📚 AS Eduqas Statistics: Unit Test Mock Paper Walkthrough | AS Eduqas统计：单元测试模拟卷解析
This article presents a complete walkthrough of a mock unit test designed for the AS Eduqas Statistics specification. Each question targets a core topic—sampling, data presentation, measures of location and spread, probability, discrete random variables, binomial distribution, normal distribution, and correlation & regression. Detailed solutions, step-by-step reasoning, and examiner tips are provided to help you consolidate understanding and boost exam confidence.
本文为AS Eduqas统计学单元测试模拟卷提供完整解析。每道题均紧扣核心主题：抽样、数据表示、集中趋势与离散程度、概率、离散随机变量、二项分布、正态分布以及相关与回归。文中包含详细解答、分步推理与考官提示，助你巩固知识、提升应试信心。
1. Sampling Methods Question | 抽样方法题目解析
A researcher wants to estimate the average daily screen time of students in a secondary school. She selects 10 students at random from each of Years 7 to 11, giving a total sample of 50 students.
一位研究者想估计一所中学学生每日平均屏幕使用时间。她从7至11年级每个年级中随机抽取10名学生，共获得50名学生的样本。
Identify the sampling method used, describe one advantage and one disadvantage of this method in the context of the study.
请指出所使用的抽样方法，并结合该研究说明此方法的一个优点和一个缺点。
The sampling method is stratified sampling. The population is divided into strata (year groups), and a simple random sample is taken from each stratum. The sample size from each stratum is proportional to the stratum size only if all strata are equal; here each year group contributes the same number (10), so it is a balanced stratified design.
该方法为分层抽样。总体被划分为若干层（年级），并从每一层中随机抽取样本。由于各年级学生数可能不同，此处每个年级抽取相同人数，属于均衡分层设计。
Advantage: Stratified sampling guarantees representation from each year group. This reduces sampling bias and ensures that the estimate reflects the behaviour of all age groups in the school, not just a single dominant cohort.
优点：分层抽样保证每个年级都有代表。这降低了抽样偏倚，确保估计结果能够反映学校内所有年龄段的行为，而不仅仅体现某一群体。
Disadvantage: Implementing the method requires an accurate list of students in each year group and extra organisational effort to select from each stratum. If some students are absent or refuse, achieving the planned numbers can be time‑consuming and costly.
缺点：实施该方法需要每个年级准确的学生名单，并且需要额外的组织工作从各层中抽取样本。若有学生缺席或拒绝参与，达到计划人数会耗时且成本较高。
2. Stem‑and‑Leaf Diagram and Outliers | 茎叶图与异常值
The following data show the distances (in kilometres) travelled by 12 employees to a new office site:
以下数据为12名员工前往新办公地点的距离（单位：公里）：
12, 15, 18, 21, 23, 25, 27, 30, 32, 35, 38, 40
(a) Construct an ordered stem‑and‑leaf diagram. Use stems representing tens and leaves representing units.
(a) 绘制一张有序茎叶图，以十位数为茎、个位数为叶。
(b) Find the median, lower quartile (Q₁) and upper quartile (Q₃).
(b) 求中位数、下四分位数(Q₁)和上四分位数(Q₃)。
(c) Identify any outliers using the 1.5 × IQR rule.
(c) 利用1.5×IQR法则识别异常值。
(a) The ordered stem‑and‑leaf diagram is:
(a) 有序茎叶图如下：
| Stem (10s) | Leaf (1s) |
| 1 | 2 5 8 |
| 2 | 1 3 5 7 |
| 3 | 0 2 5 8 |
| 4 | 0 |
Key: 1|2 represents 12 km. Data are already in ascending order.
图例：1|2表示12公里。数据已按升序排列。
(b) For n = 12 data values, the median position is the (12+1)/2 = 6.5th value, so median = (25 + 27)/2 = 26 km. The lower quartile is the median of the lower six numbers: (12,15,18,21,23,25), Q₁ = (18 + 21)/2 = 19.5 km. The upper quartile is the median of the upper six numbers: (27,30,32,35,38,40), Q₃ = (32 + 35)/2 = 33.5 km.
(b) n=12，中位数位置为第 (12+1)/2 = 6.5 个值，中位数 = (25+27)/2 = 26 km。下四分位数为前6个数的中位数：12,15,18,21,23,25，故 Q₁=(18+21)/2=19.5 km。上四分位数为后6个数的中位数：27,30,32,35,38,40，故 Q₃=(32+35)/2=33.5 km。
IQR = Q₃ − Q₁ = 33.5 − 19.5 = 14 km. Lower boundary = Q₁ − 1.5×IQR = 19.5 − 21 = −1.5. Upper boundary = Q₃ + 1.5×IQR = 33.5 + 21 = 54.5. Since all data values lie between −1.5 and 54.5, there are no outliers.
四分位距IQR = Q₃−Q₁ = 14 km。下边界 = Q₁−1.5×IQR = 19.5−21 = −1.5；上边界 = 33.5+21=54.5。由于所有数值均在−1.5与54.5之间，因此没有异常值。
3. Mean and Standard Deviation | 均值与标准差
Use the distance data from Question 2 to calculate the sample mean and the sample standard deviation. Show your working clearly.
使用第二题的距离数据计算样本均值与样本标准差，并清晰展示计算过程。
Data: 12, 15, 18, 21, 23, 25, 27, 30, 32, 35, 38, 40. Sum Σx = 12+15+18+21+23+25+27+30+32+35+38+40 = 316. Sample size n = 12, so sample mean x̄ = 316/12 ≈ 26.33 km.
数据：12, 15, 18, 21, 23, 25, 27, 30, 32, 35, 38, 40。总和 Σx = 316，n=12，样本均值 x̄ = 316/12 ≈ 26.33 km。
To compute the standard deviation, first find Σx² = 144 + 225 + 324 + 441 + 529 + 625 + 729 + 900 + 1024 + 1225 + 1444 + 1600 = 9210. Then use the formula s = √[ (Σx² − (Σx)²/n) / (n−1) ].
计算标准差：先求 Σx² = 144+225+324+441+529+625+729+900+1024+1225+1444+1600 = 9210。然后代入公式 s = √[ (Σx² − (Σx)²/n) / (n−1) ]。
(Σx)²/n = 316²/12 = 99856/12 ≈ 8321.33. So Σx² − (Σx)²/n = 9210 − 8321.33 = 888.67. Dividing by n−1 = 11 gives 888.67/11 ≈ 80.788. Taking the square root, s ≈ √80.788 ≈ 8.99 km.
(Σx)²/n = 316²/12 = 99856/12 ≈ 8321.33。因此 Σx² − (Σx)²/n = 888.67，除以 n−1=11 得 80.788，开方后 s ≈ 8.99 km。
Thus the sample standard deviation is approximately 9.0 km. The data show a typical spread of around 9 km around the mean.
因此样本标准差约为9.0 km，数据在均值附近的典型离散程度大约为9公里。
4. Probability with Venn Diagrams | 韦恩图概率问题
For two events A and B, P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Find:
对于事件A和B，已知P(A)=0.4、P(B)=0.5、P(A∩B)=0.2。求：
(i) P(A ∪ B)
(ii) P(A’|B)
(iii) Determine whether A and B are independent, giving a reason.
(iii) 判断A与B是否独立，并说明理由。
(i) Using the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.4 + 0.5 − 0.2 = 0.7.
(i) 利用加法公式：P(A∪B) = P(A)+P(B)−P(A∩B) = 0.4+0.5−0.2 = 0.7。
(ii) P(A’|B) = P(A’ ∩ B) / P(B). Since P(B) = P(A ∩ B) + P(A’ ∩ B), we have P(A’ ∩ B) = P(B) − P(A ∩ B) = 0.5 − 0.2 = 0.3. Hence P(A’|B) = 0.3 / 0.5 = 0.6.
(ii) P(A’|B) = P(A’∩B)/P(B)。由 P(B)=P(A∩B)+P(A’∩B) 可得 P(A’∩B) = 0.5−0.2=0.3，故 P(A’|B) = 0.3/0.5 = 0.6。
(iii) For independence we require P(A ∩ B) = P(A) × P(B). Here P(A) × P(B) = 0.4 × 0.5 = 0.2, which equals P(A ∩ B) = 0.2. Therefore A and B are independent. (Alternative test: P(A|B) = P(A ∩ B)/P(B) = 0.2/0.5 = 0.4 = P(A), confirming independence.)
(iii) 独立的条件为 P(A∩B)=P(A)×P(B)。计算 P(A)×P(B)=0.4×0.5=0.2，恰等于P(A∩B)=0.2，因此A与B独立。（也可验证：P(A|B)=0.2/0.5=0.4=P(A)。）
5. Discrete Random Variable | 离散随机变量
The probability distribution of a discrete random variable X is given in the table below.
离散随机变量X的概率分布见下表：
| x | 0 | 1 | 2 | 3 |
| P(X=x) | 0.1 | 0.3 | 0.4 | 0.2 |
(a) Show that the sum of probabilities is 1. (b) Calculate E(X). (c) Calculate Var(X). (d) Hence find E(3X + 2) and Var(3X + 2).
(a) 验证概率之和为1。(b) 计算E(X)。(c) 计算Var(X)。(d) 进而求E(3X+2)和Var(3X+2)。
(a) 0.1 + 0.3 + 0.4 + 0.2 = 1.0, so it is a valid probability distribution.
(a) 0.1+0.3+0.4+0.2 = 1.0，满足概率分布条件。
(b) E(X) = Σ x·P(X=x) = 0×0.1 + 1×0.3 + 2×0.4 + 3×0.2 = 0 + 0.3 + 0.8 + 0.6 = 1.7.
(b) E(X) = Σ x·P(X=x) = 0+0.3+0.8+0.6 = 1.7。
(c) E(X²) = 0²×0.1 + 1²×0.3 + 2²×0.4 + 3²×0.2 = 0 + 0.3 + 1.6 + 1.8 = 3.7. Var(X) = E(X²) − [E(X)]² = 3.7 − (1.7)² = 3.7 − 2.89 = 0.81.
(c) E(X²) = 0+0.3+1.6+1.8=3.7，Var(X)=3.7−2.89=0.81。
(d) Using linear properties: E(3X + 2) = 3E(X) + 2 = 3×1.7 + 2 = 5.1 + 2 = 7.1. Var(3X + 2) = 3² Var(X) = 9 × 0.81 = 7.29. (Adding a constant does not affect variance.)
(d) 利用线性性质：E(3X+2)=3E(X)+2=7.1；Var(3X+2)=3²Var(X)=9×0.81=7.29（加常数不影响方差）。
6. Binomial Distribution Problem | 二项分布问题
A factory produces components, and it is known that 10% are defective. A random sample of 8 components is selected. Let the random variable X represent the number of defective components in the sample.
某工厂生产的零件有10%为次品。随机抽取8个零件，设随机变量X表示样本中次品的数量。
(a) State the distribution of X, giving parameters. (b) Find P(X = 2). (c) Find P(X ≥ 1).
(a) 说出X的分布并给出参数。(b) 求P(X=2)。(c) 求P(X≥1)。
(a) X ~ B(8, 0.1). The number of trials n = 8, probability of success (defective) p = 0.1. Assumptions: trials are independent, fixed probability, each component either defective or not.
(a) X~B(8, 0.1)。试验次数n=8，成功（次品）概率p=0.1。假设各次试验独立、概率不变、每个零件非好即坏。
(b) P(X = 2) = C(8,2) × (0.1)² × (0.9)⁶. C(8,2) = 28. So P(X=2) = 28 × 0.01 × 0.531441 ≈ 28 × 0.00531441 ≈ 0.1488 (to 4 d.p.).
(b) P(X=2) = C(8,2)×(0.1)²×(0.9)⁶ = 28×0.01×0.531441 ≈ 0.1488（保留四位小数）。
(c) P(X ≥ 1) = 1 − P(X = 0). P(X=0) = (0.9)⁸ ≈ 0.4305. Hence P(X ≥ 1) ≈ 1 − 0.4305 = 0.5695.
(c) P(X≥1)=1−P(X=0)，P(X=0)=0.9⁸≈0.4305，故P(X≥1)≈0.5695。
7. Normal Distribution – Standardising Scores | 正态分布——分数标准化
The marks in an AS Statistics test are normally distributed with mean μ = 65 and standard deviation σ = 10.
某次AS统计测验成绩服从正态分布，均值μ=65，标准差σ=10。
(a) Find the probability that a randomly chosen student scores less than 70.
(a) 求随机选取一名学生成绩低于70的概率。
(b) The top 10% of students receive a distinction. Find the minimum mark needed for a distinction.
(b) 前10%的学生获得优秀，求获得优秀的最低分数。
(a) Let X ~ N(65, 10²). Standardise: Z = (X − μ)/σ. For X = 70, Z = (70 − 65)/10 = 0.5. P(X < 70) = P(Z < 0.5). Using standard normal tables, Φ(0.5) = 0.6915. So approximately 69.15% of students score less than 70.
(a) X~N(65,10²)，标准化 Z=(70−65)/10=0.5。P(X<70)=P(Z<0.5)，查标准正态表得Φ(0.5)=0.6915。因此约69.15%的学生成绩低于70。
(b) We need the value k such that P(X > k) = 0.1, i.e. P(X < k) = 0.9. The z‑score corresponding to 0.9 is approximately z = 1.2816 (or 1.28 from tables). Unstandardise: k = μ + z·σ = 65 + 1.2816 × 10 = 65 + 12.816 = 77.82. Rounding up, a mark of 78 is needed for distinction.
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8. Correlation and Linear Regression | 相关与线性回归
x (temp)
22 Published by TutorHao | AS 统计 Revision Series | aleveler.com
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