AS AQA Statistics: Unit Test Mock Paper Walkthrough | AS AQA 统计：单元测试模拟卷解析

📚 AS AQA Statistics: Unit Test Mock Paper Walkthrough | AS AQA 统计：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for AS AQA Statistics. Each question is broken down step by step, covering key topics such as descriptive statistics, probability, binomial and normal distributions, correlation, regression, sampling, and histograms. Follow along to improve your understanding and learn how to avoid common pitfalls.

本文详细解析了一份AS AQA统计单元测试模拟卷。我们将逐步剖析每一道题，涵盖描述统计、概率、二项分布与正态分布、相关与回归、抽样以及直方图等核心主题。通过阅读，可以巩固理解并学会避开常见错误。

1. Question 1: Descriptive Statistics and Box Plots | 问题1：描述统计与箱线图

The first question gives the scores of 10 students out of 50: 32, 45, 28, 39, 42, 35, 40, 38, 30, 36. You are asked to find the mean, median, and standard deviation, check for outliers using the 1.5 × IQR rule, and draw a box plot.

第一题给出了10名学生满分50的分数：32, 45, 28, 39, 42, 35, 40, 38, 30, 36。要求计算均值、中位数和标准差，用1.5×IQR规则判断离群值，并绘制箱线图。

First, calculate the mean: sum all values (365) and divide by 10. Mean x̄ = 36.5.

首先计算均值：总和为365，除以10得均值 x̄ = 36.5。

For the median, sort the data: 28, 30, 32, 35, 36, 38, 39, 40, 42, 45. With an even number of values, the median is the average of the 5th and 6th: (36 + 38)/2 = 37.

求中位数时先排序：28, 30, 32, 35, 36, 38, 39, 40, 42, 45。数据个数为偶数，中位数是第5和第6个的平均值：(36+38)/2 = 37。

To find the sample standard deviation, compute the sum of squared deviations: Σ(x – x̄)² = (32–36.5)² + … + (36–36.5)² = 260.5. Then s = √[260.5 / (10 – 1)] = √(28.944…) ≈ 5.38 (to 3 s.f.).

计算样本标准差，先求离差平方和：Σ(x – x̄)² = 260.5。然后 s = √[260.5 / (10 – 1)] = √28.944… ≈ 5.38（保留三位有效数字）。

To determine outliers, locate Q₁ and Q₃. Using the (n+1)/4 method, Q₁ lies at position 2.75: 32 + 0.75 × (35 – 32) = 34.25. Q₃ lies at position 8.25: 40 + 0.25 × (42 – 40) = 40.5. IQR = 40.5 – 34.25 = 6.25.

为判断离群值，找出Q₁与Q₃。用 (n+1)/4 方法，Q₁位于2.75位置：32+0.75×(35–32)=34.25。Q₃位于8.25位置：40+0.25×(42–40)=40.5。IQR = 6.25。

Lower fence = Q₁ – 1.5×IQR = 34.25 – 9.375 = 24.875; upper fence = 40.5 + 9.375 = 49.875. Since all scores lie between these fences, there are no outliers. The box plot uses these values: min 28, Q₁ 34.25, median 37, Q₃ 40.5, max 45.

下限 = 34.25 – 1.5×6.25 = 24.875；上限 = 40.5 + 9.375 = 49.875。所有分数均在此区间内，因此无离群值。箱线图使用最小值28、Q₁ 34.25、中位数37、Q₃ 40.5、最大值45。

2. Question 2: Probability and Tree Diagrams | 问题2：概率与树状图

This problem involves a bag with 5 red, 3 blue, and 2 green counters. Two counters are drawn without replacement. You need a tree diagram and probabilities for ‘both same colour’ and a conditional probability.

本题涉及一个装有5红、3蓝、2绿色计数块的袋子，无放回地抽取两次。需要绘制树状图，求“同色”的概率，以及条件概率。

Draw the first branch with probabilities: R 5/10, B 3/10, G 2/10. For each, branch to second draw accordingly. Without replacement, the denominators become 9. For example, P(R and R) = (5/10) × (4/9) = 20/90 = 2/9.

画出第一层分支：R 5/10，B 3/10，G 2/10。每支下继续第二层，分母变为9。例如 P(R and R) = (5/10)×(4/9) = 20/90 = 2/9。

P(both blue) = (3/10) × (2/9) = 6/90 = 1/15; P(both green) = (2/10) × (1/9) = 2/90 = 1/45. Hence P(same colour) = 2/9 + 1/15 + 1/45 = 10/45 + 3/45 + 1/45 = 14/45.

P（双蓝）= (3/10)×(2/9) = 6/90 = 1/15；P（双绿）= 2/10×1/9 = 2/90 = 1/45。因此 P（同色）= 2/9+1/15+1/45 = 10/45+3/45+1/45 = 14/45。

For the conditional part, let A be ‘at least one red’. P(A) = 1 – P(no red) = 1 – (5/10 × 4/9) = 1 – 20/90 = 70/90 = 7/9. The event ‘both red’ is a subset. So P(both red | at least one red) = P(both red) / P(A) = (2/9) / (7/9) = 2/7.

条件概率部分，设A为“至少一个红”。P(A)=1–P(无红)=1–(5/10×4/9)=1–20/90=7/9。事件“双红”是子集。故 P（双红|至少一红）= (2/9) / (7/9) = 2/7。

3. Question 3: Discrete Random Variables | 问题3：离散随机变量

The discrete random variable Y has the distribution shown below. You are given E(Y) = 2.6 and must find the missing probabilities a and b, then variance and P(Y>2).

离散随机变量Y的分布如下。已知E(Y)=2.6，需求出缺失概率a和b，再计算方差和P(Y>2)。

Using ΣP = 1 gives 0.2 + a + 0.3 + b = 1, so a + b = 0.5. From E(Y) = 2.6: 1×0.2 + 2a + 3×0.3 + 4b = 2.6 → 0.2 + 2a + 0.9 + 4b = 2.6 → 2a + 4b = 1.5.

利用总概率为1得 0.2+a+0.3+b =

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