📚 Year 12 Cambridge Statistics: Unit Test Mock Exam Analysis | 剑桥12年级统计：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for the Year 12 Cambridge Probability & Statistics 1 course. Each question is carefully analysed with step‑by‑step solutions in both English and Chinese, helping you master essential statistical concepts and examination techniques.
本文详细解析了一套针对剑桥12年级概率与统计1课程的模拟单元测试卷。每道题目均配有逐步的中英双语解答，助你掌握核心统计概念与应试技巧。
1. Descriptive Statistics: Mean and Standard Deviation | 描述统计：均值与标准差
Question: The following data set represents the test scores of 8 students: 12, 15, 18, 20, 22, 25, 28, 30. Calculate the mean, variance and standard deviation of these scores.
题目：以下数据集表示8名学生的测验成绩：12, 15, 18, 20, 22, 25, 28, 30。计算这些成绩的均值、方差和标准差。
Step 1: Find the sum of all values. Sum = 12 + 15 + 18 + 20 + 22 + 25 + 28 + 30 = 170.
步骤1：求所有值的总和。总和 = 12 + 15 + 18 + 20 + 22 + 25 + 28 + 30 = 170。
Step 2: Compute the mean (μ). Mean = 170 ÷ 8 = 21.25.
步骤2：计算均值（μ）。均值 = 170 ÷ 8 = 21.25。
Step 3: Calculate the sum of squared deviations. Subtract the mean from each score, square the result, and sum these squares: (12 − 21.25)² + (15 − 21.25)² + (18 − 21.25)² + (20 − 21.25)² + (22 − 21.25)² + (25 − 21.25)² + (28 − 21.25)² + (30 − 21.25)² = 85.5625 + 39.0625 + 10.5625 + 1.5625 + 0.5625 + 14.0625 + 45.5625 + 76.5625 = 273.5.
步骤3：计算偏差平方和。将每个成绩减去均值，平方，然后把这些平方值相加：(12 − 21.25)² + (15 − 21.25)² + (18 − 21.25)² + (20 − 21.25)² + (22 − 21.25)² + (25 − 21.25)² + (28 − 21.25)² + (30 − 21.25)² = 85.5625 + 39.0625 + 10.5625 + 1.5625 + 0.5625 + 14.0625 + 45.5625 + 76.5625 = 273.5。
Step 4: Compute the population variance (σ²). Variance = 273.5 ÷ 8 = 34.1875.
步骤4：计算总体方差（σ²）。方差 = 273.5 ÷ 8 = 34.1875。
Step 5: Compute the standard deviation (σ). Standard deviation = √34.1875 ≈ 5.847 (3 d.p.)
步骤5：计算标准差（σ）。标准差 = √34.1875 ≈ 5.847（保留三位小数）。
2. Stem-and-Leaf Diagram and Box Plot | 茎叶图与箱线图
Question: The lengths (in mm) of 14 leaves are recorded: 45, 48, 51, 52, 55, 57, 60, 62, 65, 70, 73, 75, 80, 85. (a) Construct an ordered stem‑and‑leaf diagram using a key of 4|5 = 45 mm. (b) Determine the median, lower quartile, upper quartile and interquartile range. (c) Draw a box‑and‑whisker plot and identify any outliers using the 1.5 × IQR rule.
题目：记录了14片叶子的长度（单位 mm）：45, 48, 51, 52, 55, 57, 60, 62, 65, 70, 73, 75, 80, 85。(a) 构建一个排序的茎叶图，键为 4|5 = 45 mm。(b) 求中位数、下四分位数、上四分位数和四分位距。(c) 画出箱线图，并使用 1.5 × IQR 规则识别异常值。
(a) Constructing the stem‑and‑leaf diagram. Use the tens digit as the stem and the units digit as the leaf. Order the leaves in ascending order for each stem.
(a) 构建茎叶图。以十位数为茎，个位数为叶。按升序排列每根茎的叶。
| Stem (茎) | Leaf (叶) |
| 4 | 5 8 |
| 5 | 1 2 5 7 |
| 6 | 0 2 5 |
| 7 | 0 3 5 |
| 8 | 0 5 |
Key: 4|5 means 45 mm (键：4|5 表示 45 mm)
(b) Finding quartiles. With n = 14, the median is the average of the 7th and 8th values: (60 + 62)/2 = 61. The lower quartile Q₁ is the median of the first 7 values: the 4th value = 52. The upper quartile Q₃ is the median of the last 7 values: the 11th value = 73. IQR = 73 − 52 = 21.
(b) 求四分位数。n = 14，中位数为第7与第8个值的平均：(60 + 62)/2 = 61。下四分位数 Q₁ 为前7个值的中位数：第4个值 = 52。上四分位数 Q₃ 为后7个值的中位数：第11个值 = 73。四分位距 IQR = 73 − 52 = 21。
(c) Outlier check. Lower fence = Q₁ − 1.5 × IQR = 52 − 31.5 = 20.5. Upper fence = Q₃ + 1.5 × IQR = 73 + 31.5 = 104.5. Since all values lie between 20.5 and 104.5, there are no outliers. The box‑and‑whisker plot uses the minimum (45), Q₁ (52), median (61), Q₃ (73) and maximum (85).
(c) 异常值检查。下界限 = Q₁ − 1.5 × IQR = 52 − 31.5 = 20.5。上界限 = Q₃ + 1.5 × IQR = 73 + 31.5 = 104.5。所有值均在 20.5 和 104.5 之间，因此没有异常值。箱线图的五数概括为：最小值 (45)，Q₁ (52)，中位数 (61)，Q₃ (73)，最大值 (85)。
3. Probability and Venn Diagrams | 概率与韦恩图
Question: Two events A and B satisfy P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Find (a) P(A ∪ B), (b) P(A’ ∩ B’), (c) P(A | B).
题目：两事件 A 和 B 满足 P(A) = 0.4, P(B) = 0.5，且 P(A ∩ B) = 0.2。求 (a) P(A ∪ B), (b) P(A’ ∩ B’), (c) P(A | B)。
(a) Union probability. Using the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.4 + 0.5 − 0.2 = 0.7.
(a) 并集概率。利用加法公式：P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.4 + 0.5 − 0.2 = 0.7。
(b) Complement intersection. By De Morgan’s law, P(A’ ∩ B’) = P((A ∪ B)’) = 1 − P(A ∪ B) = 1 − 0.7 = 0.3.
(b) 补集交集。由德摩根律，P(A’ ∩ B’) = P((A ∪ B)’) = 1 − P(A ∪ B) = 1 − 0.7 = 0.3。
(c) Conditional probability. P(A | B) = P(A ∩ B) / P(B) = 0.2 / 0.5 = 0.4. Notice that this equals P(A), so A and B are independent in this case.
(c) 条件概率。P(A | B) = P(A ∩ B) / P(B) = 0.2 / 0.5 = 0.4。注意该值等于 P(A)，因此 A 与 B 在此情况下独立。
4. Permutations and Combinations | 排列与组合
Question: Determine the number of distinct arrangements of the letters in the word “STATISTICS”.
题目：求单词 “STATISTICS” 中字母的不同排列总数。
Count the letters: S appears 3 times, T appears 3 times, I appears 2 times, A appears once, C appears once. Total letters = 10.
统计字母：S 出现 3 次，T 出现 3 次，I 出现 2 次，A 出现 1 次，C 出现 1 次。总字母数 = 10。
Apply the formula for permutations with repeated items. Number of arrangements = 10! / (3! × 3! × 2! × 1! × 1!) = 10! / (6 × 6 × 2) = 3628800 / 72 = 50400.
应用重复项排列公式。排列数 = 10! / (3! × 3! × 2! × 1! × 1!) = 10! / (6 × 6 × 2) = 3628800 / 72 = 50400。
Alternative method: First choose 3 positions for S out of 10 (C(10,3)), then 3 positions for T from remaining 7 (C(7,3)), then 2 for I (C(4,2)), 1 for A (C(2,1)), 1 for C (C(1,1)). Multiply: 120 × 35 × 6 × 2 × 1 = 50400.
另法：先从10个位置选3个给 S（C(10,3)），再从余下7个选3个给 T（C(7,3)），然后从4个中选2个给 I（C(4,2)），从2个中选1个给 A（C(2,1)），最后C(1,1)给 C。相乘：120 × 35 × 6 × 2 × 1 = 50400。
5. Discrete Random Variables | 离散随机变量
Question: The probability distribution of a discrete random variable X is given below.
| x | 1 | 2 | 3 | 4 |
| P(X=x) | 0.2 | a | 0.3 | b |
Given that E(X) = 2.6, find the values of a and b. Hence compute Var(X).
题目：离散随机变量 X 的概率分布如下表所示。已知 E(X) = 2.6，求 a 和 b 的值，并计算 Var(X)。
Step 1: Use sum of probabilities. 0.2 + a + 0.3 + b = 1 → a + b = 0.5. (Equation 1)
步骤1：利用概率和为1。0.2 + a + 0.3 + b = 1 → a + b = 0.5。（方程1）
Step 2: Use expectation. E(X) = 1×0.2 + 2×a + 3×0.3 + 4×b = 0.2 + 2a + 0.9 + 4b = 1.1 + 2a + 4b = 2.6 → 2a + 4b = 1.5 → a + 2b = 0.75. (Equation 2)
步骤2：利用期望。E(X) = 1×0.2 + 2×a + 3×0.3 + 4×b = 0.2 + 2a + 0.9 + 4b = 1.1 + 2a + 4b = 2.6 → 2a + 4b = 1.5 → a + 2b = 0.75。（方程2）
Step 3: Solve the system. Subtract (1) from (2): (a + 2b) − (a + b) = 0.75 − 0.5 → b = 0.25. Then a = 0.5 − 0.25 = 0.25.
步骤3：解方程组。(2) − (1) 得：(a + 2b) − (a + b) = 0.75 − 0.5 → b = 0.25。代入得 a = 0.5 − 0.25 = 0.25。
Step 4: Compute E(X²). E(X²) = 1²×0.2 + 2²×0.25 + 3²×0.3 + 4²×0.25 = 0.2 + 1.0 + 2.7 + 4.0 = 7.9.
步骤4：计算 E(X²)。E(X²) = 1²×0.2 + 2²×0.25 + 3²×0.3 + 4²×0.25 = 0.2 + 1.0 + 2.7 + 4.0 = 7.9。
Step 5: Compute Var(X). Var(X) = E(X²) − [E(X)]² = 7.9 − (2.6)² = 7.9 − 6.76 = 1.14.
步骤5：计算 Var(X)。Var(X) = E(X²) − [E(X)]² = 7.9 − (2.6)² = 7.9 − 6.76 = 1.14。
6. Binomial Distribution | 二项分布
Question: X follows a binomial distribution where X ~ B(8, 0.3). Find (a) P(X = 3), (b) P(X < 2), (c) P(3 ≤ X ≤ 5).
题目：X 服从二项分布 X ~ B(8, 0.3)。求 (a) P(X = 3), (b) P(X < 2), (c) P(3 ≤ X ≤ 5)。
(a) P(X = 3). Use the formula P(X = r) = C(n, r) × pr × (1−p)n−r. Here, C(8,3) = 56, p = 0.3, so P(X=3) = 56 × (0.3)³ × (0.7)⁵ ≈ 56 × 0.027 × 0.16807 ≈ 0.2541 (4 d.p.).
(a) P(X = 3)。使用公式 P(X = r) = C(n, r) × pr × (1−p)n−r。此处 C(8,3) = 56，p = 0.3，故 P(X=3) = 56 × (0.3)³ × (0.7)⁵ ≈ 56 × 0.027 × 0.16807 ≈ 0.2541（保留四位小数）。
(b) P(X < 2) = P(X = 0) + P(X = 1). P(X=0) = (0.7)⁸ ≈ 0.0576. P(X=1) = 8 × 0.3 × (0.7)⁷ ≈ 8 × 0.3 × 0.08235 ≈ 0.1977. Sum ≈ 0.2553.
(b) P(X < 2) = P(X = 0) + P(X = 1)。P(X=0) = (0.7)⁸ ≈ 0.0576。P(X=1) = 8 × 0.3 × (0.7)⁷ ≈ 8 × 0.3 × 0.08235 ≈ 0.1977。故总和 ≈ 0.2553。
(c) P(3 ≤ X ≤ 5) = P(X=3) + P(X=4) + P(X=5). We already have P(X=3) ≈ 0.2541. P(X=4) = C(8,4) × (0.3)⁴ × (0.7)⁴ = 70 × 0.0081 × 0.2401 ≈ 0.1361. P(X=5) = C(8,5) × (0.3)⁵ × (0.7)³ = 56 × 0.00243 × 0.343 ≈ 0.0467. Sum ≈ 0.2541 + 0.1361 + 0.0467 = 0.4369.
(c) P(3 ≤ X ≤ 5) = P(X=3) + P(X=4) + P(X=5)。已有 P(X=3) ≈ 0.2541。P(X=4) = C(8,4) × (0.3)⁴ × (0.7)⁴ = 70 × 0.0081 × 0.2401 ≈ 0.1361。P(X=5) = C(8,5) × (0.3)⁵ × (0.7)³ = 56 × 0.00243 × 0.343 ≈ 0.0467。总和 ≈ 0.2541 + 0.1361 + 0.0467 = 0.4369。
7. Normal Distribution: Probability Calculations | 正态分布：概率计算
Question: The random variable X follows a normal distribution with mean 50 and standard deviation 4, i.e., X ~ N(50, 4²). Calculate (a) P(X < 54), (b) P(X > 45), (c) P(48 < X < 55).
题目：随机变量 X 服从正态分布，均值 50，标准差 4，即 X ~ N(50, 4²)。计算 (a) P(X < 54), (b) P(X > 45), (c) P(48 < X < 55)。
(a) Standardise: z = (54 − 50) / 4 = 1.00. P(Z < 1) = 0.8413 (from standard normal table). Thus P(X < 54) = 0.8413.
(a) 标准化：z = (54 − 50) / 4 = 1.00。查标准正态表得 P(Z < 1) = 0.8413。故 P(X < 54) = 0.8413。
(b) P(X > 45) = 1 − P(X ≤ 45). z
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