SQA Higher Statistics Unit Test Mock Paper Walkthrough | SQA 高等统计单元测试模拟卷解析

📚 SQA Higher Statistics Unit Test Mock Paper Walkthrough | SQA 高等统计单元测试模拟卷解析

This article provides a full walkthrough of a mock unit test for SQA Higher Statistics. Each question is carefully selected to match the style and content of real SQA assessments, covering descriptive statistics, probability, discrete random variables, binomial and normal distributions, confidence intervals, hypothesis testing, and bivariate analysis. Detailed step-by-step solutions are presented alongside mark allocation rationale, helping Year 12 learners master exam technique and deepen conceptual understanding.

本文为 SQA 高等统计单元测试模拟卷提供完整逐题解析。所有试题均贴近真实 SQA 测评风格与范围，涵盖描述统计、概率、离散随机变量、二项分布与正态分布、置信区间、假设检验以及双变量分析。解析配有详细的步骤分采分点说明，帮助 Year 12 学生完善答题策略、巩固核心概念。

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1. Overview of the Mock Paper | 模拟卷概览

The mock paper comprises 8 compulsory questions and is designed to be completed in 90 minutes under exam conditions. A formula sheet and statistical tables (Normal, t-distribution, binomial cumulative, and PMCC critical values) are assumed to be provided. Marks range from 4 to 10 per question, with a total of 60 marks. It is strongly recommended that candidates show all working and clearly state hypotheses, test statistics, critical values and conclusions for inferential questions.

本模拟卷共 8 道必答题，设计为 90 分钟闭卷完成。假定提供公式表与统计用表（正态分布、t 分布、二项累积表和积矩相关系数临界值表）。每题分值 4 至 10 分，总分 60 分。强烈建议考生展示完整过程，并在推断性试题中明确写出假设、检验统计量、临界值与结论。

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2. Question 1: Descriptive Statistics | 问题1：描述统计

A class of 10 students obtained the following test marks:
56, 62, 71, 58, 65, 69, 74, 61, 68, 66.
Calculate the mean and the sample standard deviation. (4 marks)

某班级 10 名学生的测验成绩如下：56, 62, 71, 58, 65, 69, 74, 61, 68, 66。计算均值与样本标准差。（4 分）

Step 1: Compute the sum Σx = 56 + 62 + 71 + 58 + 65 + 69 + 74 + 61 + 68 + 66 = 650.

步骤1：求和 Σx = 650。

Step 2: Mean x̄ = Σx / n = 650 / 10 = 65.

步骤2：均值 x̄ = 650 / 10 = 65。

Step 3: Find the deviations (x − x̄ ²=36, (−7)²=49, (0)²=0, (4)²=16, (9)²=81, (−4)²=16, (3)²=9, (1)²=1. Sum of squares Σ(x − x̄)² = 81+9+36+49+0+16+81+16+9+1 = 298.

步骤3：计算离差平方： (−9)²=81, (−3)²=9, 6²=36, (−7)²=49, 0²=0, 4²=16, 9²=81, (−4)²=16, 3²=9, 1²=1。平方和 Σ(x − x̄)² = 298。

Step 4: Sample variance s² = Σ(x − x̄)² / (n − 1) = 298 / 9 ≈ 33.111. Sample standard deviation s = √33.111 ≈ 5.75 (to 2 decimal places).

步骤4：样本方差 s² = 298 / 9 ≈ 33.111，样本标准差 s = √33.111 ≈ 5.75（保留两位小数）。

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3. Question 2: Probability with Venn Diagrams | 问题2：概率与文氏图

In a survey of a sixth form, 60% of students study Mathematics (M), 45% study Physics (P), and 25% study both subjects. A student is chosen at random. Find (a) P(M ∪ P) and (b) P(P | M). (5 marks)

在某中学的调查中，60% 的学生学习数学（M），45% 学习物理（P），25% 同时学习两门科目。随机选取一名学生，求 (a) P(M ∪ P) 和 (b) P(P | M)。（5 分）

We are given: P(M) = 0.60, P(P) = 0.45, P(M ∩ P) = 0.25.

已知：P(M) = 0.60, P(P) = 0.45, P(M ∩ P) = 0.25。

(a) Using the addition rule: P(M ∪ P) = P(M) + P(P) − P(M ∩ P) = 0.60 + 0.45 − 0.25 = 0.80.

(a) 运用加法公式：P(M ∪ P) = 0.60 + 0.45 − 0.25 = 0.80。

(b) Conditional probability: P(P | M) = P(M ∩ P) / P(M) = 0.25 / 0.60 = 5/12 ≈ 0.4167.

(b) 条件概率：P(P | M) = 0.25 / 0.60 = 5/12 ≈ 0.4167。

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

The discrete random variable X has the probability distribution shown below. Find E(X) and Var(X). (5 marks)

离散随机变量 X 的概率分布如下表所示。求 E(X) 与 Var(X)。（5 分）

x	1	2	3	4
P(X=x)	0.1	0.3	0.4	0.2

on: E(X) = Σ x·P(X=x) = (1)(0.1) + (2)(0.3) + (3)(0.4) + (4)(0.2) = 0.1+0.6+1.2+0.8 = 2.7.

计算期望：E(X) = 1×0.1 + 2×0.3 + 3×0.4 + 4×0.2 = 2.7。

(2²)(0.3) + (3²)(0.4) + (4²)(0.2) = 0.1 + 4×0.3 + 9×0.4 + 16×0.2 = 0.1+1.2+3.6+3.2 = 8.1.

计算 E(X²) = 1²×0.1 + 4×0.3 + 9×0.4 + 16×0.2 = 8.1。

Then Var(X) = E(X²) − [E(X)]² = 8.1 − (2.7)² = 8.1 − 7.29 = 0.81.

故 Var(X) = 8.1 − 2.7² = 0.81。

Var(X) = Σ(x − μ)²P(X=x) = 0.81

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5. Question 4: Binomial Distribution | 问题4：二项分布

A biased coin lands heads with probability 0.3. The coin is tossed 10 times. Find the probability of (a) exactly 4 heads, (b) at least 8 heads. (7 marks)

一枚不均匀硬币正面朝上的概率为 0.3。抛掷该硬币 10 次。求 (a) 恰好 4 次正面朝上的概率，(b) 至少 8 次正面的概率。（7 分）

Let X ~ B(10, 0.3). The probability mass function is P(X=k) = C(10,k) × (0.3)^k × (0.7)^10−k.

设 X ~ B(10, 0.3)。概率质量函数为 P(X=k) = C(10,k) × 0.3^k × 0.7^10−k。

. (0.3)⁴ = 0.0081, (0.7)⁶ ≈ 0.117649. Product = 210 × 0.0081 × 0.117649 ≈ 0.2001 (4 d.p.).

(a) P(X=4) = 210 × 0.0081 × 0.117649 ≈ 0.2001（保留四位小数）。

(b) P(X ≥ 8) = P(X=8) + P(X=9) + P(X=10).
P(X=8) = C(10,8)×(0.3)⁸×(0.7)² = 45 × 0.00006561 × 0.49 ≈ 0.00145.
P(X=9) = C(10,9)×(0.3)⁹×(0.7)¹ = 10 × 0.000019683 × 0.7 ≈ 0.000138.
P(X=10) = (0.3)¹⁰ ≈ 0.0000059.
Sum ≈ 0.00145 + 0.000138 + 0.000006 = 0.00159 (to 5 d.p.).

(b) P(X ≥ 8) = P(X=8)+P(X=9)+P(X=10) ≈ 0.00159。计算各项后求和即得。

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6. Question 5: Normal Distribution | 问题5：正态分布

(a) Find the probability that a randomly selected apple weighs between 140 g and 170 g. (b) Find the weight exceeded by only 10% of the apples. (6 marks)

某果园苹果重量服从正态分布，均值 μ = 150 g，标准差 σ = 20 g。(a) 求随机选取一个苹果重量介于 140 g 与 170 g 之间的概率。(b) 求只有 10% 的苹果会超过的重量。（6 分）

Let X ~ N(150, 20²). Convert to standard normal Z = (X − μ)/σ.

令 X ~ N(150, 20²)，标准化 Z = (X − 150)/20。

(a) For x₁=140: Z₁ = (140−150

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