📚 Year 12 AQA Statistics: Unit Test Mock Exam Walkthrough | Year 12 AQA 统计：单元测试模拟卷解析
This article provides a complete walkthrough for a Year 12 AQA Statistics unit test mock exam. It covers the core AS-Level topics, including descriptive statistics, probability, binomial and normal distributions, sampling distributions, and hypothesis tests. Each section works through a typical exam-style question with detailed bilingual explanations to reinforce key techniques and exam strategy.
本文提供一份完整的 Year 12 AQA 统计单元测试模拟卷解析。覆盖描述性统计、概率、二项分布与正态分布、抽样分布以及假设检验等 AS 阶段核心主题。每个小节将讲解一道典型考题，通过详细的中英双语解析巩固关键方法与应试技巧。
1. Mock Exam Overview | 模拟卷概览
The mock paper is designed to reflect the style and demand of AQA AS Statistics. It contains ten questions spanning data presentation, probability models, discrete and continuous distributions, and both binomial and normal hypothesis testing. Marks are allocated to reflect the weighting in real assessments, with emphasis on interpretation and calculation accuracy.
这份模拟卷参照 AQA AS 统计的题型与难度设计。共包含十道大题，覆盖数据展示、概率模型、离散与连续分布，以及二项和正态假设检验。分值分配与实际考试权重一致，侧重考查解释能力与计算准确性。
Students are advised to attempt the questions under timed conditions before reviewing the solutions. Each solution presented here models the expected level of working, notation, and conclusion writing for full marks.
建议学生先限时完成模拟卷，再对照本文解析。每道题的解答均示范了取得满分所需的解题步骤、符号规范与结论表述。
2. Stem-and-Leaf Plot and Summary Statistics | 茎叶图与汇总统计
Question (Edited): The following data are the reaction times (in milliseconds) of 15 participants: 245, 261, 233, 258, 242, 267, 275, 239, 251, 264, 248, 272, 255, 237, 281. Construct a stem-and-leaf plot, find the median and interquartile range, and identify any outliers using the 1.5 × IQR rule.
题目（改编）：15 名参与者的反应时间（毫秒）数据如下：245, 261, 233, 258, 242, 267, 275, 239, 251, 264, 248, 272, 255, 237, 281。绘制茎叶图，求中位数和四分位距，并用 1.5 × IQR 法则找出离群值。
First, sort the data in ascending order: 233, 237, 239, 242, 245, 248, 251, 255, 258, 261, 264, 267, 272, 275, 281.
首先将数据按升序排列：233, 237, 239, 242, 245, 248, 251, 255, 258, 261, 264, 267, 272, 275, 281。
To construct the stem-and-leaf plot, use the hundreds and tens digits as the stem and the units digit as the leaf. Stems are 23, 24, 25, 26, 27, 28. The plot (unsorted leaves) would be: 23 | 3 7 9; 24 | 2 5 8; 25 | 1 5 8; 26 | 1 4 7; 27 | 2 5; 28 | 1. A key should be given: 23|3 represents 233 ms.
构建茎叶图时，以百位和十位数字作茎，个位数字作叶。茎为 23, 24, 25, 26, 27, 28。未排序叶图：23 | 3 7 9；24 | 2 5 8；25 | 1 5 8；26 | 1 4 7；27 | 2 5；28 | 1。须注明图例：23|3 表示 233 毫秒。
For the median, since n = 15, the median is the 8th ordered value: 255 ms. The lower quartile Q₁ is the median of the lower half (first 7 values): 242 ms. The upper quartile Q₃ is the median of the upper half (last 7 values): 267 ms. Thus IQR = Q₃ – Q₁ = 267 – 242 = 25 ms.
中位数因 n=15，为第 8 个有序值：255 ms。下四分位数 Q₁ 为前 7 个数据的中位数：242 ms。上四分位数 Q₃ 为后 7 个数据的中位数：267 ms。因此 IQR = Q₃ – Q₁ = 267 – 242 = 25 ms。
The boundaries for outliers are Q₁ – 1.5×IQR = 242 – 37.5 = 204.5, and Q₃ + 1.5×IQR = 267 + 37.5 = 304.5. All data points lie within this interval, so there are no outliers.
离群值边界为 Q₁ – 1.5×IQR = 242 – 37.5 = 204.5, 以及 Q₃ + 1.5×IQR = 267 + 37.5 = 304.5。所有数据均在该区间内，因此没有离群值。
3. Probability with Tree Diagrams | 概率与树形图
Question: A bag contains 4 red and 6 blue discs. Two discs are drawn at random without replacement. Find the probability that exactly one of the discs is red.
题目：一个袋子里有 4 个红圆盘和 6 个蓝圆盘。随机抽取两个圆盘且不放回。求恰好一个是红色的概率。
We can model this with a probability tree. On the first draw, P(Red) = 4/10 = 0.4, P(Blue) = 6/10 = 0.6.
可以用概率树来建模。第一次抽取，P(红) = 4/10 = 0.4, P(蓝) = 6/10 = 0.6。
If the first disc is red, there remain 3 red and 6 blue out of 9 discs. Then P(Red|first Red) = 3/9, P(Blue|first Red) = 6/9. If the first is blue, then P(Red|first Blue) = 4/9, P(Blue|first Blue) = 5/9.
若第一次抽到红，剩下 9 个圆盘中有 3 红 6 蓝。则 P(红|第一次红) = 3/9, P(蓝|第一次红) = 6/9。若第一次抽到蓝，则 P(红|第一次蓝) = 4/9, P(蓝|第一次蓝) = 5/9。
The event ‘exactly one red’ consists of two outcomes: (Red then Blue) or (Blue then Red). Their probabilities are (0.4 × 6/9) = 0.4 × 2/3 = 4/15 and (0.6 × 4/9) = 0.6 × 4/9 = 2.4/9 = 4/15. Sum = 8/15.
“恰好一个红” 包含两种结果：(红然后蓝) 或 (蓝然后红)。概率分别为 (0.4 × 6/9) = 0.4 × 2/3 = 4/15 和 (0.6 × 4/9) = 0.6 × 4/9 = 4/15。总和 = 8/15。
P(exactly one red) = 8/15 ≈ 0.533
P(恰好一个红) = 8/15 ≈ 0.533
4. Discrete Random Variable: Expectation and Variance | 离散随机变量：期望与方差
Question: The discrete random variable X has the probability distribution shown. Find E(X), Var(X) and E(3X – 2).
题目：离散随机变量 X 的概率分布如下表。求 E(X), Var(X) 和 E(3X – 2)。
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X=x) | 0.2 | 0.3 | 0.4 | 0.1 |
First compute the expected value: E(X) = Σ x·P(X=x) = 0×0.2 + 1×0.3 + 2×0.4 + 3×0.1 = 0 + 0.3 + 0.8 + 0.3 = 1.4.
首先计算期望：E(X) = Σ x·P(X=x) = 0×0.2 + 1×0.3 + 2×0.4 + 3×0.1 = 0 + 0.3 + 0.8 + 0.3 = 1.4。
Then E(X²) = 0²×0.2 + 1²×0.3 + 2²×0.4 + 3²×0.1 = 0 + 0.3 + 1.6 + 0.9 = 2.8. Variance Var(X) = E(X²) – [E(X)]² = 2.8 – 1.4² = 2.8 – 1.96 = 0.84.
然后 E(X²) = 0²×0.2 + 1²×0.3 + 2²×0.4 + 3²×0.1 = 0 + 0.3 + 1.6 + 0.9 = 2.8。方差 Var(X) = E(X²) – [E(X)]² = 2.8 – 1.96 = 0.84。
Using the linearity of expectation, E(3X – 2) = 3E(X) – 2 = 3×1.4 – 2 = 4.2 – 2 = 2.2.
利用期望的线性性质，E(3X – 2) = 3E(X) – 2 = 3×1.4 – 2 = 4.2 – 2 = 2.2。
5. Binomial Distribution: Calculating Probabilities | 二项分布：概率计算
Question: Let X ~ B(12, 0.25). Find P(X = 4), P(X ≤ 2), and P(3 ≤ X < 6).
题目：设 X ~ B(12, 0.25)。求 P(X = 4), P(X ≤ 2) 和 P(3 ≤ X < 6)。
The binomial probability formula is P(X = r) = nCr × pʳ × (1 – p)ⁿ⁻ʳ. Here n = 12, p = 0.25.
二项概率公式为 P(X = r) = nCr × pʳ × (1 – p)ⁿ⁻ʳ。此处 n = 12, p = 0.25。
For r = 4: P(X = 4) = 12C4 × (0.25)⁴ × (0.75)⁸. 12C4 = 495, (0.25)⁴ = 1/256 ≈ 0.00390625, (0.75)⁸ ≈ 0.100112. Product ≈ 495 × 0.00390625 × 0.100112 ≈ 0.1936 (exact rational 495 × 3⁸ / 4¹² can be used).
对于 r = 4: P(X = 4) = 12C4 × (0.25)⁴ × (0.75)⁸。12C4 = 495，(0.25)⁴ = 1/256 ≈ 0.00390625，(0.75)⁸ ≈ 0.100112。乘积约 0.1936。
For P(X ≤ 2), sum P(X = 0) + P(X = 1) + P(X = 2). P(X = 0) = (0.75)¹² ≈ 0.0317; P(X = 1) = 12 × 0.25 × (0.75)¹¹ ≈ 0.1267; P(X = 2) = 66 × (0.25)² × (0.75)¹⁰ ≈ 0.2323. Sum ≈ 0.3907.
计算 P(X ≤ 2)，累加 P(X=0)+P(X=1)+P(X=2)。P(X=0) ≈ 0.0317; P(X=1) ≈ 0.1267; P(X=2) ≈ 0.2323。总和 ≈ 0.3907。
For P(3 ≤ X < 6) = P(X=3) + P(X=4) + P(X=5). Using similar calculations: P(X=3) ≈ 0.2581, P(X=4) ≈ 0.1936, P(X=5) ≈ 0.1032. Sum ≈ 0.5549.
对于 P(3 ≤ X < 6) = P(X=3) + P(X=4) + P(X=5)。类似计算得 P(X=3)≈0.2581, P(X=4)≈0.1936, P(X=5)≈0.1032。总和 ≈ 0.5549。
6. Normal Distribution: Standardisation | 正态分布：标准化
Question: Y ~ N(80, 25). Find P(Y < 87) and P(76 < Y < 84).
题目：Y ~ N(80, 25)。求 P(Y < 87) 和 P(76 < Y < 84)。
Here Y has mean μ = 80 and variance σ² = 25, so standard deviation σ = 5. To standardise, use Z = (Y – μ)/σ ~ N(0, 1²).
此处 Y 的均值 μ = 80，方差 σ² = 25，因此标准差 σ = 5。标准化时使用 Z = (Y – μ)/σ ~ N(0, 1²)。
For Y < 87: Z = (87 - 80)/5 = 1.4. From standard normal tables, Φ(1.40) = 0.9192. Thus P(Y < 87) ≈ 0.9192.
对于 Y < 87：Z = (87 - 80)/5 = 1.4。查标准正态分布表，Φ(1.40) = 0.9192。故 P(Y < 87) ≈ 0.9192。
For 76 < Y < 84: lower bound Z₁ = (76 - 80)/5 = -0.8, upper bound Z₂ = (84 - 80)/5 = 0.8. Φ(0.80) = 0.7881, Φ(-0.80) = 1 - 0.7881 = 0.2119. Required probability = 0.7881 - 0.2119 = 0.5762.
对于 76 < Y < 84：下界 Z₁ = (76 - 80)/5 = -0.8，上界 Z₂ = (84 - 80)/5 = 0.8。Φ(0.80)=0.7881，Φ(-0.80)=1-0.7881=0.2119。所求概率 = 0.7881 - 0.2119 = 0.5762。
7. Inverse Normal Distribution | 正态分布逆问题
Question: Heights of adults are normally distributed with mean 172 cm and standard deviation 7 cm. Find the height that is exceeded by exactly 10% of the population.
题目：成人身高服从均值为 172 cm、标准差为 7 cm 的正态分布。求正好有 10% 的人超过的身高值。
Let H ~ N(172, 7²). We want h such that P(H > h) = 0.10, i.e., P(H < h) = 0.90. The z-value corresponding to a cumulative probability of 0.90 is found from tables: z ≈ 1.2816.
设 H ~ N(172, 7²)。需求 h 使得 P(H > h) = 0.10，即 P(H < h) = 0.90。查表得当累积概率 0.90 时 z ≈ 1.2816。
Unstandardising: h = μ + zσ = 172 + 1.2816 × 7 ≈ 172 + 8.971 = 180.97 cm. Thus approximately 181.0 cm.
反标准化：h = μ + zσ = 172 + 1.2816 × 7 ≈ 172 + 8.971 = 180.97 cm。因此约为 181.0 cm。
It is good practice to sketch the normal curve and mark the upper 10% tail to confirm the direction of the inequality.
绘制正态曲线并标出右侧 10% 尾部以核对不等号方向，是良好的答题习惯。
8. Distribution of the Sample Mean | 样本均值的分布
Question: The weight of a packet of chips is normally distributed with mean 200 g and standard deviation 4 g. A random sample of 16 packets is selected. Find the probability that the sample mean weight is less than 198.5 g.
题目：薯片每包重量服从均值为 200 g、标准差为 4 g 的正态分布。随机抽取 16 包，求样本均值重量少于 198.5 g 的概率。
If X ~ N(200, 4²), then for a sample of size n = 16, the sample mean X̄ ~ N(μ, σ²/n) = N(200, 16/16) = N(200, 1). Standard deviation of the sample mean is σ/√n = 4/4 = 1.
若 X ~ N(200, 4²)，则对于样本量 n=16，样本均值 X̄ ~ N(μ, σ²/n) = N(200, 16/16) = N(200, 1)。样本均值的标准差为 σ/√n = 4/4 = 1。
Compute Z = (198.5 – 200)/1 = -1.5. Φ(-1.50) = 1 – Φ(1.50) = 1 – 0.9332 = 0.0668. Thus P(X̄ < 198.5) ≈ 0.0668.
计算 Z = (198.5 – 200)/1 = -1.5。Φ(-1.50) = 1 – Φ(1.50) = 1 – 0.9332 = 0.0668。故 P(X̄ < 198.5) ≈ 0.0668。
9. Binomial Hypothesis Test (One-Tailed) | 二项假设检验（单尾）
Question: A manufacturer claims that at least 80% of their components pass a quality test. A random sample of 20 components is tested, and 13 pass. Test at the 5% significance level whether there is evidence that the proportion is less than claimed.
题目：某制造商声称其至少 80% 的元件通过质检。随机抽取 20 个元件测试，其中 13 个通过。在 5% 显著性水平下，检验是否有证据表明真实比例低于声称值。
Define X ~ B(20, p). H₀: p = 0.8, H₁: p < 0.8 (one-tailed test). Test statistic: observed number of passes = 13.
定义 X ~ B(20, p)。原假设 H₀: p = 0.8，备择假设 H₁: p < 0.8（单尾检验）。检验统计量：观测通过数 = 13。
We need to find the critical region: the largest value of X such that P(X ≤ k) ≤ 0.05 under H₀. Using binomial tables or cumulative probabilities for B(20, 0.8): P(X ≤ 13) ≈ 0.032 (let’s verify with known values: B(20,0.8) P(X≤13) = 0.0321). Since 0.0321 < 0.05, the critical region is X ≤ 13.
需找到拒绝域：满足 H₀ 下 P(X ≤ k) ≤ 0.05 的最大 k 值。查 B(20, 0.8) 表：P(X ≤ 13) ≈ 0.0321。由于 0.0321 < 0.05，拒绝域为 X ≤ 13。
Our observed value 13 falls in the critical region, so we reject H₀ at the 5% significance level. There is sufficient evidence to suggest that the true proportion of passing components is less than 80%.
我们的观测值 13 落入拒绝域，因此在 5% 显著性水平下拒绝 H₀。有足够证据表明通过元件的真实比例低于 80%。Published by TutorHao | Year 12 统计 Revision Series | aleveler.com
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