无任何限制；(b) 恰好包含2名女士；(c) 至少包含1名女士。若随机选取，求在已知至少选中1名女士的条件下恰好选中2名女士的概率。

Part (a): Total number of ways = number of combinations of 4 from 11: ¹¹C₄ = 11!/(4!7!) = 330.

(a) 部分：总方法数 = 从11人中任选4人的组合数：¹¹C₄ = 330。

Part (b): Exactly 2 women means choose 2 women from 5 and 2 men from 6. Number of ways = ⁵C₂ × ⁶C₂ = 10 × 15 = 150.

(b) 部分：恰好2名女士意味着从5名女士中选2人，从6名男士中选2人。方法数 = ⁵C₂ × ⁶C₂ = 10 × 15 = 150。

Part (c): At least 1 woman = total ways minus ways with zero women. Zero women means all 4 are men: ⁶C₄ = 15. So ways with at least 1 woman = 330 – 15 = 315.

(c) 部分：至少1名女士 = 总方法数减去没有女士的方法数。没有女士即全部4人为男士：⁶C₄ = 15。所以至少1名女士的方法数 = 330 – 15 = 315。

Conditional probability: Let A be the event ‘exactly 2 women’, B be the event ‘at least 1 woman’. We need P(A|B) = P(A∩B)/P(B). Here A ∩ B = A, because ‘exactly 2 women’ already satisfies ‘at least 1 woman’. So P(A|B) = P(A)/P(B) = (150/330) / (315/330) = 150/315 = 10/21.

条件概率：设事件A为“恰好2名女士”，事件B为“至少1名女士”。需求P(A|B) = P(A∩B)/P(B)。因为A含于B，A∩B = A。因此P(A|B) = (150/330) / (315/330) = 150/315 = 10/21。

Always structure conditional probability as ‘given’ over ‘total of the restricted sample space’. Using combinations is usually more reliable than constructing a probability tree for such selections without replacement.

条件概率总是将“给定事件”作为限制后的样本空间。对于此类不放回选取，使用组合通常比画概率树更为可靠。

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3. Discrete Random Variable and Expectation | 离散随机变量与期望

Mock Question 3: The discrete random variable Y has probability distribution given by P(Y=y) = ky² for y = 1, 2, 3, 4, where k is a constant. (i) Find the value of k. (ii) Calculate E(Y) and Var(Y). (iii) Find E(3Y – 2) and Var(3Y – 2).

模拟题3：离散随机变量Y的概率分布为P(Y=y) = ky²，y = 1,2,3,4，其中k为常数。(i) 求k的值。(ii) 计算E(Y)和Var(Y)。(iii) 求E(3Y – 2)和Var(3Y – 2)。

(i) The sum of all probabilities must equal 1: k(1² + 2² + 3² + 4²) = k(1+4+9+16) = 30k = 1, so k = 1/30.

(i) 所有概率之和必须等于1：k(1²+2²+3²+4²) = 30k = 1，因此k = 1/30。

Construct a probability distribution table:

y	1	2	3	4
P(Y=y)	1/30	4/30	9/30	16/30

(ii) E(Y) = Σ y·P(Y=y) = 1×(1/30) + 2×(4/30) + 3×(9/30) + 4×(16/30) = (1 + 8 + 27 + 64)/30 = 100/30 = 10/3 ≈ 3.33. E(Y²) = Σ y²·P(Y=y) = 1²×(1/30) + 2²×(4/30) + 3²×(9/30) + 4²×(16/30) = (1 + 16 + 81 + 256)/30 = 354/30 = 11.8. Var(Y) = E(Y²) – [E(Y)]² = 11.8 – (10/3)² = 11.8 – 100/9 = (106.2 – 100)/9 = 6.2/9 = 62/90 = 31/45 ≈ 0.689.

(ii) E(Y) = 100/30 = 10/3 ≈ 3.33。E(Y²) = 354/30 = 11.8。Var(Y) = 11.8 – (10/3)² = 11.8 – 100/9 = 31/45 ≈ 0.689。

(iii) Using linearity of expectation: E(3Y – 2) = 3E(Y) – 2 = 3×(10/3) – 2 = 10 – 2 = 8. Var(3Y – 2) = 3² Var(Y) = 9 × (31/45) = 279/45 = 6.2.

(iii) 利用期望的线性性：E(3Y – 2) = 3E(Y) – 2 = 10 – 2 = 8。Var(3Y – 2) = 3²Var(Y) = 9×(31/45) = 6.2。

Important: Adding a constant shifts the mean but does not affect variance. Multiplication by a constant scales the variance by the square of that constant.

关键点：加常数会使期望平移，但不影响方差；乘以常数会使方差乘以该常数的平方。

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4. Binomial Distribution Calculations | 二项分布计算

Mock Question 4: A biased coin is tossed 12 times. The probability of obtaining a head on any toss is 0.3. Let X be the number of heads. (i) State the distribution of X. (ii) Find P(X = 5). (iii) Find P(X ≤ 2). (iv) Find P(3 < X ≤ 8).

模拟题4：一枚不均匀的硬币抛掷12次，每次出现正面的概率为0.3。设X为正面次数。(i) 指出X的分布。(ii) 求P(X = 5)。(iii) 求P(X ≤ 2)。(iv) 求P(3 < X ≤ 8)。

(i) X ~ B(12, 0.3). The trials are independent and each has a fixed probability of success 0.3.

(i) X ~ B(12, 0.3)。各次试验相互独立，每次成功的概率固定为0.3。

(ii) P(X = 5) = ¹²C₅ × (0.3)⁵ × (0.7)⁷. Using calculator or formula, ¹²C₅ = 792, hence P(X = 5) = 792 × (0.3⁵) × (0.7⁷) ≈ 0.1585 (4 d.p.).

(ii) P(X = 5) = ¹²C₅ × (0.3)⁵ × (0.7)⁷。计算得792 × 0.00243 × 0.0823543 ≈ 0.1585（保留四位小数）。

(iii) P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2). P(X=0) = (0.7)¹² ≈ 0.0138; P(X=1) = 12×(0.3)×(0.7)¹¹ ≈ 0.0712; P(X=2) = ⁶⁶C₂×(0.3)²×(0.7)¹⁰ ≈ 0.1678. Sum ≈ 0.2528. Alternatively, use cumulative binomial tables or GDC.

(iii) P(X ≤ 2) = P(X=0)+P(X=1)+P(X=2)。分别计算后相加≈0.2528。也可使用累积二项分布表或图形计算器。

(iv) P(3 < X ≤ 8) = P(X ≤ 8) – P(X ≤ 3). Using table, P(X ≤ 8) ≈ 0.9983, P(X ≤ 3) ≈ 0.4925, giving 0.5058. Always express the interval correctly; ‘3 < X’ means X ≥ 4.

(iv) P(3 < X ≤ 8) = P(X ≤ 8) – P(X ≤ 3)。查表得差约0.5058。注意区间表达：“3 < X”意味着X ≥ 4。

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5. Normal Distribution Calculations | 正态分布计算

Mock Question 5: The mass of a sample of eggs is normally distributed with mean 58 g and standard deviation 3.5 g. (i) Find the probability that a randomly selected egg has a mass greater than 62 g. (ii) Find the probability that its mass lies between 55 g and 61 g. (iii) Find the mass exceeded by the heaviest 10% of eggs.

模拟题5：某批鸡蛋的质量服从正态分布，均值为58克，标准差为3.5克。(i) 求随机选取的一枚鸡蛋质量大于62克的概率。(ii) 求质量介于55克和61克之间的概率。(iii) 求最重10%的鸡蛋所超过的质量界限。

(i) Let M ~ N(58, 3.5²). Standardize: Z = (62 – 58) / 3.5 = 4/3.5 ≈ 1.1429. P(M > 62) = P(Z > 1.1429) = 1 – Φ(1.14) ≈ 1 – 0.8729 = 0.1271 (using standard normal table).

(i) 设M ~ N(58, 3.5²)。标准化：Z = (62 – 58)/3.5 ≈ 1.1429。P(M > 62) = 1 – Φ(1.14) ≈ 0.1271。

(ii) P(55 < M < 61) = P((55-58)/3.5 < Z < (61-58)/3.5) = P(-0.8571 < Z < 0.8571) = Φ(0.86) – (1 – Φ(0.86)) = 2×Φ(0.86) – 1 ≈ 2×0.8051 – 1 = 0.6102.

(ii) P(55 < M < 61) = P(-0.8571 < Z < 0.8571) = Φ(0.86) – (1 – Φ(0.86)) = 2×0.8051 – 1 = 0.6102。

(iii) Let m be the mass such that P(M > m) = 0.10. Then P(M ≤ m) = 0.90 → Z = 1.2816 (from table). Hence m = 58 + 1.2816 × 3.5 ≈ 62.49 g.

(iii) 设m为满足P(M > m) = 0.10的质量，则P(M ≤ m) = 0.90 → Z = 1.2816，m ≈ 58 + 1.2816×3.5 = 62.49克。

Standardisation is a critical step: always subtract the mean and divide by the standard deviation. Be careful to use the correct tail when reading the normal table.

标准化是关键步骤：总是先减去均值再除以标准差。查表时必须区分左尾、右尾区域。

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6. Confidence Interval for the Population Mean | 总体均值的置信区间

Mock Question 6: A random sample of 40 chocolate bars has a mean weight of 102.5 g. The population standard deviation is known to be 3.2 g. Construct a 95% confidence interval for the population mean weight. Interpret the interval.

模拟题6：从一批巧克力棒中随机抽取40条，测得样本均重102.5克，已知总体标准差为3.2克。构建总体平均质量的95%置信区间并解释其含义。

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