AS Edexcel Statistics Mock Exam Solutions & Analysis | AS Edexcel 统计模拟卷解析

📚 AS Edexcel Statistics Mock Exam Solutions & Analysis | AS Edexcel 统计模拟卷解析

This article walks you through a complete AS Edexcel Statistics mock paper, providing detailed solutions and examiner-style commentary for every question. Covering grouped frequency, probability, discrete random variables, binomial and normal distributions, and correlation–regression, the walkthrough sharpens both your conceptual understanding and exam technique.

本文将逐题解析一份适用于 AS Edexcel 统计的完整模拟卷，涵盖分组频率、概率、离散随机变量、二项分布与正态分布、相关与回归等核心主题，并附有详细的解题步骤与考官视角的点评，帮助你巩固知识、提升应试技巧。

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1. Mock Exam Overview | 模拟卷概述

The mock paper consists of six compulsory questions, totalling 60 marks, to be completed in 75 minutes. Question styles mirror those seen in AS Edexcel Statistics units, mixing calculation, interpretation and diagrammatic work. A calculator is allowed, and statistical tables are provided where needed.

模拟卷共有六道必答题，满分 60 分，建议用时 75 分钟。题型贴近 AS Edexcel 统计真题，融合计算、解释与图示。考试允许使用计算器，需要时提供统计表。

The following table summarises the topic coverage and mark allocation. Use this to plan your revision focus and time management during practice.

下表总结了各题考点与分值，可用于复习重点规划和限时训练中的时间分配。

Question	Topic	Marks
1	Grouped frequency, histogram, mean & standard deviation	12
2	Probability without replacement, tree diagrams, conditional probability	10
3	Discrete random variable, expectation and variance	8
4	Binomial distribution, probability calculations	10
5	Normal distribution, standardisation, inverse normal	10
6	Linear regression, interpretation of gradient	10

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2. Question 1: Grouped Frequency & Histogram | 第1题：分组频率表与直方图

A botanist measures the heights of 40 young sunflower plants. The results are summarised in the grouped frequency table below. (a) Calculate an estimate for the mean height. (b) Find the median class and estimate the median. (c) Compute the standard deviation. (d) Draw a histogram to represent the data.

一位植物学家测量了 40 株向日葵幼苗的高度，数据如下表所示。(a) 计算平均高度的估计值；(b) 找出中位数组并估计中位数；(c) 计算标准差；(d) 绘制直方图。

Height (cm)	Frequency (f)
0 ≤ h < 10	5
10 ≤ h < 20	8
20 ≤ h < 30	12
30 ≤ h < 50	10
50 ≤ h < 70	5

Part (a) – Mean: First, determine the midpoint x for each class, then compute fx and Σfx. The estimated mean is Σfx / Σf. Midpoints: 5, 15, 25, 40, 60. Thus Σfx = 5×5 + 8×15 + 12×25 + 10×40 + 5×60 = 25 + 120 + 300 + 400 + 300 = 1145. Σf = 40. Mean = 1145 ÷ 40 = 28.625 cm.

(a) 平均值：先求各组中点 x，再计算 fx 与 Σfx。估计均值 = Σfx / Σf。中点：5, 15, 25, 40, 60。Σfx = 5×5 + 8×15 + 12×25 + 10×40 + 5×60 = 1145。Σf = 40。均值 = 1145 ÷ 40 = 28.625 cm。

Part (b) – Median: Cumulative frequencies are 5, 13, 25, 35, 40. The median position is (40+1)/2 = 20.5th value, which lies in the 20 ≤ h < 30 class. Using interpolation: median = 20 + ( (20.5 – 13) / 12 ) × 10 = 20 + (7.5/12)×10 ≈ 26.25 cm.

(b) 中位数：累积频率为 5, 13, 25, 35, 40。中位数位置为 (40+1)/2 = 20.5，位于 20 ≤ h < 30 组内。采用线性插值：中位数 = 20 + ( (20.5 – 13) / 12 ) × 10 ≈ 26.25 cm。

Part (c) – Standard deviation: Use the formula s = √[ Σf x² / n – (x̄)² ] for grouped data (with divisor n). Compute Σf x² = 5×5² + 8×15² + 12×25² + 10×40² + 5×60² = 5×25 + 8×225 + 12×625 + 10×1600 + 5×3600 = 125 + 1800 + 7500 + 16000 + 18000 = 43425. Then variance = 43425/40 – 28.625² = 1085.625 – 819.390625 = 266.234375. So standard deviation s = √266.234375 ≈ 16.32 cm.

(c) 标准差：使用公式 s = √[ Σf x² / n – (x̄)² ]。Σf x² = 5×5² + 8×15² + 12×25² + 10×40² + 5×60² = 43425。方差 = 43425/40 – 28.625² ≈ 266.234。标准差 = √266.234 ≈ 16.32 cm。

Part (d) – Histogram: Since class widths are unequal, plot frequency density = frequency / class width. Widths: 10, 10, 10, 20, 20. Densities: 5/10=0.5, 8/10=0.8, 12/10=1.2, 10/20=0.5, 5/20=0.25. Draw bars with these heights on a properly scaled axes; the area of each bar is proportional to frequency.

(d) 直方图：由于组距不相等，需使用频率密度 = 频率 ÷ 组距。组距：10, 10, 10, 20, 20。频率密度：0.5, 0.8, 1.2, 0.5, 0.25。绘制时以频率密度为纵轴，横轴为高度，每个直条的面积与频率成正比。

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3. Question 2: Probability Without Replacement & Tree Diagrams | 第2题：不放回概率与树状图

A bag contains 3 red (R) and 5 blue (B) balls. Two balls are drawn at random without replacement. (a) Draw a fully labelled tree diagram showing all probabilities. (b) Find the probability that both balls are red. (c) Find the probability that exactly one is blue. (d) Given the second ball is red, find the probability that the first ball was blue.

袋中有 3 个红球和 5 个蓝球。随机抽取两球且不放回。(a) 画出完整的树状图并标出所有概率；(b) 求两球均为红色的概率；(c) 求恰好一个蓝球的概率；(d) 已知第二球为红色，求第一球为蓝色的概率。

Tree diagram: First draw: P(R) = 3/8, P(B) = 5/8. Second draw branches: if first R, then P(R|R)=2/7, P(B|R)=5/7; if first B, then P(R|B)=3/7, P(B|B)=4/7. Multiply along branches.

树状图：第一次抽取：P(R) = 3/8, P(B) = 5/8。第二次分支：若第一次为 R，则 P(R|R)=2/7, P(B|R)=5/7；若第一次为 B，则 P(R|B)=3/7, P(B|B)=4/7。沿分支相乘。

(b) Both red: P(RR) = (3/8)×(2/7) = 6/56 = 3/28 ≈ 0.1071.

(b) 两红：P(RR) = (3/8)×(2/7) = 6/56 = 3/28。

(c) Exactly one blue means RB or BR. P(RB) = (3/8)×(5/7)=15/56. P(BR) = (5/8)×(3/7)=15/56. Total = 30/56 = 15/28 ≈ 0.5357.

(c) 恰好一个蓝球：即 RB 或 BR。P(RB)=15/56，P(BR)=15/56，和为 30/56 = 15/28。

(d) Conditional: P(first B | second R) = P(B ∩ second R) / P(second R). P(second R) = P(RR) + P(BR) = 6/56 + 15/56 = 21/56. P(B ∩ second R) = P(BR) = 15/56. Therefore probability = (15/56) / (21/56) = 15/21 = 5/7.

(d) 条件概率：P(第一次 B | 第二次 R) = P(B ∩ 第二次 R) / P(第二次 R) = (15/56) / (21/56) = 15/21 = 5/7。

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4. Question 3: Discrete Random Variable, Expectation & Variance | 第3题：离散随机变量、期望与方差

The probability distribution of a discrete random variable X is given below. (a) Find the value of k. (b) Calculate E(X) and Var(X). (c) Hence find E(2X + 3) and Var(2X + 3).

离散随机变量 X 的概率分布如下。(a) 求 k 的值；(b) 计算 E(X) 和 Var(X)；(c) 进而求 E(2X + 3) 与 Var(2X + 3)。

x	0	1	2	3
P(X=x)	0.1	0.2	k	0.3

(a) Find k: Sum of probabilities = 1 ⇒ 0.1 + 0.2 + k + 0.3 = 1 ⇒ k = 0.4.

(a) k = 0.4：概率总和为 1。

(b) Expectation: E(X) = Σ x P(X=x) = 0×0.1 + 1×0.2 + 2×0.4 + 3×0.3 = 0 + 0.2 + 0.8 + 0.9 = 1.9. E(X²) = 0²×0.1 + 1²×0.2 + 2²×0.4 + 3²×0.3 = 0 + 0.2 + 1.6 + 2.7 = 4.5. Var(X) = E(X²) – [E(X)]² = 4.5 – 1.9² = 4.5 – 3.61 = 0.89.

(b) 期望与方差：E(X) = 1.9；E(X²) = 4.5；Var(X) = 4.5 – 3.61 = 0.89。

(c) Linear transformation: E(2X+3) = 2E(X)+3 = 2×1.9+3 = 6.8. Var(2X+3) = 2² Var(X) = 4 × 0.89 = 3.56.

(c) 线性变换：E(2X+3)=6.8，Var(2X+3)=3.56。

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5. Question 4: Binomial Distribution | 第4题：二项分布

A biased coin lands heads with probability 0.3. The coin is tossed 10 times. Let X be the number of heads. (a) State the distribution of X. (b) Find P(X = 4). (c) Find P(X ≤ 3). (d) Find P(X > 7).

一枚不均匀硬币出现正面的概率为 0.3。抛掷该硬币 10 次，记 X 为正面朝上的次数。(a) 写出 X 的分布；(b) 求 P(X=4)；(c) 求 P(X≤3)；(d) 求 P(X>7)。

(a) Distribution: X ~ B(10, 0.3).

(b) Exactly 4 heads: P(X=4) = ¹⁰C₄ × 0.3⁴ × 0.7⁶. ¹

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