📚 Year 11 Cambridge Statistics: Unit Test Mock Paper Analysis | 剑桥Year 11统计：单元测试模拟卷解析
This article provides a detailed walkthrough of a Year 11 Cambridge Statistics unit test mock paper. Each question is broken down with step-by-step solutions, highlighting key statistical concepts and common pitfalls. The goal is to deepen your understanding of descriptive statistics, probability, distributions, and data interpretation in preparation for the actual exam.
本文详细解析了一套剑桥Year 11统计单元测试模拟卷。每道题目均配有分步解答，突出关键统计概念和常见易错点，旨在加深你对描述性统计、概率、分布及数据解释的理解，为正式考试做好充分准备。
1. Question 1: Mean, Median, Mode and Range | 第1题：均值、中位数、众数和极差
The following data set shows the number of hours spent on revision by 12 students in one week: 3, 5, 2, 4, 6, 5, 7, 4, 5, 6, 8, 3. Calculate the mean, median, mode, range, and interquartile range (IQR).
以下数据集显示了12名学生一周内用于复习的小时数：3, 5, 2, 4, 6, 5, 7, 4, 5, 6, 8, 3。请计算均值、中位数、众数、极差和四分位距（IQR）。
Step 1: Order the data. Arrange the data in ascending order: 2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 8.
步骤1：排序。 将数据按升序排列：2, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 8。
Step 2: Mean. Sum all values: 2+3+3+4+4+5+5+5+6+6+7+8 = 58. Divide by n = 12: mean = 58 ÷ 12 ≈ 4.83 hours.
步骤2：均值。 所有数值求和：2+3+3+4+4+5+5+5+6+6+7+8 = 58。除以 n = 12：均值 = 58 ÷ 12 ≈ 4.83 小时。
Step 3: Median. With 12 values (even), the median is the average of the 6th and 7th values in the ordered list. Both are 5. Therefore, median = 5 hours.
步骤3：中位数。 有12个数值（偶数），中位数是排序后第6和第7个数值的平均值。两个都是5。因此，中位数 = 5 小时。
Step 4: Mode. The value 5 appears most frequently (three times). Mode = 5 hours.
步骤4：众数。 数值5出现的次数最多（三次）。众数 = 5 小时。
Step 5: Range. Range = maximum – minimum = 8 – 2 = 6 hours.
步骤5：极差。 极差 = 最大值 – 最小值 = 8 – 2 = 6 小时。
Step 6: Interquartile range (IQR). The lower quartile Q₁ is the median of the lower half (first 6 values): 2,3,3,4,4,5. Q₁ = (3+4)/2 = 3.5. The upper quartile Q₃ is the median of the upper half (last 6 values): 5,5,6,6,7,8. Q₃ = (6+6)/2 = 6. IQR = Q₃ – Q₁ = 6 – 3.5 = 2.5 hours.
步骤6：四分位距（IQR）。 下四分位数 Q₁ 是下半部分（前6个数值）的中位数：2,3,3,4,4,5。Q₁ = (3+4)/2 = 3.5。上四分位数 Q₃ 是上半部分（后6个数值）的中位数：5,5,6,6,7,8。Q₃ = (6+6)/2 = 6。IQR = Q₃ – Q₁ = 6 – 3.5 = 2.5 小时。
Answer Summary: Mean ≈ 4.83, Median = 5, Mode = 5, Range = 6, IQR = 2.5 (all in hours).
答案总结： 均值 ≈ 4.83，中位数 = 5，众数 = 5，极差 = 6，IQR = 2.5（单位均为小时）。
2. Question 2: Grouped Frequency – Estimated Mean and Histogram | 第2题：分组频数——估计均值和直方图
The table below shows the distances travelled (in km) by 50 delivery vans in one day. Estimate the mean distance and construct a histogram to represent the distribution.
下表显示了50辆送货车一天内的行驶距离（公里）。估计平均距离，并绘制直方图表示该分布。
| Distance (km) | 0–10 | 10–20 | 20–30 | 30–40 | 40–60 |
| Frequency | 8 | 14 | 12 | 10 | 6 |
Step 1: Find midpoints. For each class interval, the midpoint is (lower boundary + upper boundary)/2. For 0–10: 5, for 10–20: 15, for 20–30: 25, for 30–40: 35, for 40–60: 50.
步骤1：求组中值。 每个区间的中点 = (下限 + 上限) / 2。0–10: 5，10–20: 15，20–30: 25，30–40: 35，40–60: 50。
Step 2: Multiply each midpoint by its frequency. (5×8) + (15×14) + (25×12) + (35×10) + (50×6) = 40 + 210 + 300 + 350 + 300 = 1200.
步骤2：各中点乘以对应的频数。 (5×8) + (15×14) + (25×12) + (35×10) + (50×6) = 40 + 210 + 300 + 350 + 300 = 1200。
Step 3: Estimated mean. Mean estimate = total (Σf·x) ÷ Σf = 1200 ÷ 50 = 24 km.
步骤3：估计均值。 估计均值 = 总和 (Σf·x) ÷ Σf = 1200 ÷ 50 = 24 公里。
Step 4: Histogram construction. Since the class intervals are not all uniform (the last class has width 20), we must use frequency density = frequency ÷ class width. For 0–10: f.d. = 8/10 = 0.8; 10–20: 14/10 = 1.4; 20–30: 12/10 = 1.2; 30–40: 10/10 = 1.0; 40–60: 6/20 = 0.3. Draw bars with these heights on the appropriate class width. The horizontal axis is distance (km), vertical axis is frequency density.
步骤4：绘制直方图。 由于组距不完全一致（最后一组宽度为20），必须使用频数密度 = 频数 ÷ 组距。0–10: 频数密度 = 8/10 = 0.8；10–20: 14/10 = 1.4；20–30: 12/10 = 1.2；30–40: 10/10 = 1.0；40–60: 6/20 = 0.3。以这些高度在相应的组距上绘制矩形。横轴为距离（公里），纵轴为频数密度。
3. Question 3: Cumulative Frequency and Quartiles | 第3题：累积频数与四分位数
Using the data from Question 2, construct a cumulative frequency table, then draw a cumulative frequency curve. From the graph, estimate the median, the lower quartile, the upper quartile, and the number of vans that travelled less than 35 km.
使用第2题的数据，构建累积频数表，然后画出累积频数曲线。由图像估计中位数、下四分位数、上四分位数，以及行驶距离少于35公里的车辆数。
Step 1: Cumulative frequency table. Add up frequencies successively.
步骤1：累积频数表。 依次累加频数。
| Distance ≤ (upper boundary) | 10 | 20 | 30 | 40 | 60 |
| Cumulative frequency | 8 | 22 | 34 | 44 | 50 |
Step 2: Plot points at upper class boundaries (10,8), (20,22), (30,34), (40,44), (60,50). Join with a smooth curve starting from (0,0).
步骤2：描点 在组上限处：(10,8), (20,22), (30,34), (40,44), (60,50)。用光滑曲线连接，并从 (0,0) 开始。
Step 3: Estimate median. Half of total frequency is 25. Draw a horizontal line from 25 to the curve, then read down to the distance axis. Approximate median ≈ 22 km.
步骤3：估计中位数。 总频数的一半为25。从25画水平线与曲线相交，向下读出距离值。估计中位数约 22 公里。
Step 4: Lower quartile (Q₁). ¼ of 50 = 12.5. From 12.5 on the cumulative frequency axis, read across → curve → down. Q₁ ≈ 15 km.
步骤4：下四分位数（Q₁）。 50 的 ¼ = 12.5。从累积频数轴 12.5 处画水平线交曲线，向下得 Q₁ ≈ 15 公里。
Step 5: Upper quartile (Q₃). ¾ of 50 = 37.5. Read across from 37.5 gives Q₃ ≈ 33 km.
步骤5：上四分位数（Q₃）。 50 的 ¾ = 37.5。从37.5画线得 Q₃ ≈ 33 公里。
Step 6: Number of vans less than 35 km. From 35 km on the horizontal axis, go up to the curve, then across to cumulative frequency axis: approximately 40 vans.
步骤6：少于35公里的车辆数。 从横轴35公里处向上交曲线，再读累积频数：大约40辆。
4. Question 4: Probability – Two Dice | 第4题：概率——两枚骰子
Two fair six-sided dice are thrown. Find the probability that: (a) the sum of the scores is 7; (b) at least one die shows a 5; (c) the sum is 8 given that exactly one die shows an even number.
投掷两枚均匀六面骰子。求下列事件的概率：(a) 点数之和为7；(b) 至少有一枚骰子显示为5；(c) 在恰好一枚骰子显示为偶数的条件下，点数之和为8。
Step 1: Sample space. Total outcomes = 6 × 6 = 36.
步骤1：样本空间。 总结果数 = 6 × 6 = 36。
(a) Sum of 7. Favorable pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes. P(sum=7) = 6/36 = 1/6.
(a) 和为7。 有利结果：(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6种。P(和为7) = 6/36 = 1/6。
(b) At least one 5. Use complement: probability of no 5s = (5/6)×(5/6) = 25/36. Thus, P(at least one 5) = 1 – 25/36 = 11/36.
(b) 至少一个5。 用补集：无5的概率 = (5/6)×(5/6) = 25/36。因此 P(至少一个5) = 1 – 25/36 = 11/36。
(c) Sum is 8 given exactly one even. First, identify outcomes with exactly one even number. An outcome is (odd, even) or (even, odd). There are 3 odds (1,3,5) and 3 evens (2,4,6), so number = 3×3 + 3×3 = 18. Now, sum = 8 within this reduced sample space. List all pairs with exactly one even and sum 8: (2,6) even-even – not allowed, (3,5) odd-odd sum 8 – no even; (4,4) even-even; (5,3) odd-odd; (6,2) even-even. So no pair where exactly one is even and sum 8. Check systematically: possible sum 8 pairs: (2,6), (3,5), (4,4), (5,3), (6,2). All have either both even or both odd. Hence, probability = 0/18 = 0.
(c) 在恰好一枚偶数时和为8。 首先，找出恰好一枚偶数的结果。(奇,偶) 或 (偶,奇)。奇数有3个，偶数有3个，所以数量 = 3×3 + 3×3 = 18。现在在缩小的样本空间中求和为8。列出恰好一枚偶数且和为8的组合：检查(2,6)均偶； (3,5)均奇； (4,4)均偶； (5,3)均奇； (6,2)均偶。没有恰好一枚偶数的情形。因此概率 = 0。
5. Question 5: Probability Tree – Without Replacement | 第5题：概率树图——不放回抽取
A bag contains 5 red beads and 3 blue beads. Two beads are drawn at random without replacement. Using a tree diagram, find the probability that: (a) both beads are red; (b) exactly one bead is blue; (c) the second bead is red given that the first was blue.
一个袋子里有5颗红珠子和3颗蓝珠子。随机不放回地抽取两颗珠子。用树状图求以下概率：(a) 两颗都是红色；(b) 恰好一颗是蓝色；(c) 已知第一颗是蓝色，第二颗是红色的概率。
Step 1: Tree structure. First draw: P(R) = 5/8, P(B) = 3/8. Second draw (without replacement). If first R: remaining 4R, 3B, so P(R|R) = 4/7, P(B|R) = 3/7. If first B: remaining 5R, 2B, so P(R|B) = 5/7, P(B|B) = 2/7.
步骤1：树状结构。 第一次抽取：P(红) = 5/8, P(蓝) = 3/8。第二次抽取（不放回）。若第一次红：剩余4红3蓝，P(红|红) = 4/7, P(蓝|红) = 3/7。若第一次蓝：剩余5红2蓝，P(红|蓝) = 5/7, P(蓝|蓝) = 2/7。
(a) Both red: (5/8) × (4/7) = 20/56 = 5/14.
(a) 两颗皆红： (5/8) × (4/7) = 20/56 = 5/14。
(b) Exactly one blue: Two paths: (R then B) + (B then R). P(RB) = (5/8)×(3/7) = 15/56; P(BR) = (3/8)×(5/7) = 15/56. Total = 30/56 = 15/28.
(b) 恰好一颗蓝： 两条路径：(红然后蓝) + (蓝然后红)。P(红蓝) = 15/56；P(蓝红) = 15/56。合计 = 30/56 = 15/28。
(c) Second red given first blue: From the tree, P(second R | first B) = 5/7. No further calculation needed.
(c) 第一颗蓝条件下第二颗红： 由树图，P(第二红 | 第一蓝) = 5/7。无需额外计算。
6. Question 6: Permutations and Combinations | 第6题：排列与组合
A committee of 4 students is to be selected from 7 boys and 5 girls. Find the number of ways: (a) the committee can be formed if there is no restriction; (b) the committee contains exactly 2 girls; (c) the 4 students are arranged in a row for a photograph, with the two girls standing together.
要从7名男生和5名女生中选出一个4人委员会。求以下情况的方法数：(a) 无任何限制时组成委员会的方法数；(b) 委员会中恰好有2名女生；(c) 4名学生排成一排拍照，且两名女生必须站在一起。
(a) No restriction: Choose any 4 from 12 students. Number of combinations = ₁₂C₄ = 495.
(a) 无限制： 从12名学生中任选4人。组合数 = ₁₂C₄ = 495。
(b) Exactly 2 girls: Choose 2 girls from 5: ₅C₂ = 10; choose 2 boys from 7: ₇C₂ = 21. Multiply: 10 × 21 = 210 ways.
(b) 恰好2名女生： 从5名女生中选2名：₅C₂ = 10；从7名男生中选2名：₇C₂ = 21。相乘：10 × 21 = 210 种。
(c) Arrangement with girls together: First, form a committee of 2 girls and 2 boys (210 ways as above). Treat the two girls as one “block”, so we have 3 items to arrange (block of girls + 2 boys). These 3 items can be arranged in 3! = 6 ways. Inside the block, the two girls can be arranged in 2! = 2 ways. Total arrangements = 210 × 6 × 2 = 2520.
(c) 女生站在一起的排列： 首先，组成有2女2男的委员会（如上，210种）。把两名女生视为一个整体，这样我们有3个物体（女生整体+2名男生）。3个物体排列有3! = 6种方式。整体内部两名女生可排列为2! = 2种。总排列数 = 210 × 6 × 2 = 2520。
7. Question 7: Binomial Distribution | 第7题：二项分布
A biased coin lands heads with probability 0.3. It is tossed 8 times. Find the probability that: (a) exactly 3 heads appear; (b) at most 2 heads appear. Use the binomial formula.
一枚不均匀硬币正面朝上的概率为0.3。抛掷该硬币8次。求以下概率：(a) 恰好出现3次正面；(b) 至多出现2次正面。使用二项公式。
(a) Exactly 3 heads: X ~ B(8, 0.3). P(X=3) = ₈C₃ × (0.3)³ × (0.7)⁵. ₈C₃ = 56, (0.3)³ = 0.027, (0.7)⁵ ≈ 0.16807. Multiply: 56 × 0.027 × 0.16807 ≈ 0.2541 (4 d.p.).
(a) 恰好3次正面： X ~ B(8, 0.3)。P(X=3) = ₈C₃ × (0.3)³ × (0.7)⁵。₈C₃ = 56，(0.3)³ = 0.027，(0.7)⁵ ≈ 0.16807。乘积 ≈ 0.2541（4位小数）。
(b) At most 2 heads: P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2). P(X=0) = (0.7)⁸ ≈ 0.057648. P(X=1) = ₈C₁ × 0.3 × (0.7)⁷ = 8 × 0.3 × 0.0823543 ≈ 0.19765. P(X=2) = ₈C₂ × (0.3)² × (0.7)⁶ = 28 × 0.09 × 0.117649 ≈ 0.29648. Sum ≈ 0.55178 ≈ 0.5518.
(b) 至多2次正面： P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)。P(X=0) = (0.7)⁸ ≈ 0.057648。P(X=1) = ₈C₁ × 0.3 × (0.7)⁷ ≈ 0.19765。P(X=2) = ₈C₂ × (0.3)² × (0.7)⁶ ≈ 0.29648。总和 ≈ 0.5518。
8. Question 8: Normal Distribution – Standardisation | 第8题：正态分布——标准化
The lengths of a batch of metal rods are normally distributed with mean μ = 25.0 cm and standard deviation σ = 0.4 cm. Find the proportion of rods that have length: (a) less than 24.6 cm; (b) between 24.8 cm and 25.5 cm; (c) the length exceeded by the longest 10% of rods. Use the standard normal table.
一批金属棒的长度服从正态分布，均值 μ = 25.0 cm，标准差 σ = 0.4 cm。求以下比例的金属棒：(a) 长度小于24.6 cm；(b) 长度介于24.8 cm和25.5 cm之间；(c) 长度超过最长的10%的临界值。使用标准正态表。
(a) Less than 24.6 cm: Standardise: z = (24.6 – 25.0) ÷ 0.4 = –1.0. P(Z < –1.0) = 1 – Φ(1.0) ≈ 1 – 0.8413 = 0.1587.
(a) 小于24.6 cm： 标准化：z = (24.6 – 25.0) ÷ 0.4 = –1.0。P(Z < –1.0) = 1 – Φ(1.0) ≈ 0.1587。
(b) Between 24.8 and 25.5: z₁ = (24.8 – 25.0)/0.4 = –0.5; z₂ = (25.5 – 25.0)/0.4 = 1.25. P(–0.5 < Z < 1.25) = Φ(1.25) – Φ(–0.5) = 0.8944 – (1 – 0.6915) = 0.8944 – 0.3085 = 0.5859.
(b) 在24.8和25.5之间： z₁ = –0.5, z₂ = 1.25。P(–0.5 < Z < 1.25) = Φ(1.25) – [1 – Φ(0.5)] = 0.8944 – 0.3085 =
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