A-Level Cambridge Statistics: Unit Test Mock Paper Walkthrough | A-Level 剑桥统计：单元测试模拟卷解析

📚 A-Level Cambridge Statistics: Unit Test Mock Paper Walkthrough | A-Level 剑桥统计：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for Cambridge A-Level Statistics, covering core topics from Probability & Statistics 1 and 2. Each question is broken down with step-by-step solutions, exam tips, and bilingual explanations to help you master key concepts and boost your confidence.

本文详细解析了一份剑桥 A-Level 统计学的单元测试模拟卷，涵盖概率与统计 1 和 2 的核心主题。每道题均有逐步求解、考试技巧与中英双语解释，帮助你掌握关键概念并增强应考信心。

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1. Probability and Counting Principles | 概率与计数原理

A committee of 4 is selected at random from 7 men and 5 women. Find the probability that the committee contains at least one woman.

从 7 名男性和 5 名女性中随机选出 4 人组成委员会。求委员会中至少有一名女性的概率。

Total selections = ¹²C₄ = 495. The complement of 'at least one woman' is 'all men'.

总选择数 = ¹²C₄ = 495。'至少一名女性'的补集为'全是男性'。

Number of all-men committees = ⁷C₄ = 35.

全为男性的委员会数目 = ⁷C₄ = 35。

P(at least one woman) = 1 - (35/495) = 460/495 = 92/99.

P(至少一名女性) = 1 - (35/495) = 460/495 = 92/99。

Always check whether the complement is faster to compute – in 'at least' problems it often is.

务必检查补集是否更易计算——在涉及'至少'的问题中往往如此。

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2. Discrete Random Variable | 离散随机变量

x = 1,2,3,4 with P(X=x) = 0.2, 0.3, k, 0.1. Find k, E(X) and Var(X), and evaluate E(2X+3).

随机变量 X 的分布为：x = 1,2,3,4，对应概率 0.2, 0.3, k, 0.1。求 k, E(X), Var(X) 并计算 E(2X+3)。

Sum of probabilities: 0.2 + 0.3 + k + 0.1 = 1 → k = 0.4.

概率之和为 1：0.2 + 0.3 + k + 0.1 = 1 → k = 0.4。

E(X) = 1(0.2) + 2(0.3) + 3(0.4) + 4(0.1) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4.

E(X) = 1×0.2 + 2×0.3 + 3×0.4 + 4×0.1 = 2.4。

E(X²) = 1²(0.2) + 2²(0.3) + 3²(0.4) + 4²(0.1) = 0.2 + 1.2 + 3.6 + 1.6 = 6.6.

E(X²) = 1²×0.2 + 2²×0.3 + 3²×0.4 + 4²×0.1 = 6.6。

Var(X) = E(X²) - [E(X)]² = 6.6 - 2.4² = 6.6 - 5.76 = 0.84.

Var(X) = 0.84。

Using linearity: E(2X+3) = 2E(X) + 3 = 2(2.4) + 3 = 4.8 + 3 = 7.8.

利用线性性质：E(2X+3) = 2E(X) + 3 = 7.8。

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3. Binomial Distribution | 二项分布

X ~ B(15, 0.4). Find P(X = 6), P(X ≤ 4) and state the shape of the distribution.

X ~ B(15, 0.4)。求 P(X = 6), P(X ≤ 4) 并说明分布的形状。

P(X = 6) = ¹⁵C₆ (0.4)⁶(0.6)⁹. Compute stepwise: ¹⁵C₆ = 5005, (0.4)⁶ ≈ 0.004096, (0.6)⁹ ≈ 0.010077, product ≈ 0.2066.

P(X = 6) = ¹⁵C₆(0.4)⁶(0.6)⁹。分步计算：¹⁵C₆ = 5005，(0.4)⁶ ≈ 0.004096，(0.6)⁹ ≈ 0.010077，乘积 ≈ 0.2066。

6¹³)≈0.0219, P(X=3)=455(0.4³)(0.6¹²)≈0.0659, P(X=4)=1365(0.4⁴)(0.6¹¹)≈0.1268. Sum ≈ 0.2198.

P(X ≤ 4) = 逐项求和 ≈ 0.2198。

Since p = 0.4 < 0.5, the distribution is positively skewed (longer tail to the right).

由于 p = 0.4 < 0.5，分布呈正偏态（右侧尾部更长）。

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4. Poisson Distribution | 泊松分布

Y ~ Po(3.5). Calculate P(Y = 2), P(Y ≥ 4) and find the most likely number of occurrences.

Y ~ Po(3.5)。计算 P(Y = 2), P(Y ≥ 4) 并找出最可能出现的次数。

P(Y = 2) = (e⁻³·⁵ × 3.5²)/2! = (0.030197 × 12.25)/2 ≈ 0.1850.

P(Y = 2) ≈ 0.1850。

)≈0.1850, P(3)=(3.5³/6)e⁻³·⁵≈(42.875/6)×0.0302≈0.2158. Sum = 0.5367, so P(Y ≥ 4) ≈ 0.4633.

P(Y ≥ 4) = 1 - [P(0)+P(1)+P(2)+P(3)] ≈ 0.4633。

The mode is floor(λ) = 3, with P(Y=3) highest at 0.2158.

众数为 λ 向下取整 = 3，P(Y=3) 最高为 0.2158。

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5. Normal Distribution Calculations | 正态分布计算

X ~ N(50, 8²). Find P(X < 45), P(55 < X < 62), and the value of c such that P(X > c) = 0.1.

X ~ N(50, 8²)。求 P(X < 45), P(55 < X < 62)，以及满足 P(X > c) = 0.1 的 c 值。

)/8. For 45, Z = (45 - 50)/8 = -0.625. P(Z < -0.625) = 1 - Φ(0.625) ≈ 1 - 0.7340 = 0.2660.

标准化：Z = (45-50)/8 = -0.625。P(Z < -0.625) = 1 - Φ(0.625) ≈ 0.2660。

, Z₂ = (62-50)/8 = 1.5. P(0.625 < Z < 1.5) = Φ(1.5) - Φ(0.625) ≈ 0.9332 - 0.7340 = 0.1992.

介于 55 和 62 之间：Z 值 0.625 和 1.5，概率差 ≈ 0.1992。

For P(X > c) = 0.1, P(Z > z) = 0.1 → z ≈ 1.2816. Then c = 50 + 1.2816×8 ≈ 60.25.

由 P(X > c) = 0.1 知右侧尾部 z = 1.2816，c = 50 + 1.2816×8 ≈ 60.25。

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6. Normal Approximation to Binomial | 二项分布的正态近似

X ~ B(200, 0.3). Use a normal approximation with continuity correction to estimate P(X ≤ 50).

X ~ B(200, 0.3)。使用正态近似并作连续校正，估计 P(X ≤ 50)。

Check conditions: np = 200×0.3 = 60 > 5, nq = 200×0.7 = 140 > 5, valid.

检验条件：np=60 > 5, nq=140 > 5，可近似。

μ = np = 60, σ² = npq = 60×0.7 = 42, so σ = √42 ≈ 6.4807.

均值 μ = 60，方差 42，σ ≈ 6.4807。

With continuity correction, P(X ≤ 50) ≈ P(Y < 50.5) for Y ~ 42). Z = (50.5 - 60)/6.4807 ≈ -1.466. P(Z < -1.466) = 1 - Φ(1.466) ≈ 1 - 0.9286 = 0.0714.

连续校正：P(X ≤ 50) ≈ P(Y < 50.5)，Z = -1.466，概率 ≈ 0.0714。

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7. Stem-and-Leaf and Box Plots | 茎叶图与箱线图

A dataset of 15 values: 12, 15, 21, 22, 23, 25, 28, 30, 32, 35, 38, 42, 48, 55, 60. Construct a stem-and-leaf plot, find the five-number summary, and draw a box plot. Identify any outliers.

一组 15 个数据：12,15,21,22,23,25,28,30,32,35,38,42,48,55,60。绘制茎叶图，求五数概括，画箱线图，并识别异常值。

Stem-and-leaf (stem=tens): 1|2,5 ; 2|1,2,3,5,8 ; 3|0,2,5,8 ; 4|2,8 ; 5|5 ; 6|0.

茎叶图（茎为十位）：1|2,5 ; 2|1,2,3,5,8 ; 3|0,2,5,8 ; 4|2,8 ; 5|5 ; 6|0。

(23) = 22.5, Median (8th) = 30, Q₃ (12th) = average of 12th(38) and 13th(42) = 40, Max=60.

五数概括：最小值 12，下四分位数 22.5，中位数 30，上四分位数 40，最大值 60。

rs. Upper fence = Q₃ + 1.5×IQR = 40 + 26.25 = 66.25, so 60 is not an outlier. No outliers.

IQR = 17.5。下界限 -3.75，上界限 66.25，无异常值。

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8. Cumulative Frequency and Percentiles | 累积频率与百分位数

Grouped data: 0-10 (freq 5), 10-20 (12), 20-30 (18), 30-40 (10), 40-50 (5). Estimate the median and 80th percentile using linear interpolation.

分组数据：0-10 (频数5), 10-20 (12), 20-30 (18), 30-40 (10), 40-50 (5)。用线性插值估计中位数和第80百分位数。

20, class width 10, freq 18. Median = 20 + ((25.5 - 17)/18)×10 = 20 + (8.5/18)×10 ≈ 24.72.

总频数 50。中位数位置 25.5，位于 20-30 组，计算公式：20 + ((25.5-17)/18)×10 ≈ 24.72。

40th value, falls in 30-40 class (CF before=35). P₈₀ = 30 + ((40-35)/10)×10 = 30 + 5 = 35.

第80百分位数位置 40，位于 30-40 组，P₈₀ = 30 + ((40-35)/10)×10 = 35。

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9. Probability Trees and Conditional Probability | 概率树图与条件概率

Bag A has 4 red and 3 blue balls; Bag B has 5 red and 2 blue. A ball is drawn at random from A and placed into B. Then a ball is drawn from B. Given that the final ball is red, find the probability the transferred ball was blue.

A袋有4红3蓝；B袋有5红2蓝。从A随机取出一球放入B，随后从B取一球。已知最终取出红球，求转移球是蓝色的概率。

Tree: Transfer Red (4/7) → B has 6R,2B → P(Red|Red trans)=6/8=3/4. Transfer Blue (3/7) → B has 5R,3B → P(Red|Blue trans)=5/8.

树图：转移红球概率4/7，若转移红则B中6红2蓝，取红概率3/4。转移蓝球概率3/7，B中5红3蓝，取红概率5/8。

Total P(final Red) = (4/7)×(3/4) + (3/7)×(5/8) = (12/28)+(15/56) = (24/56)+(15/56)=39/56.

Published by TutorHao | A-Level 统计 Revision Series | aleveler.com

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