📚 CAIE AS Statistics: Cross-disciplinary Comprehensive Question Training | CAIE AS 统计学：跨学科综合题型训练
Cross-disciplinary problems in CAIE AS Statistics bridge theoretical concepts with real-world applications, from biology to economics. This article provides a series of integrated exercises that train you to apply probability distributions, hypothesis testing, correlation and regression in varied contexts. Each example includes a detailed bilingual walkthrough to sharpen your exam skills.
CAIE AS 统计学中的跨学科题目将理论概念与现实应用相结合，涵盖生物、经济等领域。本文提供一系列综合练习，训练你将概率分布、假设检验、相关与回归应用于不同情境的能力。每个例题均配有双语详细解析，助你提升应试技巧。
1. Biology: Normal Distribution of Leaf Lengths | 生物学：叶片长度的正态分布
The lengths of leaves from a certain plant species are normally distributed with mean μ = 8.5 cm and standard deviation σ = 1.2 cm. A biologist selects leaves at random.
某种植物的叶片长度服从正态分布，均值 μ = 8.5 cm，标准差 σ = 1.2 cm。一位生物学家随机选取叶片。
(i) Find the probability that a randomly chosen leaf is longer than 10 cm.
(i) 求随机选取一片叶子的长度超过 10 cm 的概率。
(ii) If a sample of 10 leaves is taken, find the probability that at least 8 are shorter than 9 cm. State any assumptions.
(ii) 若抽取 10 片叶子组成样本，求至少有 8 片长度短于 9 cm 的概率，并说明所需假设。
Solution (i): Standardise: Z = (10 – 8.5)/1.2 = 1.25. Using the normal table, Φ(1.25) = 0.8944. Hence P(X > 10) = 1 – 0.8944 = 0.1056.
解析 (i)：标准化：Z = (10 – 8.5)/1.2 = 1.25。查标准正态表，Φ(1.25) = 0.8944，因此 P(X > 10) = 1 – 0.8944 = 0.1056。
Solution (ii): First compute P(X < 9). Z = (9 – 8.5)/1.2 ≈ 0.4167. Φ(0.4167) ≈ 0.6616. Let p = 0.6616. Assume leaves are independent, so the number Y of leaves shorter than 9 cm follows B(10, 0.6616). We require P(Y ≥ 8) = P(8)+P(9)+P(10).
解析 (ii)：先计算单叶短于 9 cm 的概率：Z ≈ 0.4167，Φ(0.4167) ≈ 0.6616，取 p = 0.6616。假设叶片间独立，则样本中短于 9 cm 的叶片数 Y ~ B(10, 0.6616)。需求 P(Y ≥ 8) = P(8)+P(9)+P(10)。
Using the binomial formula:
利用二项公式：
P(Y=8) = C(10,8) p⁸ (1-p)² ≈ 45 × 0.6616⁸ × 0.3384² ≈ 0.156
P(Y=9) = C(10,9) p⁹ (1-p) ≈ 10 × 0.6616⁹ × 0.3384 ≈ 0.070
P(Y=10) = p¹⁰ ≈ 0.6616¹⁰ ≈ 0.016
Sum ≈ 0.156 + 0.070 + 0.016 = 0.242. Thus the probability that at least 8 of the 10 leaves are shorter than 9 cm is about 0.242 (3 s.f.).
总和约为 0.242。因此 10 片叶子中至少有 8 片短于 9 cm 的概率约为 0.242（三位有效数字）。
2. Economics: Correlation and Regression – Inflation vs Unemployment | 经济学：相关与回归——通货膨胀率与失业率
An economist collects annual data from six countries, recording unemployment rate (x, %) and inflation rate (y, %). The aim is to investigate a possible linear relationship.
一位经济学家收集了六个国家的年度数据，记录了失业率 (x, %) 与通胀率 (y, %)，以探究两者之间可能存在的线性关系。
The data are summarised below:
数据汇总如下：
| Country | x (unemployment) | y (inflation) |
|---|---|---|
| A | 4.2 | 2.1 |
| B | 5.8 | 3.0 |
| C | 6.5 | 3.4 |
| D | 7.1 | 3.8 |
| E | 8.0 | 4.5 |
| F | 9.4 | 5.2 |
n = 6, Σx = 41.0, Σy = 22.0, Σx² = 295.90, Σy² = 87.30, Σxy = 160.84.
n = 6, Σx = 41.0, Σy = 22.0, Σx² = 295.90, Σy² = 87.30, Σxy = 160.84。
(i) Calculate the product moment correlation coefficient r.
(i) 计算积矩相关系数 r。
(ii) Find the equation of the regression line of y on x.
(ii) 求 y 对 x 的回归直线方程。
Solution (i): Sxy = Σxy – (Σx Σy)/n = 160.84 – (41.0×22.0)/6 = 160.84 – 150.333 = 10.507. Sxx = Σx² – (Σx)²/n = 295.90 – 41.0²/6 = 295.90 – 280.1667 = 15.7333. Syy = Σy² – (Σy)²/n = 87.30 – 22.0²/6 = 87.30 – 80.6667 = 6.6333.
解析 (i)：Sxy = Σxy – (Σx Σy)/n = 160.84 – (41.0×22.0)/6 = 10.507。Sxx = Σx² – (Σx)²/n = 15.7333。Syy = Σy² – (Σy)²/n = 6.6333。
r = Sxy / √(Sxx Syy) = 10.507 / √(15.7333 × 6.6333) = 10.507 / √104.37 ≈ 10.507 / 10.216 ≈ 1.028… Wait, this exceeds 1. Let’s correct data to ensure r < 1. Adjusted raw data: use consistent x, y to give r ≈ 0.98. I’ll recalculate: Instead choose x 3,5,6,7,8,10; y 1.5,2.4,3.0,3.6,4.2,5.0. Σx=39, Σy=19.7, Σx²=299, Σy²=75.61, Σxy=145.5. Then Sxy=145.5 – (39*19.7)/6 =145.5 – 128.05=17.45, Sxx=299 – 39²/6=299 – 253.5=45.5, Syy=75.61 – 19.7²/6=75.61 – 64.683=10.927. r=17.45/√(45.5*10.927)=17.45/√497.3≈17.45/22.30=0.783. So r=0.783. good. Need to update table to match. I will replace table with new numbers: Country A: x=3.0, y=1.5; B:5.0,2.4; C:6.0,3.0; D:7.0,3.6; E:8.0,4.2; F:10.0,5.0. Then Σx=39, Σy=19.7. Σx²=299, Σy²=75.61, Σxy=145.5. And edit.
为了得到合理的 r，修正数据如下：
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