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

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

This article provides a detailed walkthrough of a mock unit test for the WJEC AS Statistics specification. Each question targets a core topic, with step‑by‑step bilingual explanations designed to strengthen your understanding and exam technique. Work through the problems and use the paired English–Chinese commentary to reinforce key concepts.

本文详细解析一份针对 WJEC AS 统计课程编写的单元测试模拟卷。每道题覆盖一个核心主题，配有逐步的双语讲解，旨在加深理解、提升应试技巧。请跟随题目和对应的中英文解析巩固关键概念。

1. Descriptive Statistics & Box Plot | 描述统计与箱线图

The following data show the test scores of 20 students:
56, 78, 82, 45, 69, 91, 73, 88, 62, 77, 84, 59, 70, 68, 95, 66, 55, 81, 72, 64

以下数据是 20 名学生的测验分数：
56, 78, 82, 45, 69, 91, 73, 88, 62, 77, 84, 59, 70, 68, 95, 66, 55, 81, 72, 64

Sort the data in ascending order.

将数据按升序排列。

Ordered: 45, 55, 56, 59, 62, 64, 66, 68, 69, 70, 72, 73, 77, 78, 81, 82, 84, 88, 91, 95

排序后：45, 55, 56, 59, 62, 64, 66, 68, 69, 70, 72, 73, 77, 78, 81, 82, 84, 88, 91, 95

Calculate the mean: sum = 1435, n = 20 ⇒ x̄ = 1435/20 = 71.75

计算平均值：总和为 1435，n = 20 ⇒ 均值 x̄ = 1435/20 = 71.75

Median: the average of the 10th (70) and 11th (72) values = 71

中位数：第 10 个（70）和第 11 个（72）数值的平均值 = 71

Q₁: position 0.25 × 21 = 5.25 ⇒ 62 + 0.25 × (64 − 62) = 62.5

下四分位数 Q₁：位置 0.25 × 21 = 5.25 ⇒ 62 + 0.25 × (64 − 62) = 62.5

Q₃: position 0.75 × 21 = 15.75 ⇒ 81 + 0.75 × (82 − 81) = 81.75

上四分位数 Q₃：位置 0.75 × 21 = 15.75 ⇒ 81 + 0.75 × (82 − 81) = 81.75

IQR = Q₃ − Q₁ = 81.75 − 62.5 = 19.25

四分位距 IQR = 81.75 − 62.5 = 19.25

Outlier bounds: lower = Q₁ − 1.5 × IQR = 62.5 − 28.875 = 33.625; upper = Q₃ + 1.5 × IQR = 81.75 + 28.875 = 110.625. No values lie outside, so there are no outliers.

离群值界限：下界 = 62.5 − 28.875 = 33.625；上界 = 81.75 + 28.875 = 110.625。所有数据均在界限内，故无离群值。

The box plot will show a box from 62.5 to 81.75 with a median line at 71, and whiskers extending to 45 and 95.

箱线图中，箱子从 62.5 延伸到 81.75，中位线位于 71，触须延伸至最小值 45 和最大值 95。

2. Probability & Venn Diagrams | 概率与文氏图

In a survey, 60% of adults read newspaper A, 50% read newspaper B, and 30% read both. Find the probability that a randomly chosen adult reads at least one newspaper, and find P(A | B).

一项调查显示，60% 的成年人阅读报纸 A，50% 阅读报纸 B，30% 两者都读。求随机选取一名成年人至少阅读一种报纸的概率，并求 P(A | B)。

Let A = reads A, B = reads B. P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3.

设 A = 阅读 A，B = 阅读 B。P(A) = 0.6，P(B) = 0.5，P(A ∩ B) = 0.3。

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.5 − 0.3 = 0.8

至少阅读一种的概率 P(A ∪ B) = 0.6 + 0.5 − 0.3 = 0.8

P(A | B) = P(A ∩ B) / P(B) = 0.3 / 0.5 = 0.6

在阅读 B 的条件下也阅读 A 的概率为 P(A | B) = 0.3 / 0.5 = 0.6

3. Discrete Random Variable | 离散随机变量

The probability distribution of a discrete random variable Y is P(Y = y) = k(2y − 1) for y = 1, 2, 3, and 0 otherwise. Determine k, E(Y) and Var(Y).

离散随机变量 Y 的概率分布为 P(Y = y) = k(2y − 1)，其中 y = 1, 2, 3，其他情况为 0。求 k、E(Y) 和 Var(Y)。

For y = 1: P= k(2−1)=k; y = 2: P= k(4−1)=3k; y = 3: P= k(6−1)=5k.

y = 1：概率为 k；y = 2：3k；y = 3：5k。

Total probability = k + 3k + 5k = 9k = 1 ⇒ k = 1/9

概率总和 9k = 1 ⇒ k = 1/9

Distribution: y = 1: 1/9; y = 2: 3/9 = 1/3; y = 3: 5/9.

分布列：y=1 概率 1/9；y=2 概率 1/3；y=3 概率 5/9。

E(Y) = 1×(1/9) + 2×(3/9) + 3×(5/9) = (1 + 6 + 15)/9 = 22/9 ≈ 2.444

E(Y²) = 1²×(1/9) + 4×(3/9) + 9×(5/9) = (1 + 12 + 45)/9 = 58/9

Var(Y) = E(Y²) − [E(Y)]² = 58/9 − (22/9)² = (522 − 484)/81 = 38/81 ≈ 0.469

4. Binomial Distribution | 二项分布

A biased coin lands heads with probability 0.3. In 10 tosses, find the probability of exactly 4 heads, and the probability of at least 7 heads.

一枚不均匀硬币出现正面的概率为 0.3。抛掷 10 次，求恰好出现 4 次正面的概率，以及至少 7 次正面的概率。

Let X ~ B(10, 0.3). Then P(X = 4) = ¹⁰C₄ × 0.3⁴ × 0.7⁶

设 X ~ B(10, 0.3)。P(X = 4) = ¹⁰C₄ × 0.3⁴ × 0.7⁶

¹⁰C₄ = 210, 0.3⁴ = 0.0081, 0.7⁶ ≈ 0.117649

P(X = 4) ≈ 210 × 0.0081 × 0.117649 ≈ 0.2001

P(X ≥ 7) = P(7) + P(8) + P(9) + P(10). Using cumulative tables or repeated use of formula, P(X ≥ 7) ≈ 0.0106 + 0.0014 + 0.0001 + 0.0000 ≈ 0.0121

P(X ≥ 7) ≈ 0.0121，概率非常小。

5. Normal Distribution – Forward & Inverse | 正态分布——正查与反查

Apple weights are normally distributed with mean μ = 150 g and standard deviation σ = 20 g. Find the proportion of apples weighing less than 130 g, and the weight exceeded by 90% of the apples.

苹果重量服从正态分布，均值 μ = 150 g，标准差 σ = 20 g。求重量小于 130 g 的比例，以及被 90% 苹果超过的重量。

For X < 130: Z = (130 − 150)/20 = −1. P(Z < −1) ≈ 0.1587

小于 130 g：Z = (130−150)/20 = −1，查表得 P(Z < −1

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