📚 Year 12 CAIE Statistics: Unit Test Mock Paper Analysis | Year 12 CAIE 统计：单元测试模拟卷解析
This article provides a detailed walkthrough of a unit test mock paper for Year 12 CAIE Statistics (S1). We break down each question to highlight key concepts, common pitfalls, and efficient problem-solving strategies, helping you build confidence for the actual exam.
本文为 Year 12 CAIE 统计（S1）单元测试模拟卷提供详细解析。我们逐题拆解，突出关键概念、常见错误和高效解题策略，帮助你建立考试信心。
1. Mock Paper Structure & Topic Coverage | 模拟卷结构与考点分布
The mock paper consists of 7 compulsory questions worth a total of 50 marks, designed for a 75‑minute session. The questions cover data representation (stem‑and‑leaf, box plots), grouped frequency estimation, probability and tree diagrams, permutations, discrete random variables, binomial and geometric distributions, normal distribution, and a full hypothesis test.
模拟卷由7道必答题组成，总分50分，限时75分钟。题目涵盖数据表示（茎叶图、箱线图）、分组频数估算、概率与树图、排列、离散随机变量、二项分布与几何分布、正态分布以及完整的假设检验。
Question 1 is a straightforward data summary, Question 2 explores probability with conditional events, Question 3 applies permutations, Question 4 tests discrete random variables, Question 5 deals with the binomial distribution, Question 6 brings in geometric and normal distributions, and Question 7 is a hypothesis test. This structure mirrors the balance of a real S1 paper.
第1题是简单的数据汇总，第2题考查条件概率，第3题应用排列，第4题测试离散随机变量，第5题涉及二项分布，第6题结合几何分布与正态分布，第7题为假设检验。这种结构很好地模拟了真实S1试卷的平衡性。
2. Stem‑and‑Leaf Diagram and Box Plot | 茎叶图与箱线图解析
Question 1a: The lengths (in mm) of 24 leaves are recorded: 32, 35, 38, 41, 43, 44, 47, 49, 51, 52, 53, 55, 56, 58, 59, 61, 63, 65, 67, 68, 71, 74, 78, 82. Construct a stem‑and‑leaf diagram and a box‑and‑whisker plot, then identify any outliers.
第1a题：记录了24片叶子的长度（mm）：32, 35, 38, 41, 43, 44, 47, 49, 51, 52, 53, 55, 56, 58, 59, 61, 63, 65, 67, 68, 71, 74, 78, 82。构建茎叶图和箱线图，并识别任何异常值。
Common error – forgetting the key: A stem‑and‑leaf diagram must include a key, e.g., 3 | 2 means 32 mm. The leaves are ordered, so for stem 3 the leaves are 2,5,8.
常见错误——遗漏图例：茎叶图必须包含图例，例如 3 | 2 表示 32 mm。叶必须排序，因此茎3的叶为 2,5,8。
| 3 | 2 5 8 |
| 4 | 1 3 4 7 9 |
| 5 | 1 2 3 5 6 8 9 |
| 6 | 1 3 5 7 8 |
| 7 | 1 4 8 |
| 8 | 2 |
For the box plot, the five‑number summary is needed. Since n = 24 (even), the median is the average of the 12th and 13th values: (55+56)/2 = 55.5. Q₁ is the median of the first 12 values: (44+47)/2 = 45.5. Q₃ is the median of the last 12 values: (67+68)/2 = 67.5. The IQR = 67.5 − 45.5 = 22. Outlier boundaries: Q₁ − 1.5×IQR = 45.5 − 33 = 12.5 (no lower outlier), Q₃ + 1.5×IQR = 67.5 + 33 = 100.5 (no upper outlier). The whiskers extend to the minimum (32) and maximum (82).
对于箱线图，需要五数概括。由于 n=24（偶数），中位数是第12和13个值的平均值：(55+56)/2=55.5。Q₁ 是前12个值的中位数：(44+47)/2=45.5。Q₃ 是后12个值的中位数：(67+68)/2=67.5。IQR = 67.5 − 45.5 = 22。异常值界限：Q₁ − 1.5×IQR = 12.5（无低异常值），Q₃ + 1.5×IQR = 100.5（无高异常值）。须线延伸至最小值 32 和最大值 82。
3. Central Tendency with Grouped Data | 分组数据的集中趋势
Question 1b continues with a grouped frequency table for the masses of 40 apples: 100–120 (5), 120–130 (8), 130–140 (12), 140–150 (10), 150–170 (5). Estimate the mean and standard deviation, and explain why your answers are estimates.
第1b题继续提供一个分组频数表用于40个苹果的质量：100–120(5), 120–130(8), 130–140(12), 140–150(10), 150–170(5)。估算均值和标准差，并解释为何答案是估计值。
We use mid‑points: 110, 125, 135, 145, 160. The estimated mean is x̄ = Σfx / Σf. Σf = 40, Σfx = 5×110 + 8×125 + 12×135 + 10×145 + 5×160 = 550 + 1000 + 1620 + 1450 + 800 = 5420. Hence x̄ = 5420/40 = 135.5 g. For the standard deviation, use the formula s = √[ (Σfx² / (n−1)) − (n/(n−1))x̄² ] or the working formula. Σfx² = 5×110² + 8×125² + 12×135² + 10×145² + 5×160² = 60500 + 125000 + 218700 + 210250 + 128000 = 742450. Then s² = (742450 − 40×135.5²)/39 = (742450 − 40×18360.25)/39 = (742450 − 734410)/39 = 8040/39 ≈ 206.15. s ≈ √206.15 ≈ 14.36 g.
我们使用组中值：110, 125, 135, 145, 160。估算均值 x̄ = Σfx / Σf。Σf=40, Σfx=5×110+8×125+12×135+10×145+5×160=550+1000+1620+1450+800=5420。因此 x̄=5420/40=135.5 g。对于标准差，使用公式 s=√[(Σfx²/(n−1))−(n/(n−1))x̄²] 或工作公式。Σfx²=5×110²+8×125²+12×135²+10×145²+5×160²=60500+125000+218700+210250+128000=742450。然后 s²=(742450−40×135.5²)/39=(742450−734410)/39=8040/39≈206.15，s≈14.36 g。
The answers are estimates because we assumed all observations in a class fall exactly at the midpoint; the true raw data are unknown. When calculating s² for a sample, the divisor is n−1, not n — this is a frequent mistake.
答案是估计值，因为我们假设每个组内的所有观测值都恰好落在组中值上；真正的原始数据是未知的。在计算样本方差 s² 时，分母是 n−1 而不是 n——这是一个常见错误。
4. Probability and Conditional Events | 概率与条件事件
Question 2: In a school, 45% of students study Art, 30% study Biology, and 12% study both. Find the probability that a randomly chosen student studies Art or Biology, and the conditional probability that a student studies Art given they study Biology. Then draw a Venn diagram.
第2题：一所学校中，45%的学生学习艺术，30%学习生物，12%同时学习两者。求随机选择一名学生学习艺术或生物的概率，以及在学习生物的条件下学习艺术的条件概率。然后画出韦恩图。
Let A = Art, B = Biology. P(A) = 0.45, P(B) = 0.30, P(A∩B) = 0.12. P(A∪B) = P(A) + P(B) − P(A∩B) = 0.45 + 0.30 − 0.12 = 0.63. The conditional probability P(A|B) = P(A∩B) / P(B) = 0.12 / 0.30 = 0.4. Many candidates confuse P(A∩B) with P(A∪B) or invert the fraction when calculating conditional probability.
设 A=艺术，B=生物。P(A)=0.45, P(B)=0.30, P(A∩B)=0.12。P(A∪B)=P(A)+P(B)−P(A∩B)=0.45+0.30−0.12=0.63。条件概率 P(A|B)=P(A∩B)/P(B)=0.12/0.30=0.4。许多考生会将 P(A∩B) 与 P(A∪B) 混淆，或在计算条件概率时颠倒分子分母。
The Venn diagram has two overlapping circles. The region “A only” is 0.33, “B only” is 0.18, the intersection is 0.12, and outside is 0.37. Always label the diagram and check that probabilities sum to 1.
韦恩图包含两个重叠的圆。仅 A 区域为 0.33，仅 B 区域为 0.18，交集为 0.12，外部为 0.37。务必标记图示并检查概率总和为1。
5. Permutations in Context | 排列的实际应用
Question 3: The digits 1, 2, 3, 4, 5, 6 are used to form six‑digit numbers without repetition. How many of these numbers are even? How many are greater than 400000? And how many have the digit 3 in the middle position?
第3题：用数字1, 2, 3, 4, 5, 6构成无重复的六位数。这些数中有多少个是偶数？有多少个大于400000？有多少个数字3在中间位置？
For an even number, the last digit must be even (2, 4 or 6): 3 choices. The remaining 5 positions can be arranged in 5! ways. So total even = 3 × 120 = 360. A common slip is to forget that the first digit can be any of the remaining digits, so the multiplier is still 5!.
对于偶数，末位必须是偶数（2,4或6）：3种选择。剩余5个位置以 5! 种方式排列。因此偶数总数 = 3×120 = 360。常见的错误是忘记首位可以是剩余数字中的任何一个，因此乘数仍为 5!。
For numbers greater than 400000, the first digit must be 4, 5 or 6. If the first is 4, the remaining 5 digits arrange in 5! = 120 ways. Same for 5 and 6, giving 3 × 120 = 360 numbers. For the digit 3 in the middle (the 3rd position is fixed as 3), the other 5 places organise in 5! = 120 ways.
对于大于400000的数，首位必须是4、5或6。若首位为4，其余5位按 5! = 120 排列。5和6同理，共 3×120 = 360 个数。数字3在中间（第3位固定为3）时，其余5个位置按 5! = 120 种方式排列。
6. Discrete Random Variables and Expectation | 离散随机变量与期望
Question 4: The discrete random variable Y has the following probability distribution: y = 1, 2, 3, 4; P(Y=y) = 0.1, 0.3, a, b. Given that E(Y) = 2.8, find a and b. Then calculate Var(Y) and E(3Y²).
第4题：离散随机变量Y的概率分布如下：y=1,2,3,4；P(Y=y)=0.1, 0.3, a, b。已知 E(Y)=2.8，求 a 和 b。然后计算 Var(Y) 和 E(3Y²)。
First, sum of probabilities = 1: 0.1 + 0.3 + a + b = 1 ⇒ a + b = 0.6. Using E(Y) = Σ y·p(y): 1×0.1 + 2×0.3 + 3×a + 4×b = 2.8 ⇒ 0.1 + 0.6 + 3a + 4b = 2.8 ⇒ 3a + 4b = 2.1. Solve the simultaneous equations: from a = 0.6 − b, substitute: 3(0.6 − b) + 4b = 2.1 ⇒ 1.8 − 3b + 4b = 2.1 ⇒ 1.8 + b = 2.1 ⇒ b = 0.3, a = 0.3. So distribution: 0.1, 0.3, 0.3, 0.3.
首先，概率总和为1：0.1+0.3+a+b=1 ⇒ a+b=0.6。利用 E(Y)=Σ y·p(y)：1×0.1+2×0.3+3a+4b=2.8 ⇒ 0.1+0.6+3a+4b=2.8 ⇒ 3a+4b=2.1。解联立方程：由 a=0.6−b 代入得：3(0.6−b)+4b=2.1 ⇒ 1.8−3b+4b=2.1 ⇒ b=0.3, a=0.3。因此分布为 0.1, 0.3, 0.3, 0.3。
E(Y²) = Σ y²·p(y) = 1×0.1 + 4×0.3 + 9×0.3 + 16×0.3 = 0.1 + 1.2 + 2.7 + 4.8 = 8.8. Then Var(Y) = E(Y²) − [E(Y)]² = 8.8 − (2.8)² = 8.8 − 7.84 = 0.96. E(3Y²) = 3 × E(Y²) = 26.4. Never forget that Var(Y) uses the square of the expected value, not 2×E(Y).
E(Y²)=Σ y²·p(y)=1×0.1+4×0.3+9×0.3+16×0.3=0.1+1.2+2.7+4.8=8.8。Var(Y)=E(Y²)−[E(Y)]²=8.8−7.84=0.96。E(3Y²)=3×E(Y²)=26.4。切勿忘记方差公式中使用的是期望值的平方，而不是 2×E(Y)。
7. Binomial Distribution Calculations | 二项分布与概率计算
Question 5: In a manufacturing process, 4% of bolts are defective. A random sample of 20 bolts is taken. Find the probability that exactly 2 are defective, that at most 1 is defective, and the expected number of defectives.
第5题：在一道制造工序中，4%的螺栓有缺陷。随机抽取20个螺栓，求恰好有2个缺陷的概率、至多1个缺陷的概率以及缺陷品的期望数量。
Let X ~ B(20, 0.04). P(X = k) = ²⁰Cₖ (0.04)ᵏ (0.96)²⁰⁻ᵏ. For exactly 2, k = 2: ²⁰C₂ × 0.04² × 0.96¹⁸. ²⁰C₂ = 190. Using a calculator, P(X = 2) ≈ 0.146. For at most 1, find P(X=0) + P(X=1). P(X=0) = 0.96²⁰ ≈ 0.442; P(X=1) = 20 × 0.04 × 0.96¹⁹ ≈ 0.368; sum ≈ 0.810.
设 X ~ B(20, 0.04)。P(X=k)= ²⁰Cₖ (0.04)ᵏ (0.96)²⁰⁻ᵏ。对于 k=2：²⁰C₂ × 0.04² × 0.96¹⁸，²⁰C₂=190。使用计算器，P(X=2)≈0.146。对于至多1个，求 P(X=0)+P(X=1)。P(X=0)=0.96²⁰≈0.442；P(X=1)=20×0.04×0.96¹⁹≈0.368；合计约0.810。
The expected number E(X) = np = 20 × 0.04 = 0.8. Avoid using n as the sample size for a hypergeometric scenario – here sampling is from a large population so binomial is appropriate.
期望数量 E(X) = np = 20 × 0.04 = 0.8。避免在超几何情形下使用 n 作为样本量——此处是从大总体中抽样，因此二项分布适用。
8. Geometric Distribution and Waiting Time | 几何分布与等待时间
Question 6a: In a game, the probability of rolling a six on a fair die is 1/6. The die is rolled repeatedly until a six appears. Find the probability that the first six occurs on the 4th roll, and calculate the expected number of rolls needed.
第6a题：在一游戏中，公平骰子掷出六点的概率为 1/6。反复掷骰子直到出现一个六点。求首次六点出现在第4次掷骰子的概率，并计算所需的期望掷骰子次数。
Let G ~ Geo(1/6). Probability first success on trial 4: P(G = 4) = (5/6)³ × (1/6) = (125/216)×(1/6) = 125/1296 ≈ 0.0965. The expected number of trials is 1/p = 6. Do not misapply the binomial ‘n’ here; geometric distribution deals with the number of trials up to the first success.
设 G ~ Geo(1/6)。首次成功在第4次试验的概率：P(G=4) = (5/6)³ × 1/6 = 125/1296 ≈ 0.0965。期望试验次数为 1/p = 6。不要误用二项分布的 ‘n’；几何分布处理的是直到首次成功的试验次数。
9. Normal Distribution and Standardisation | 正态分布与标准化
Question 6b: The mass of jam in a jar follows a normal distribution with mean 340 g and standard deviation 8 g. Find the probability that a randomly chosen jar contains less than 330 g. A jar is rejected if its mass is more than 2 standard deviations away from the mean. What proportion is acceptable?
第6b题：一罐果酱的质量服从正态分布，均值为340 g，标准差为8 g。求随机选取的一罐果酱质量低于330 g 的概率。如果一罐果酱的质量距离均值超过2个标准差，则被拒收。可接受的比例是多少？
Let M ~ N(340, 8²). For P(M < 330), standardise: Z = (330 − 340)/8 = −1.25. Use Φ(−1.25) = 1 − Φ(1.25) ≈ 1 − 0.8944 = 0.1056. The probability is about 0.106.
设 M
Published by TutorHao | Year 12 统计 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply