AS CCEA Statistics: Unit Test Mock Exam Analysis | AS CCEA 统计：单元测试模拟卷解析

📚 AS CCEA Statistics: Unit Test Mock Exam Analysis | AS CCEA 统计：单元测试模拟卷解析

This article provides a comprehensive walk-through of a mock exam for the AS CCEA Statistics unit, covering descriptive measures, probability, discrete and special distributions, sampling, and bivariate analysis. Each section presents a typical exam-style question followed by a detailed solution, enabling you to revise core concepts and sharpen problem-solving skills.

本文对一份 AS CCEA 统计单元测试模拟卷进行详尽解析，涵盖描述统计量、概率、离散及特殊分布、抽样与双变量分析。每个小节给出典型考题并配以中英双语的详细解答，帮助巩固核心概念并提升解题能力。

1. Descriptive Statistics: Mean, Median, IQR & Standard Deviation | 描述统计：均值、中位数、四分位距与标准差

Question: The weights (in kg) of 9 students are: 52, 48, 55, 60, 47, 53, 58, 51, 49. Calculate the mean, median, interquartile range (IQR) and sample standard deviation.

问题： 9 名学生的体重（kg）如下：52, 48, 55, 60, 47, 53, 58, 51, 49。计算均值、中位数、四分位距（IQR）和样本标准差。

First, sort the data: 47, 48, 49, 51, 52, 53, 55, 58, 60. Mean = (47+48+49+51+52+53+55+58+60)/9 = 473/9 ≈ 52.56 kg. Median is the 5th value = 52 kg. For quartiles: Q1 median of lower half (47,48,49,51) is (48+49)/2 = 48.5; Q3 median of upper half (53,55,58,60) is (55+58)/2 = 56.5. IQR = 56.5 – 48.5 = 8 kg.

先将数据排序：47, 48, 49, 51, 52, 53, 55, 58, 60。均值 = (47+48+49+51+52+53+55+58+60)/9 = 473/9 ≈ 52.56 kg。中位数为第 5 个值 = 52 kg。四分位数：Q1 是下半部 (47,48,49,51) 的中位数 = (48+49)/2 = 48.5；Q3 是上半部 (53,55,58,60) 的中位数 = (55+58)/2 = 56.5。IQR = 56.5 – 48.5 = 8 kg。

To find the sample standard deviation, compute Σx = 473 and Σx² = 47²+48²+49²+51²+52²+53²+55²+58²+60² = 2209+2304+2401+2601+2704+2809+3025+3364+3600 = 25017. Sample variance s² = [Σx² – (Σx)²/n] / (n-1) = [25017 – (473²)/9] / 8. Since 473² = 223729 and 223729/9 ≈ 24858.78, the numerator is 25017 – 24858.78 ≈ 158.22. Then s² ≈ 158.22/8 ≈ 19.7775, so s ≈ √19.7775 ≈ 4.45 kg.

计算样本标准差：Σx = 473，Σx² = 25017。样本方差 s² = [Σx² – (Σx)²/n] / (n-1) = [25017 – 473²/9] / 8。473² = 223729，÷9 ≈ 24858.78，两者差约 158.22，除以 8 得 19.7775，开方后 s ≈ 4.45 kg。

2. Probability with Venn Diagrams | 概率与韦恩图

Question: In a group of 30 students, 18 study Chemistry, 15 study Biology, and 6 study neither. A student is chosen at random. Find the probability that the student studies both subjects, and the probability that the student studies Chemistry but not Biology.

问题： 30 名学生中，18 人学化学，15 人学生物，6 人两门都不学。随机选一名学生，求该生同时学习两门课的概率，以及学习化学但未学生物的概率。

Let x be the number studying both. Then only Chemistry = 18 – x, only Biology = 15 – x, and neither = 6. Sum: (18-x) + x + (15-x) + 6 = 30 → 39 – x = 30 → x = 9. Probability of both = 9/30 = 0.3. Probability of Chemistry only = (18 – 9)/30 = 9/30 = 0.3.

设同时学习的人数为 x，则只学化学为 18 – x，只学生物为 15 – x，都不学为 6。总和：(18-x) + x + (15-x) + 6 = 30 → 39 – x = 30 → x = 9。两门都学的概率 = 9/30 = 0.3。只学化学的概率 = (18 – 9)/30 = 9/30 = 0.3。

3. Discrete Random Variables & Expectation | 离散随机变量与期望

Question: The probability distribution of a discrete random variable X is given. x: 1, 2, 3, 4; P(X=x): 0.2, k, 0.3, 0.1. Given that E(X) = 2.3, find the value of k and calculate Var(X).

问题： 离散随机变量 X 的分布列为：x 取 1, 2, 3, 4，对应概率 P(X=x) 为 0.2, k, 0.3, 0.1。已知 E(X) = 2.3，求 k 的值并计算 Var(X)。

E(X) = 1×0.2 + 2×k + 3×0.3 + 4×0.1 = 0.2 + 2k + 0.9 + 0.4 = 1.5 + 2k = 2.3 → 2k = 0.8 → k = 0.4. Next, E(X²) = 1²×0.2 + 2²×0.4 + 3²×0.3 + 4²×0.1 = 0.2 + 1.6 + 2.7 + 1.6 = 6.1. Var(X) = E(X²) –

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