Edexcel Year 13 Statistics: Unit Test Mock Paper Walkthrough | Edexcel 高三统计单元测试模拟卷解析

📚 Edexcel Year 13 Statistics: Unit Test Mock Paper Walkthrough | Edexcel 高三统计单元测试模拟卷解析

This walkthrough breaks down a representative mock paper for the Year 13 Edexcel Statistics component (Paper 3: Statistics & Mechanics). The mock covers the core A2 topics: normal distributions, sampling distributions, normal approximations, hypothesis testing for means, regression, correlation, and hypothesis tests on correlation. By working through these model solutions, you will reinforce the key techniques and examiner expectations, building confidence for the real examination.

本解析逐题拆解一份具有代表性的 Edexcel 高三统计模拟卷（试卷三：统计与力学）。模拟卷涵盖 A2 核心主题：正态分布、抽样分布、正态近似、均值假设检验、回归、相关以及相关性的假设检验。通过研读这些范例解答，你将巩固关键技巧和考官所期望的答题规范，为实考积累信心。

1. Mock Paper Overview | 模拟卷概览

The mock paper is structured like a real Edexcel Statistics paper, blending short-answer questions with multi-step contextual problems. Time management is crucial: allocate about 1 minute per mark. Questions often build from simple probability calculations to hypothesis testing and interpretation, so reading the full scenario before jumping in helps avoid missing details.

本模拟卷的编排贴近真实的 Edexcel 统计试卷，将简答题与多步骤情境题相结合。时间管理至关重要：建议按每分钟一分的节奏分配答题时间。题目经常从简单的概率计算逐步过渡到假设检验与解读，因此在动笔前通读完整的情景有助于避免遗漏细节。

2. Working with the Normal Distribution | 正态分布计算

To find a probability such as P(X > 55) when X ~ N(50, 4²), first standardise: Z = (55 – 50) / 4 = 1.25. The standard normal table gives P(Z < 1.25) = 0.8944, so P(X > 55) = 1 – 0.8944 = 0.1056. Always sketch the normal curve to visualise the tail area.

若要计算 P(X > 55)，其中 X ~ N(50, 4²)，首先标准化：Z = (55 – 50) / 4 = 1.25。标准正态表给出 P(Z < 1.25) = 0.8944，因此 P(X > 55) = 1 – 0.8944 = 0.1056。务必画一条正态曲线来直观显示尾部面积。

Z = (X – μ) / σ

When a percentile is given, e.g., find x such that P(X < x) = 0.9, use the inverse normal function. The standard normal quantile is z = Φ⁻¹(0.9) ≈ 1.2816. Then x = μ + z σ = 50 + 1.2816 × 4 = 55.13. On a calculator, this is accessed via the Inverse Normal menu with area 0.9, μ = 50, σ = 4.

当给出分位数，例如求 x 使 P(X < x) = 0.9，使用逆正态函数。标准正态分位数为 z = Φ⁻¹(0.9) ≈ 1.2816，则 x = μ + z σ = 50 + 1.2816 × 4 = 55.13。在计算器上，通过逆正态菜单输入面积 0.9、μ = 50、σ = 4 即可求得。

3. Inverse Normal for Unknown Parameters | 逆正态求未知参数

Some exam questions give two probability conditions while both μ and σ are unknown. For instance, P(X < 15) = 0.2 and P(X > 35) = 0.1. Set up standardised equations: (15 – μ)/σ = Φ⁻¹(0.2) ≈ -0.8416 and (35 – μ)/σ = Φ⁻¹(0.9) ≈ 1.2816 (since P(X > 35)=0.1 implies P(X < 35)=0.9). Solving the two simultaneous equations yields μ and σ.

某些考题会在 μ 和 σ 均未知时给出两个概率条件。例如 P(X < 15) = 0.2 且 P(X > 35) = 0.1。建立标准化方程：(15 – μ)/σ = Φ⁻¹(0.2) ≈ -0.8416，(35 – μ)/σ = Φ⁻¹(0.9) ≈ 1.2816（因为 P(X > 35)=0.1 意味着 P(X < 35)=0.9）。联立这两个方程即可解出 μ 和 σ。

Subtract the first from the second: (35 – μ) – (15 – μ) = σ(1.2816 – (-0.8416)) ⇒ 20 = σ × 2.1232 ⇒ σ ≈ 9.42. Substitute back into the first equation: (15 – μ) = -0.8416 × 9.42, giving μ ≈ 22.93. Always check that the resulting probabilities are consistent.

用第二个方程减去第一个：(35 – μ) – (15 – μ) = σ(1.2816 – (-0.8416)) ⇒ 20 = σ × 2.1232 ⇒ σ ≈ 9.42。代回第一个方程，(15 – μ) = -0.8416 × 9.42，得 μ ≈ 22.93。最后务必检查所得概率是否一致。

4. Sample Means and the Central Limit Theorem | 样本均值与中心极限定理

If a random sample of size n is taken from a population with mean μ and variance σ², the sample mean X̄ has mean μ and variance σ²/n. For a normal population, X̄ is exactly normal; for non-normal populations with large n (typically n ≥ 30), the Central Limit Theorem ensures X̄ is approximately normal.

若从均值为 μ、方差为 σ² 的总体中抽取容量为 n 的随机样本，样本均值 X̄ 的均值为 μ，方差为 σ²/n。若总体服从正态分布，X̄ 也精确服从正态分布；对于非正态总体且 n 较大（一般 n ≥ 30），中心极限定理保证 X̄ 近似服从正态分布。

Example: The weight of apples has μ = 150 g and σ = 20 g. For a random sample of 25 apples, X̄ ~ N(150, 20²/25) i.e. N(150, 16). The standard error is σ/√n = 4 g. To find P(X̄ > 155), compute Z = (155 – 150)/4 = 1.25, giving a probability of about 0.1056.

例如：苹果重量 μ = 150 g, σ = 20 g，随机抽取 25 个苹果，则 X̄ ~ N(150, 20²/25) 即 N(150, 16)。标准误为 σ/√n = 4 g。计算 P(X̄ > 155)，Z = (155 – 150)/4 = 1.25，概率约为 0.1056。

5. Normal Approximation to the Binomial | 二项分布的正态近似

When X ~ B(n, p) and both np and nq are > 5 (some texts use 10), the distribution of X can be approximated by Y ~ N(np, npq). The continuity correction adjusts for the discrete nature: P(X ≥ r) ≈ P(Y > r – 0.5), P(X ≤ r) ≈ P(Y < r + 0.5), and P(X = r) ≈ P(r - 0.5 < Y < r + 0.5).

当 X ~ B(n, p) 且 np 与 nq 均大于 5（有些教材用 10），X 的分布可由 Y ~ N(np, npq) 近似。连续性校正针对离散特性进行调整：P(X ≥ r) ≈ P(Y > r – 0.5)，P(X ≤ r) ≈ P(Y < r + 0.5)，而 P(X = r) ≈ P(r - 0.5 < Y < r + 0.5)。

Suppose X ~ B(100, 0.35). Then np = 35, npq = 22.75, σ ≈ 4.77. To find P(X ≥ 40), apply the continuity correction: P(X ≥ 40) ≈ P(Y > 39.5). Z = (39.5 – 35)/4.77 ≈ 0.944, giving a tail probability of about 0.1726. Without the correction, the result would be noticeably less accurate.

假设 X ~ B(100, 0.35)，np = 35, npq = 22.75，σ ≈ 4.77。求 P(X ≥ 40)，使用连续性校正：P(X ≥ 40) ≈ P(Y > 39.5)。Z = (39.5 – 35)/4.77 ≈ 0.944，尾部概率约为 0.1726。若不进行校正，结果的准确性会明显下降。

6. Hypothesis Test for a Population Mean | 总体均值的假设检验

When the population standard deviation σ is known, the test statistic for the mean is Z = (X

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