📚 WJEC Year 12 Statistics Unit Test Mock Paper Walkthrough | WJEC 12年级统计单元测试模拟卷解析
This walkthrough takes you through a complete WJEC Year 12 Statistics unit test mock paper. We cover typical questions, step-by-step solutions, common pitfalls, and examiner tips. The mock paper is designed to reflect the style and difficulty of real WJEC assessments, blending probability, discrete and continuous distributions, sampling, and hypothesis testing. Use this resource to sharpen your technique and build confidence before your exam.
本文带领大家完整解析一份 WJEC 12年级统计单元测试模拟卷。我们将涵盖典型考题、逐步解题过程、常见错误和考官提示。这份模拟卷参照真实 WJEC 考试的风格与难度设计，融合了概率、离散与连续分布、抽样以及假设检验等内容。请利用这份资源打磨解题技巧，在考前建立信心。
1. Overview of the Mock Paper | 模拟卷概览
The mock paper contains seven questions, spanning the main topics of the WJEC AS Statistics specification. Question 1 tests basic probability and Venn diagrams. Question 2 focuses on the binomial distribution. Question 3 deals with the Poisson distribution. Question 4 explores normal distribution calculations and inverse normal. Question 5 covers the sampling distribution of the mean. Question 6 is a hypothesis test for a binomial proportion. Question 7 is a chi-squared test for independence. The total marks add up to 75, with roughly equal weighting across topics. You should aim to complete the paper in 1 hour 30 minutes.
模拟卷包含七道题目，覆盖了 WJEC AS 统计课程的主要知识点。第 1 题考查基础概率与维恩图。第 2 题聚焦二项分布。第 3 题涉及泊松分布。第 4 题探索正态分布计算与逆正态。第 5 题涵盖样本均值的抽样分布。第 6 题是关于二项比例的单尾假设检验。第 7 题是独立性卡方检验。卷面总分 75 分，各知识点权重大致均衡。你应在 1 小时 30 分钟内完成全卷。
Before diving into detailed solutions, gather your formula booklet, calculator, and statistical tables. Many errors arise from misreading the requirements for continuity corrections or using the wrong tail in hypothesis tests. We will highlight these moments as they appear.
在深入详细解答之前，请备好公式手册、计算器和统计用表。很多错误都源于误读连续性校正要求，或在假设检验中选错尾部方向。我们将在出现这些要点时加以强调。
2. Question 1: Probability and Venn Diagrams | 第1题：概率与维恩图
Question: For two events A and B, P(A) = 0.6, P(B) = 0.5, and P(A ∩ B) = 0.3. Find (a) P(A ∪ B), (b) P(A' ∩ B), (c) P(A | B). State whether A and B are independent, giving a reason.
题目：对于两个事件 A 和 B，已知 P(A) = 0.6，P(B) = 0.5，P(A ∩ B) = 0.3。求 (a) P(A ∪ B)，(b) P(A' ∩ B)，(c) P(A | B)。判断 A 与 B 是否独立，并说明理由。
Solution (a): Use the addition rule for union. P(A ∪ B) = P(A) + P(B) - P(A ∩ 0.6 + 0.5 - 0.3 = 0.8. Always check your answer does not exceed 1. Here 0.8 is plausible.
解答 (a)：使用并集加法公式。P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.6 + 0.5 - 0.3 = 0.8。务必检查结果不超过 1，此处 0.8 合理。
Solution (b): P(A' ∩ B) is the probability that B occurs but A does not. From the Venn diagram logic, P(A' ∩ B) = P(B) - P(A ∩ B) = 0.5 - 0.3 = 0.2. Alternatively, you can find P(A' ∩ B) = P(B) × (1 - P(A | B)), but the direct subtraction is faster and less prone to error.
解答 (b)：P(A' ∩ B) 表示 B 发生而 A 不发生的概率。根据维恩图逻辑，P(A' ∩ B) = P(B) - P(A ∩ B) = 0.5 - 0.3 = 0.2。也可以使用 P(B) × (1 - P(A | B)) 计算，但直接相减更快且不易出错。
Solution (c): Conditional probability P(A | B) = P(A ∩ B) / P(B) = 0.3 / 0.5 = 0.6. Notice that P(A | because the occurrence of B does not change the probability of A. Alternatively, check P(A ∩ B) = P(A) × P(B) = 0.6 × 0.5 = 0.3, which holds true. Hence the events are independent.
解答 (c)：条件概率 P(A | B) = P(A ∩ B) / P(B) = 0.3 / 0.5 = 0.6。注意到 P(A | B) = P(A) = 0.6。这立刻表明 A 与 B 独立，因为 B 的发生没有改变 A 的概率。另一种检验方式：P(A ∩ B) = P(A) × P(B) = 0.6 × 0.5 = 0.3，等式成立，因此事件独立。
3. Question 2: Binomial Distribution | 第2题：二项分布
nce of X. In a second part, four independent trials of this experiment are observed. Find the probability that exactly two of these trials result in X = 3.
题目：随机变量 X 服从二项分布 B(8, 0.35)。求 (a) P(X = 3)，(b) P(X ≥ 2)，(c) X 的均值与方差。第二部分：现对该实验进行四次独立观测，求恰好有两次观测结果满足 X = 3 的概率。
86. So P(X = 3) ≈ 0.2786. Using calculator functions or tables is recommended to reduce rounding errors.
第 (a) 部分：P(X = 3) = C(8,3) × (0.35)³ × (0.65)⁵。C(8,3) = 56。计算 (0.35)³ = 0.042875，(0.65)⁵ ≈ 0.11603。相乘得 56 × 0.042875 × 0.11603 ≈ 0.2786。因此 P(X = 3) ≈ 0.2786。建议使用计算器函数或查表以减少舍入误差。
≈ 1 - 0.1692 = 0.8308. This high probability makes sense because with n=8 and p=0.35 we expect about 2.8 successes.
第 (b) 部分：P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]。P(X = 0) = (0.65)⁸ ≈ 0.0319。P(X = 1) = C(8,1) × 0.35 × (0.65)⁷ = 8 × 0.35 × 0.0490 ≈ 0.1373。两者之和为 0.1692。故 P(X ≥ 2) ≈ 1 - 0.1692 = 0.8308。这个高概率符合预期，因为 n=8, p=0.35 下期望成功次数约为 2.8。
np = 8 × 0.35 = 2.8. Variance σ² = np(1-p) = 8 × 0.35 × 0.65 = 1.82. Standard deviation = √1.82 ≈ 1.349.
第 (c) 部分：二项分布的均值 μ = np = 8 × 0.35 = 2.8。方差 σ² = np(1-p) = 8 × 0.35 × 0.65 = 1.82。标准差 = √1.82 ≈ 1.349。
Second part: Let Y be the number of trials (out of 4) that yield X = 3. Then Y ~ 5204. Multiplying gives 6 × 0.07763 × 0.5204 ≈ 0.242. Thus the probability is about 0.242.
第二部分：设 Y 为四次独立观测中出现 X = 3 的次数，则 Y ~ B(4, 0.2786)。需求 P(Y = 2) = C(4,2) × (0.2786)² × (0.7214)²。C(4,2)=6，(0.2786)² ≈ 0.07763，(0.7214)² ≈ 0.5204。相乘得 6 × 0.07763 × 0.5204 ≈ 0.242。因此概率约为 0.242。
4. Question 3: Poisson Distribution | 第3题：泊松分布
Question: The number of flaws in a metre of cloth follows a Poisson distribution with mean 2.4. Find the probability that a randomly chosen metre contains (a) no flaws, (b) fewer than 3 flaws, (c) exactly 4 flaws, given that it contains at least 2 flaws. Also find the probability that in 5 metres there are exactly 10 flaws, assuming flaws occur independently per metre.
题目：每米布料上的瑕疵数服从均值为 2.4 的泊松分布。求随机选取的一米布料 (a) 无瑕疵的概率，(b) 少于 3 个瑕疵的概率，(c) 已知至少含 2 个瑕疵的条件下，恰好有 4 个瑕疵的概率。再求在 5 米布料中共有 10 个瑕疵的概率，假设各米之间独立。
Solution (a): Let X ~ Po(2.4). P(X = 0) = e⁻²·⁴ × 2.4⁰ / 0! = e⁻²·⁴. Using e⁻²·⁴ ≈ 0.0907. So probability of no flaws is 0.0907.
解答 (a)：设 X ~ Po(2.4)。P(X = 0) = e⁻²·⁴ × 2.4⁰ / 0! = e⁻²·⁴。e⁻²·⁴ 约为 0.0907，故无瑕疵概率为 0.0907。
)² / 2 = 0.0907 × 5.76 / 2 ≈ 0.2613. Sum = 0.0907 + 0.2177 + 0.2613 = 0.5697. About 0.570.
解答 (b)：P(X < 3) = P(X=0)+P(X=1)+P(X=2)。P(X=1) = e⁻²·⁴ × 2.4 ≈ 0.2177。P(X=2) = e⁻²·⁴ × (2.4)² / 2 ≈ 0.2613。三项之和为 0.5697，约 0.570。
Solution (c): P(X = 4 | X ≥ 2) = P(X = 4) / P(X ≥ 2). First, P(X = 4) = e⁻²·⁴ × (2.4)⁴ / 4! 1)] = 1 - (0.0907+0.2177) = 0.6916. So conditional probability = 0.1254 / 0.6916 ≈ 0.1814.
解答 (c)：P(X=4 | X≥2) = P(X=4) / P(X≥2)。首先 P(X=4) = e⁻²·⁴ × (2.4)⁴ / 4! ≈ 0.1254。P(X≥2) = 1 - (0.0907+0.2177) = 0.6916。条件概率为 0.1254 / 0.6916 ≈ 0.1814。
For 5 metres: Total flaws T ~ Po(5 × 2.4) = Po(12). P(T = 10) = e⁻¹² × 12¹⁰ / 10!. Using calculator, 12¹⁰ = 6.1917e10, 10! get P(T=10) ≈ 0.1048. The Poisson model aggregates cleanly when intervals are independent.
对于 5 米布料：总瑕疵数 T ~ Po(5 × 2.4) = Po(12)。P(T = 10) = e⁻¹² × 12¹⁰ / 10!。使用计算器，12¹⁰ = 6.1917×10¹⁰，10! = 3628800，e⁻¹² ≈ 6.144×10⁻⁶。计算得 P(T=10) ≈ 0.1048。在区间独立的条件下，泊松模型可以简单叠加。
5. Question 4: Normal Distribution and Inverse Normal | 第4题：正态分布与逆正态
Question: The masses of bags of flour are normally distributed with mean 1.50 kg and standard deviation 0.04 kg. (a) Find the probability that a randomly selected bag weighs less than 1.47 kg. (b) Determine the mass exceeded by 95% of bags. (c) A quality check selects the lightest 2.5% of bags for reprocessing. Below what mass does a bag get reprocessed?
题目：袋装面粉的质量服从正态分布，均值为 1.50 kg，标准差为 0.04 kg。(a) 求随机抽取一袋质量小于 1.47 kg 的概率。(b) 求超过 95% 袋子的质量值。(c) 质检将最轻的 2.5% 的袋子送去重新加工，低于多少质量的袋子会被重加工？
Thus P(X < 1.47) = 0.2266. Always check you use the correct tail; drawing a sketch helps.
第 (a) 部分：标准化：Z = (1.47 - 1.50) / 0.04 = -0.75。查表得 Φ(-0.75)，利用对称性，Φ(-0.75) = 1 - Φ(0.75) ≈ 1 - 0.7734 = 0.2266。所以 P(X < 1.47) = 0.2266。务必检查所用尾部是否正确，画图有助于判断。
P(X > k) = 0.95, meaning P(X < k) = 0.05. Find z such that Φ(z) = 0.05. From tables, the 5% z ≈ -1.6449. Then k = μ + zσ = 1.50 + (-1.6449)×0.04 = 1.50 - 0.0658 = 1.4342 kg. Thus 95% of bags weigh more than 1.434 kg.
第 (b) 部分：需求值 k 使得 P(X > k) = 0.95，即 P(X < k) = 0.05。查表找到 Φ(z)=0.05 对应 z ≈ -1.6449。然后 k = μ + zσ = 1.50 + (-1.6449)×0.04 = 1.4342 kg。因此 95% 的袋子质量大于 1.434 kg。
Part (c): Lightest 2.5% out 1.422 kg are reprocessed. This inverse normal technique is standard in quality control problems.
第 (c) 部分：最轻的 2.5% 对应下尾概率 0.025，z = Φ⁻¹(0.025) = -1.96。质量阈值 = 1.50 + (-1.96)×0.04 = 1.4216 kg。因此低于约 1.422 kg 的袋子会被重加工。这种逆正态方法是质量控制问题中的标准操作。
6. Question 5: Sampling Distribution of the Mean | 第5题：样本均值的抽样分布
dom sample of 16 bags is taken. (a) State the distribution of the sample mean. (b) Find the probability that the sample mean exceeds 1.51 kg. (c) Determine symmetric limits around μ that contain the sample mean with probability 0.90.
题目：沿用前述面粉总体（μ = 1.50, σ = 0.04），随机抽取 16 袋作为样本。(a) 陈述样本均值的分布。(b) 求样本均值超过 1.51 kg 的概率。(c) 求关于 μ 对称的区间，使得样本均值落在此区间的概率为 0.90。
Part (a): Since the population is normal, sample mean X̄ ~ N(μ, σ²/n) = N(1.50, 0.04²/16) = N(1.50, 0.0001). So standard error = σ/√n = 0.04/4 = 0.01 kg.
第 (a) 部分：由于总体服从正态分布，样本均值 X̄ ~ N(μ, σ²/n) = N(1.50, 0.04²/16) = N(1.50, 0.0001)。标准误 = σ/√n = 0.04/4 = 0.01 kg。
Part (b): Standardise: Z = (1.51 - 1.50) / 0.01 = 1.0. P(X̄ = 0.1587. Notice how the smaller standard error makes the mean more concentrated around μ.
第 (b) 部分：标准化：Z = (1.51 - 1.50) / 0.01 = 1.0。P(X̄ > 1.51) = P(Z > 1.0) = 1 - Φ(1.0) ≈ 1 - 0.8413 = 0.1587。注意更小的标准误使得样本均值更集中于 μ 周围。
Part (c): For 90% 6, 1.5164). So with probability 0.90 the sample mean lies between 1.4836 kg and 1.5164 kg.
第 (c) 部分：对于 90% 中心区间，双侧尾概率各为 0.05，z = ±1.6449。区间：1.50 ± 1.6449 × 0.01 = (1.4836, 1.5164)。因此样本均值以 0.90 的概率落在 1.4836 kg 与 1.5164 kg 之间。
7. Question 6: Hypothesis Testing for a Binomial Proportion | 第6题：二项比例的假设检验
Question: A company claims that 30% of customers prefer a new packaging. To test this, a random sample of 20 customers is taken, and 3 of them prefer the new packaging. Test at the 5% significance level whether there is evidence that the true proportion is less than 0.30. State your hypotheses, test statistic, critical region and conclusion clearly.
题目：某公司声称有 30% 的顾客偏好新包装。为检验该说法，随机抽取 20 名顾客，其中 3 人偏好新包装。以 5% 的显著性水平检验是否有证据表明真实比例低于 0.30。请清晰陈述假设、检验统计量、拒绝域及结论。
es: H₀: p = 0.30, H₁: p < 0.30 (one-tailed test). Significance level α = 0.05. Under H₀, X ~ B(20, 0.30). Test statistic is the observed number of successes, x = 3.
假设：H₀: p = 0.30，H₁: p < 0.30（单尾检验）。显著性水平 α = 0.05。在 H₀ 下，X ~ B(20, 0.30)。检验统计量为观测成功次数 x = 3。
P(X=3). P(X=0) = 0.7²⁰ ≈ 0.0008, P(X=1) = 20×0.3×0.7¹⁹ ≈ 0.0068, P(X=2) = 190×0.09×0.7¹⁸ ≈ 0.0278, P(X=3) = 1140×0.027×0.7¹⁷ ≈ 0.0716. Sum = 0.1070. Since 0.1070 > 0.05, P(X ≤ 3) is not ≤ 0.05. Check P(X ≤ 2) = 0.0008+0.0068+0.0278 = 0.0354 ≤ 0.05. So critical region is X ≤ 2. By observing X=3, we do not fall into the critical region. Therefore we do not reject H₀. There is insufficient evidence at the 5% level to support that the true proportion is less than 0.30.
确定拒绝域：找到最大的 c 使得 P(X ≤ c) ≤ 0.05。查表或用计算器：P(X ≤ 3) = 0.1070 > 0.05。P(X ≤ 2) = 0.0354 ≤ 0.05。故拒绝域为 X ≤ 2。观测到 x=3，未落入拒绝域。因此不拒绝 H₀。在 5% 显著性水平下，没有充分证据支持真实比例低于 0.30。
ction rule clearly. Here p-value = P(X ≤ 3) = 0.1070 > 0.05, confirming the same conclusion.
重要提示：许多学生直接比较 p 值与 α 而不明确写出拒绝域。在 WJEC 考试中，务必清晰说明拒绝规则。此处 p 值 = P(X ≤ 3) = 0.1070 > 0.05，结论一致。
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