📚 Year 12 AQA Statistics: Unit Test Mock Paper Analysis | AQA Year 12 统计：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 12 AQA Statistics, covering key topics such as sampling methods, data representation, measures of central tendency and dispersion, probability, the binomial distribution, and hypothesis testing. Each section presents a typical exam-style question together with a full solution and commentary, helping students consolidate their understanding and avoid common pitfalls.
本文对一份 AQA Year 12 统计单元测试模拟卷进行逐题解析，涵盖抽样方法、数据呈现、集中趋势与离散度量、概率、二项分布以及假设检验等核心考点。每一小节都展示一道典型考题和完整解答，并配以点评，帮助学生巩固知识并避开常见错误。
1. Sampling Methods | 抽样方法
A researcher wants to estimate the average amount of time Year 12 students spend on homework per week. The school has 600 Year 12 students divided into 20 tutor groups of varying sizes. The researcher plans to select 5 tutor groups at random and then survey every student in those groups. Identify this sampling technique and give one advantage and one disadvantage.
一名研究者想估计 Year 12 学生每周花在家庭作业上的平均时间。该校共有 600 名 Year 12 学生，分布在 20 个规模不等的辅导小组中。研究者计划随机选择 5 个小组，然后调查这些小组中的每一位学生。请说明这种抽样技术的名称，并给出一个优点和一个缺点。
This is an example of cluster sampling. The population is divided into clusters (tutor groups), a random sample of clusters is selected, and all members of the chosen clusters are surveyed. One advantage is that it is often quicker and cheaper than simple random sampling when the population is widely spread, because you only need access to a few clusters. A disadvantage is that members within the same cluster may be more alike, which can increase sampling error compared with simple random sampling of the same total size.
这是整群抽样。先将总体划分成群（辅导小组），随机抽取部分群，然后对选中群的所有成员进行调查。优点是当总体分布较广时通常比简单随机抽样更快、更经济，因为只需要接触少数几个群。缺点是同一群内的个体可能较为相似，相比同样样本量的简单随机抽样可能会增加抽样误差。
2. Types of Data | 数据类型
Classify the following variables as qualitative or quantitative, and for quantitative variables state whether they are discrete or continuous: (a) the brand of mobile phone used, (b) the number of text messages sent in a day, (c) the time taken to run 100 metres.
将下列变量分类为定性变量或定量变量，对于定量变量还需说明是离散的还是连续的：(a) 使用的手机品牌，(b) 一天内发送的短信条数，(c) 跑 100 米所用的时间。
(a) The brand of mobile phone is a qualitative variable because it describes a category rather than a numerical measurement. (b) The number of text messages is quantitative and discrete, as it can only take whole number values (0, 1, 2, …) and arises from counting. (c) The time taken to run 100 metres is quantitative and continuous, since it is measured on a continuous scale and can, in principle, take any value within a given range.
(a) 手机品牌是定性变量，因为它描述的是类别而非数值。(b) 短信条数是定量且离散的，因为它只能取整数值（0, 1, 2, …），由计数产生。(c) 跑 100 米所用时间是定量且连续的，因为它是在连续尺度上测量的，理论上可以取某一范围内的任意值。
3. Histogram and Frequency Density | 直方图与频率密度
The table summarises the times, in minutes, taken by 80 students to complete a puzzle.
| Time (t minutes) | Frequency |
|---|---|
| 0 ≤ t < 5 | 10 |
| 5 ≤ t < 10 | 18 |
| 10 ≤ t < 20 | 28 |
| 20 ≤ t < 30 | 16 |
| 30 ≤ t < 60 | 8 |
Calculate the frequency density for the 30 ≤ t < 60 interval and explain why frequency density is used on the vertical axis of a histogram when class widths are unequal.
下表汇总了 80 名学生完成一个拼图所用时间的分钟数。计算 30 ≤ t < 60 这一区间的频率密度，并解释为什么在组距不相等时直方图的纵轴要使用频率密度。
For the interval 30 ≤ t < 60, the class width is 60 − 30 = 30 minutes. The frequency is 8. Frequency density = frequency ÷ class width = 8 ÷ 30 = 0.267 (to 3 decimal places). Frequency density is used in a histogram with unequal class widths so that the area of each bar is proportional to the frequency it represents. If we plotted raw frequencies on the vertical axis, wider intervals would appear disproportionately large, distorting the visual impression of the distribution.
区间 30 ≤ t < 60 的组距是 60 − 30 = 30 分钟，频数为 8。频率密度 = 频数 ÷ 组距 = 8 ÷ 30 = 0.267（保留三位小数）。在组距不等的直方图中使用频率密度，是为了让每个直条的面积与其所代表的频数成正比。如果在纵轴上直接绘制频数，较宽的区间就会显得不成比例地大，从而扭曲分布的视觉效果。
4. Mean and Standard Deviation from a Frequency Table | 从频数表求均值与标准差
The heights, in cm, of 50 plants were measured. The results are summarised as Σ x = 1850 and Σ x² = 69 350. Calculate the mean and the standard deviation of the heights.
测量了 50 株植物的高度，单位为 cm。结果摘要为 Σ x = 1850，Σ x² = 69 350。计算高度的均值与标准差。
Mean μ = Σ x / n = 1850 / 50 = 37 cm. The variance σ² = (Σ x² / n) − (μ)² = (69350 / 50) − 37² = 1387 − 1369 = 18. Therefore, the standard deviation σ = √18 ≈ 4.24 cm (to 3 significant figures).
均值 μ = Σ x / n = 1850 / 50 = 37 cm。方差 σ² = (Σ x² / n) − (μ)² = (69350 / 50) − 37² = 1387 − 1369 = 18。因此标准差 σ = √18 ≈ 4.24 cm（保留三位有效数字）。
5. Probability and Venn Diagrams | 概率与韦恩图
Events A and B are such that P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.7. Find P(A ∩ B). Hence determine whether A and B are independent.
事件 A 和 B 满足 P(A) = 0.5，P(B) = 0.4，P(A ∪ B) = 0.7。求 P(A ∩ B)，并由此判断 A 和 B 是否独立。
Using the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B), so 0.7 = 0.5 + 0.4 − P(A ∩ B). Solving gives P(A ∩ B) = 0.9 − 0.7 = 0.2. For independence we require P(A ∩ B) = P(A) × P(B). Here P(A) × P(B) = 0.5 × 0.4 = 0.2, which equals P(A ∩ B). Therefore A and B are independent.
利用加法公式：P(A ∪ B) = P(A) + P(B) − P(A ∩ B)，得 0.7 = 0.5 + 0.4 − P(A ∩ B)，解得 P(A ∩ B) = 0.9 − 0.7 = 0.2。独立性的条件是 P(A ∩ B) = P(A) × P(B)，这里 P(A) × P(B) = 0.5 × 0.4 = 0.2，恰好等于 P(A ∩ B)。因此 A 与 B 独立。
6. Discrete Random Variables and Expectation | 离散随机变量与期望
The probability distribution of a discrete random variable X is given by P(X = x) = k(4 − x) for x = 1, 2, 3. Find the value of k, and calculate E(X) and Var(X).
离散随机变量 X 的概率分布为 P(X = x) = k(4 − x)，x = 1, 2, 3。求 k 的值，并计算 E(X) 和 Var(X)。
First, determine k: ∑ P(X = x) = k(4−1) + k(4−2) + k(4−3) = k(3 + 2 + 1) = 6k = 1, so k = 1/6. The distribution is: P(X=1) = 3/6 = 0.5, P(X=2) = 2/6 = 1/3, P(X=3) = 1/6. E(X) = 1×(3/6) + 2×(2/6) + 3×(1/6) = (3 + 4 + 3)/6 = 10/6 ≈ 1.667. E(X²) = 1²×(3/6) + 2²×(2/6) + 3²×(1/6) = (3 + 8 + 9)/6 = 20/6 ≈ 3.333. Var(X) = E(X²) − [E(X)]² = 20/6 − (10/6)² = 20/6 − 100/36 = (120 − 100)/36 = 20/36 = 5/9 ≈ 0.556.
先求 k：由 ∑ P(X = x) = 1 得 k(3 + 2 + 1) = 6k = 1，故 k = 1/6。分布为 P(X=1)=3/6=0.5，P(X=2)=2/6≈0.333，P(X=3)=1/6≈0.167。E(X)=1×0.5+2×(1/3)+3×(1/6)=10/6≈1.667。E(X²)=1×0.5+4×(1/3)+9×(1/6)=20/6≈3.333。Var(X)=E(X²)−[E(X)]²=20/6−(10/6)²=20/36=5/9≈0.556。
7. Binomial Distribution – Probability Calculation | 二项分布——概率计算
A fair die is rolled 12 times. Find the probability that the number 6 appears exactly 3 times.
抛掷一枚均匀的骰子 12 次。求数字 6 恰好出现 3 次的概率。
Let X be the number of times a 6 appears. Then X ~ B(12, 1/6). P(X = 3) = ₁₂C₃ × (1/6)³ × (5/6)⁹. ₁₂C₃ = 12!/(3!9!) = (12×11×10)/(3×2×1) = 220. So P(X = 3) = 220 × (1/6)³ × (5/6)⁹ ≈ 220 × (1/216) × (1953125/10077696) … A more streamlined calculation gives 220 × (5⁹/6¹²). 5⁹ = 1953125, 6¹² = 2176782336. Therefore P ≈ 220 × 1953125 / 2176782336 ≈ 429687500 / 2176782336 ≈ 0.197 (3 s.f.).
设 X 为出现 6 的次数，则 X ~ B(12, 1/6)。P(X = 3) = ₁₂C₃ × (1/6)³ × (5/6)⁹。₁₂C₃ = 220。因此 P(X = 3) = 220 × (1/6)³ × (5/6)⁹ ≈ 220 × (1953125 / 2176782336) ≈ 0.197（保留三位有效数字）。
8. Binomial Distribution – Hypothesis Testing | 二项分布——假设检验
A manufacturer claims that only 10% of its light bulbs are defective. A customer suspects the proportion is higher and tests a random sample of 20 bulbs, finding 5 defectives. Test at the 5% significance level whether there is evidence to support the customer’s suspicion. State your hypotheses, the p‑value and your conclusion clearly.
某制造商声称其灯泡的次品率仅为 10%。一位顾客怀疑真实次品率更高，并随机抽取 20 只灯泡进行检验，发现 5 只次品。在 5% 的显著性水平下检验是否有证据支持顾客的怀疑。请清楚地写出假设、p 值以及结论。
Let p be the probability that a bulb is defective. H₀: p = 0.1, H₁: p > 0.1 (one‑tailed test). Under H₀, X ~ B(20, 0.1). The observed number of defectives is 5. The p‑value is P(X ≥ 5). Using tables or the formula: P(X ≥ 5) = 1 − P(X ≤ 4). From cumulative binomial tables with n=20, p=0.1, P(X ≤ 4) = 0.9568 (commonly found in AQA formula booklet). Hence p‑value ≈ 1 − 0.9568 = 0.0432. Since 0.0432 < 0.05, we reject H₀. There is sufficient evidence at the 5% level to support the customer’s suspicion that the proportion of defective bulbs is greater than 10%.
设 p 为灯泡的次品概率。H₀: p = 0.1，H₁: p > 0.1（单尾检验）。在零假设下 X ~ B(20, 0.1)，观测到次品数 5。p 值 = P(X ≥ 5) = 1 − P(X ≤ 4)。查二项分布累积表，n=20, p=0.1 时 P(X ≤ 4) = 0.9568，故 p 值 ≈ 0.0432。由于 0.0432 < 0.05，拒绝 H₀。在 5% 显著性水平下有足够证据支持顾客的怀疑，即次品率高于 10%。
9. Critical Region and Critical Value | 拒绝域与临界值
For the hypothesis test in the previous question, find the critical region for the number of defective bulbs at the 5% significance level. Hence determine the smallest number of defectives that would lead to the rejection of the manufacturer’s claim.
对于上一题中的假设检验，求在 5% 显著性水平下关于次品个数的拒绝域，并由此确定导致拒绝制造商声称的最小次品数量。
We need the smallest value c such that P(X ≥ c) ≤ 0.05 under H₀: X ~ B(20, 0.1). From binomial tables, P(X ≥ 5) = 1 − P(X ≤ 4) = 1 − 0.9568 = 0.0432 ≤ 0.05, while P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − 0.8670 = 0.1330 > 0.05. Therefore the critical region is X ≥ 5. The smallest number of defectives that would lead to rejection is 5. This matches the conclusion from the p‑value method.
我们需要找到最小的 c 使得 H₀ 下 P(X ≥ c) ≤ 0.05。查表得 P(X ≥ 5) = 0.0432 ≤ 0.05，而 P(X ≥ 4) = 0.1330 > 0.05。所以拒绝域为 X ≥ 5。导致拒绝的最小次品数量是 5。这与 p 值法的结论一致。
10. The Normal Distribution – Probability | 正态分布——概率
The lengths of bolts produced by a machine are normally distributed with mean 50 mm and standard deviation 0.8 mm. A bolt is acceptable if its length lies between 48.6 mm and 51.0 mm. Find the proportion of bolts that are acceptable.
某机器生产的螺栓长度服从正态分布，均值为 50 mm，标准差为 0.8 mm。若长度在 48.6 mm 到 51.0 mm 之间则为合格。求合格螺栓的比例。
Let L ~ N(50, 0.8²). We need P(48.6 < L < 51.0). Standardise: z₁ = (48.6 − 50) / 0.8 = −1.75, z₂ = (51.0 − 50) / 0.8 = 1.25. Using standard normal tables: Φ(1.25) = 0.8944, Φ(−1.75) = 1 − Φ(1.75) = 1 − 0.9599 = 0.0401. Then P = 0.8944 − 0.0401 = 0.8543. So approximately 85.4% of bolts are acceptable.
设 L ~ N(50, 0.8²)，需要求 P(48.6 < L < 51.0)。标准化：z₁ = (48.6 − 50)/0.8 = −1.75，z₂ = (51.0 − 50)/0.8 = 1.25。查标准正态分布表：Φ(1.25) = 0.8944，Φ(−1.75) = 1 − Φ(1.75) = 1 − 0.9599 = 0.0401。概率 = 0.8944 − 0.0401 = 0.8543。因此约 85.4% 的螺栓合格。
11. The Normal Distribution – Finding the Mean | 正态分布——求均值
A second machine produces bolts whose lengths are normally distributed with standard deviation 0.6 mm. It is known that 10% of its bolts are longer than 50.8 mm. Find the mean length of bolts from this machine.
另一台机器生产的螺栓长度服从标准差为 0.6 mm 的正态分布。已知 10% 的螺栓长度超过 50.8 mm。求该机器生产的螺栓的平均长度。
Let X ~ N(μ, 0.6²). We are given P(X > 50.8) = 0.10. Standardising: P(Z > (50.8 − μ)/0.6) = 0.10. From tables, the z‑value that cuts off 10% in the right tail is approximately 1.2816 (or use 1.28 as commonly acceptable in AQA). So (50.8 − μ)/0.6 = 1.2816. Hence 50.8 − μ = 1.2816 × 0.6 = 0.76896, giving μ = 50.8 − 0.76896 = 50.03104 ≈ 50.0 mm (3 s.f.). (If using z = 1.28, μ ≈ 50.8 − 0.768 = 50.032 ≈ 50.0 mm.) The mean length is approximately 50.0 mm.
设 X ~ N(μ, 0.6²)，已知 P(X > 50.8) = 0.10。标准化：P(Z > (50.8 − μ)/0.6) = 0.10。查表得右侧尾部概率 0.10 对应的 z 值约为 1.2816（AQA 常用 1.28 亦可）。因此 (50.8 − μ)/0.6 = 1.2816，解得 μ = 50.8 − 1.2816×0.6 = 50.8 − 0.76896 ≈ 50.0 mm（三位有效数字）。平均长度约为 50.0 mm。
12. Combining Binomial and Normal Approximations | 二项分布的正态近似
A large company has 2000 employees. It is believed that 45% of them are in favour of a new working hours policy. Using a normal approximation with a continuity correction, estimate the probability that fewer than 870 employees are in favour.
一家大公司有 2000 名员工。据信其中 45% 赞成一项新的工时政策。使用带连续校正的正态近似，估计赞成人数少于 870 的概率。
Let Y be the number in favour. Y ~ B(2000, 0.45). Mean μ = np = 2000×0.45 = 900. Variance σ² = npq = 2000×0.45×0.55 = 495, so σ = √495 ≈ 22.25. We need P(Y < 870). With continuity correction, use P(Y ≤ 869.5). Standardise: z = (869.5 − 900) / 22.25 ≈ −30.5 / 22.25 ≈ −1.371. From standard normal tables, Φ(−1.37) ≈ 0.0853 (using 1.37, the exact value is about 0.0853). Therefore the approximate probability is 0.0853 (about 8.5%).
设 Y 为赞成人数，Y ~ B(2000, 0.45)。均值 μ = np = 900，方差 σ² = npq = 2000×0.45×0.55 = 495，σ ≈ 22.25。需计算 P(Y < 870)。连续校正后使用 P(Y ≤ 869.5)。标准化：z = (869.5 − 900)/22.25 ≈ −30.5/22.25 ≈ −1.371。查标准正态表得 Φ(−1.37) ≈ 0.0853。因此近似概率约为 0.0853（8.5%）。
Published by TutorHao | Statistics Revision Series | aleveler.com
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