Year 12 CCEA Statistics: Mock Paper Walkthrough | 英国CCEA AS统计：模拟试卷精讲

📚 Year 12 CCEA Statistics: Mock Paper Walkthrough | 英国CCEA AS统计：模拟试卷精讲

This article provides a detailed walkthrough of a typical CCEA AS Statistics unit test mock paper. Every question tackles a core topic from the Year 12 specification, including stem-and-leaf diagrams, probability trees, normal and binomial distributions, regression, sampling, and hypothesis testing. Use this walkthrough to strengthen your exam technique and deepen your understanding of reasoning with data.

本文详细解析一份典型的 CCEA AS 统计单元测试模拟卷。每道题都覆盖 Year 12 课程的核心主题：茎叶图、概率树、正态分布、二项分布、回归、抽样与假设检验。借助本解析，你可以打磨解题技巧，并加深对数据推理的理解。

1. Stem-and-Leaf Diagram & Box Plot Analysis | 茎叶图与箱线图分析

Question: An unordered stem‑and‑leaf diagram shows the test scores of 15 students. Key: 1 | 2 means 12. Stem 1: 2 5 8; Stem 2: 1 3 4 4 6; Stem 3: 0 2 5 7; Stem 4: 1 2.

题目：一张未经排序的茎叶图给出了 15 名学生的测验成绩。茎叶图单位：1 | 2 表示 12。茎 1：2 5 8；茎 2：1 3 4 4 6；茎 3：0 2 5 7；茎 4：1 2。

First, write the ordered list: 12, 15, 18, 21, 23, 24, 24, 26, 30, 32, 35, 37, 41, 42.

首先，将所有数据按升序排列：12, 15, 18, 21, 23, 24, 24, 26, 30, 32, 35, 37, 41, 42。

Median position is (15 + 1)/2 = 8th value, so Q₂ = 26. Lower quartile Q₁ is the median of the lower seven values: the 4th value Q₁ = 21. Upper quartile Q₃ is the 12th value Q₃ = 37. IQR = Q₃ − Q₁ = 16.

中位数的位置为 (15+1)/2 = 第 8 个值，故 Q₂ = 26。下四分位数 Q₁ 是下半部分七个数的中位数，第 4 个值 Q₁ = 21。上四分位数 Q₃ 是第 12 个值 Q₃ = 37。四分位距 IQR = Q₃ − Q₁ = 16。

1.5 × IQR = 24

Outlier boundaries: lower fence = Q₁ − 1.5×IQR = 21 − 24 = −3; upper fence = Q₃ + 1.5×IQR = 37 + 24 = 61. All data lie within these fences, so there are no outliers.

离群值界限：下围栏 = Q₁ − 1.5×IQR = 21 − 24 = −3；上围栏 = Q₃ + 1.5×IQR = 37 + 24 = 61。所有数据均落在围栏之内，因此没有离群值。

For the box plot, draw a scale, mark the minimum (12), Q₁ (21), median (26), Q₃ (37), and maximum (42). Draw a box from Q₁ to Q₃ with a vertical line at the median, then add whiskers to the extremes.

绘制箱线图时，先画出数轴，标出最小值 12、Q₁ 21、中位数 26、Q₃ 37 和最大值 42。从 Q₁ 到 Q₃ 画一矩形，并在中位数处画竖线，再从矩形向两端延伸须线至最小值和最大值。

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2. Probability Tree Diagrams | 概率树图

Question: A bag contains 5 red (R) and 3 blue (B) balls. Two balls are drawn without replacement. Find P(both red), P(at least one blue), and P(second red | first red).

题目：一个袋子中有 5 个红球和 3 个蓝球。不放回地抽取两次，每次一球。计算 P(两球皆红)、P(至少一蓝) 以及 P(第二次为红 | 第一次为红)。

Draw a tree: first draw P(R) = 5/8, P(B) = 3/8. If first R, second draw P(R) = 4/7, P(B) = 3/7. If first B, second draw P(R) = 5/7, P(B) = 2/7.

画树图：第一次抽球 P(R) = 5/8，P(B) = 3/8。若第一次为 R，第二次 P(R) = 4/7，P(B) = 3/7；若第一次为 B，第二次 P(R) = 5/7，P(B) = 2/7。

P(both red) = (5/8) × (4/7) = 20/56 = 5/14.

两球全红的概率 = (5/8) × (4/7) = 20/56 = 5/14。

P(at least one blue) = 1 − P(both red) = 1 − 5/14 = 9/14. Alternatively, add the three favourable paths: RB + BR + BB.

至少一蓝的概率 = 1 − P(两红) = 1 − 5/14 = 9/14。也可将三条有利路径相加：RB + BR + BB。

P(second red | first red) = P(both red) / P(first red) = (5/14) ÷ (5/8) = (5/14) × (8/5) = 8/14 = 4/7, which matches the second‑stage probability on the red branch.

在第一次为红的条件下第二次为红的概率 = P(两红) / P(第一次红) = (5/14) ÷ (5/8) = 4/7，与树图上第二条红枝的概率一致。

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3. Normal Distribution Calculations | 正态分布计算

Question: Apple weights follow N(150, 12²). Find P(145 < X < 160) and the weight exceeded by only 10% of apples.

题目：苹果重量服从正态分布 N(150, 12²)。求 P(145 < X < 160)，以及只有 10% 的苹果超过的重量。

Standardise: Z = (X − μ) / σ. For X = 145, Z = (145 − 150) / 12 = −0.4167. For X = 160, Z = (160 − 150) / 12 = 0.8333.

标准化：Z = (X − μ)/σ。当 X = 145 时，Z = (145 − 150)/12 ≈ −0.4167；当 X = 160 时，Z = (160 − 150)/12 ≈ 0.8333。

P(145 < X < 160) = Φ(0.8333) − Φ(−0.4167)

Using standard normal tables: Φ(0.8333) ≈ 0.7978, Φ(−0.4167) = 1 − Φ(0.4167) ≈ 1 − 0.6616 = 0.3384. Required probability ≃ 0.7978 − 0.3384 = 0.4594.

查标准正态表：Φ(0.8333) ≈ 0.7978，Φ(−0.4167) = 1 − Φ(0.4167) ≈ 1 − 0.6616 = 0.3384。所求概率 ≈ 0.7978 − 0.3384 = 0.4594。

For the upper 10% tail, find z such that Φ(z) = 0.90. Table gives z ≈ 1.2816. Then x = μ + zσ = 150 + 1.2816×12 ≈ 165.4 g.

求右侧尾部 10% 对应的重量：查表使 Φ(z) = 0.90，得 z ≈ 1.2816。则 x = μ + zσ = 150 + 1.2816×12 ≈ 165.4 g。

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4. Bivariate Data & Regression Line | 双变量数据与回归直线

Question: Data on revision hours (x) and test score (y) for six students: (2,45), (3,52), (4,55), (5,60), (6,68), (7,75). Find the regression line y = a + bx, interpret slope, and predict y when x = 5.5 hours.

题目：六名学生复习小时数 (x) 与测验成绩 (y) 的数据： (2,45), (3,52), (4,55), (5,60), (6,68), (7,75)。求 y 对 x 的回归直线 y = a + bx，解释斜率，并预测复习 5.5 小时时的成绩。

Compute summaries: n = 6, Σx = 27, Σy = 355, Σxy = 1673, Σx² = 139, Σy² = 21523.

计算基本统计量：n = 6，Σx = 27，Σy = 355，Σxy = 1673，Σx² = 139，Σy² = 21523。

Sxy = Σxy − (Σx)(Σy)/n = 1673 − (27×355)/6 = 1673 − 1597.5 = 75.5

Sxx = 139 − (27²)/6 = 139 − 121.5 = 17.5

Slope b = Sxy / Sxx = 75.5 / 17.5 ≈ 4.314. Intercept a = (Σy − b Σx)/n = (355 − 4.314×27)/6 = (355 − 116.478)/6 ≈ 39.75. Regression line: y = 39.75 + 4.314x.

斜率 b = Sxy/Sxx = 75.5/17.5 ≈ 4.314。截距 a = (355 − 4.314×27)/6 ≈ 39.75。回归直线为 y = 39.75 + 4.314x。

Interpretation: For each extra hour of revision, the test score increases by approximately 4.31 marks. At x = 5.5, predicted y = 39.75 + 4.314×5.5 ≈ 63.5.

解释：每增加一小时复习时间，测验成绩平均提高约 4.31 分。当 x = 5.5 时，预测成绩 y ≈ 63.5。

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5. Binomial Distribution | 二项分布

Question: A biased coin has P(Head) = 0.3. It is tossed 10 times. Find P(exactly 4 heads), P(at most 3 heads), E(X) and Var(X).

题目：一枚偏币出现正面的概率为 0.3，投掷 10 次。求恰好出现 4 次正面的概率、至多出现 3 次正面的概率、期望值与方差。

Let X ~ B(10, 0.3). Probability mass: P(X = k) = ¹⁰Cₖ × (0.3)ᵏ × (0.7)¹⁰⁻ᵏ.

设 X ~ B(10, 0.3)，概率函数为 P(X = k) = ¹⁰Cₖ × (0.3)ᵏ × (0.7)¹⁰⁻ᵏ。

For k = 4: ¹⁰C₄ = 210. P(X = 4) = 210 × (0.3)⁴ × (0.7)⁶ = 210 × 0.0081 × 0.117649 ≈ 0.2001.

当 k = 4：¹⁰C₄ = 210，P(X = 4) = 210 × 0.0081 × 0.117649 ≈ 0.2001。

P(X ≤ 3) = P(0) + P(1) + P(2) + P(3). Using binomial tables or a cumulative formula, P(X ≤ 3) ≈ 0.6496.

P(X ≤ 3) = P(0)+P(1)+P(2)+P(3)，查二项累积表或计算可得 ≈ 0.6496。

Mean E(X) = np = 10 × 0.3 = 3. Variance Var(X) = np(1−p) = 10 × 0.3 × 0.7 = 2.1.

期望值 E(X) = np = 3。方差 Var(X) = np(1−p) = 2.1。

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6. Stratified Sampling | 分层抽样

Question: A school has 600 students: 200 in Year 12, 250 in Year 13, and 150 in Year 14. Describe how to select a stratified sample of 80 students and state one advantage.

题目：某校有 600 名学生：Year 12 200 人，Year 13 250 人，Year 14 150 人。描述如何选取一个数量为 80 的分层样本，并说明分层抽样的一个优点。

Calculate the stratum sizes proportional to population: Year 12: (200/600)×80 ≈ 26.67 → 27; Year 13: (250/600)×80 = 33.33 → 33; Year 14: (150/600)×80 = 20. The total becomes 80.

按比例计算各层样本数：Year 12 应取 (200/600)×80 ≈ 26.67，四舍五入为 27；Year 13 取 33；Year 14 取 20，合计 80 人。

Obtain a numbered list of all students in each stratum. Then use a random number generator to select the required number within each year group. This ensures every student in the population has an equal chance of being selected within their stratum.

为每层学生建立编号清单，在层内用随机数发生器抽取所需人数。这样保证了人口中每一层内的个体都有等概率被选入。

Advantage: Stratified sampling yields a more representative sample than simple random sampling, especially when the variable of interest is related to the strata. It reduces sampling error for the overall estimate.

优点：分层抽样比简单随机抽样更能保证样本的代表性，尤其是当研究变量与分层因素相关时，能减少整体估计的抽样误差。

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7. Hypothesis Test for a Proportion | 比例假设检验

Question: A company claims at least 40% of customers prefer its new product. In a survey of 30 customers, 9 preferred the product. Test at the 5% significance level.

题目：某公司声称至少有 40% 的顾客偏爱其新产品。一份 30 人的调查中只有 9 人偏爱。以 5% 显著性水平检验该声明。

State hypotheses: H₀: p = 0.4, H₁: p < 0.4 (one‑tailed). Under H₀, X ~ B(30, 0.4). Observed test statistic x = 9.

建立假设：H₀: p = 0.4，H₁: p < 0.4（左尾检验）。在 H₀ 下，检验统计量 X ~ B(30, 0.4)，观测值 x = 9。

Compute the p‑value: P(X ≤ 9 | p = 0.4). Using cumulative binomial tables or calculation, P(X ≤ 9) ≈ 0.0709.

计算 p 值：P(X ≤ 9 | p = 0.4)，查表得 ≈ 0.0709。

Since p‑value = 0.0709 > 0.05, we do not reject H₀. There is insufficient evidence to say the true proportion is less than 40%. The company’s claim cannot be rejected based on this sample.

因为 p 值 0.0709 > 0.05，不拒绝 H₀。没有充分证据表明真实偏爱比例低于 40%。基于该样本，尚不能推翻公司的声明。

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