📚 CCEA Year 12 Statistics: Cross-disciplinary Integrated Problem Solving | CCEA 12年级统计：跨学科综合题型训练
In CCEA Year 12 Statistics, you are expected to apply statistical techniques to real-world scenarios spanning many subjects. This cross-disciplinary approach not only deepens your conceptual understanding but also prepares you for exam questions that embed statistics within biology, business, psychology and more. This article provides integrated problem-solving training, with each section presenting a contextual problem, a step-by-step solution and an interpretation — all delivered in paired English and Chinese explanations. Topics covered include binomial, Poisson and normal distributions, hypothesis testing, correlation, regression, chi-squared tests and confidence intervals.
在 CCEA 12 年级统计课程中，你需要将统计方法应用于横跨多个学科的真实场景。这种跨学科的学习方式不仅能加深你对概念的理解，也能帮助你应对那些将统计学嵌入生物、商业、心理学等背景的考试题目。本文提供综合题型训练，每节给出一个情境问题、分步解法和结果解读，全部以中英配对的形式呈现。涵盖的主题包括二项分布、泊松分布、正态分布、假设检验、相关与回归、卡方检验以及置信区间。
1. Biology: Binomial Distribution – Seed Germination | 生物学：二项分布——种子萌发
A biologist claims that a new seed treatment boosts the germination rate to 85%. A researcher plants 15 treated seeds. Let X be the number of seeds that germinate. Assuming the claim is true, find the probability that at least 13 seeds germinate.
一位生物学家声称新的种子处理可以将萌发率提高到 85%。一位研究员种植了 15 颗处理过的种子。设 X 为萌发的种子数。假设该说法正确，求至少 13 颗种子萌发的概率。
X ~ B(15, 0.85)
X follows a binomial distribution with n = 15 and p = 0.85.
X 服从二项分布，参数 n = 15，p = 0.85。
The probability that exactly k seeds germinate is given by P(X = k) = C(15, k) × 0.85ᵏ × 0.15¹⁵⁻ᵏ. We need P(X ≥ 13) = P(13) + P(14) + P(15).
恰好 k 颗种子萌发的概率为 P(X = k) = C(15, k) × 0.85ᵏ × 0.15¹⁵⁻ᵏ。我们需要 P(X ≥ 13) = P(13) + P(14) + P(15)。
C(15,13) = 105, so P(13) = 105 × 0.85¹³ × 0.15² ≈ 0.2856.
C(15,14) = 15, so P(14) = 15 × 0.85¹⁴ × 0.15¹ ≈ 0.2312.
C(15,15) = 1, so P(15) = 0.85¹⁵ ≈ 0.0874.
Summing these gives approximately 0.6042.
C(15,13) = 105，P(13) ≈ 0.2856；C(15,14) = 15，P(14) ≈ 0.2312；C(15,15) = 1，P(15) ≈ 0.0874。三者之和约为 0.6042。
There is a 60.4% chance of obtaining at least 13 germinated seeds if the true germination rate is 85%. This outcome is quite likely, so it would not provide strong evidence against the biologist's claim in a hypothesis test.
若真实萌发率为 85%，则观察到至少 13 颗种子萌发的概率为 60.4%。这一结果相当常见，因此在假设检验中不会构成反对该说法的有力证据。
2. Business: Poisson Distribution – Call Centre Arrivals | 商业：泊松分布——呼叫中心来电
A customer service hotline receives an average of 4.2 calls per minute during peak hours. The arrivals are modelled by a Poisson distribution. Find the probability that in a randomly chosen peak minute the centre receives exactly 2 calls, and the probability that it receives more than 6 calls.
某客服热线在高峰时段平均每分钟接到 4.2 通电话，来电次数可用泊松分布建模。求在随机选择的一个高峰分钟内，恰好接到 2 通电话的概率，以及接到超过 6 通电话的概率。
X ~ Po(4.2)
te; X follows a Poisson distribution with mean λ = 4.2. Then P(X = 2) = (e⁻⁴·² × 4.2²) / 2! .
设 X 为一分钟内接到的电话数，X 服从均值为 λ = 4.2 的泊松分布。则 P(X = 2) = (e⁻⁴·² × 4.2²) / 2!。
g a calculator, e⁻⁴·² ≈ 0.0150, 4.2² = 17.64, so P(X = 2) = (0.0150 × 17.64) / 2 ≈ 0.1323.
使用计算器，e⁻⁴·² ≈ 0.0150，4.2² = 17.64，因此 P(X = 2) = (0.0150 × 17.64) / 2 ≈ 0.1323。
From the Poisson cumulative table with λ = 4.2, P(X ≤ 6) ≈ 0.8669, thus P(X > 6) ≈ 0.1331.
对于 P(X > 6)，使用 P(X > 6) = 1 − P(X ≤ 6)。查泊松累积分布表，λ = 4.2 时 P(X ≤ 6) ≈ 0.8669，因此 P(X > 6) ≈ 0.1331。
The call centre can expect about a 13.2% chance of exactly 2 calls and a 13.3% chance of more than 6 calls in any given peak minute. Such calculations assist in workforce planning.
该热线在任何给定高峰分钟内，恰好接到 2 通电话的概率约 13.2%，接到超过 6 通电话的概率约 13.3%。这类计算有助于人员排班规划。
3. Psychology: Chi-Squared Test for Independence – Gender and Colour Choice | 心理学：卡方独立性检验——性别与颜色偏好
A psychologist surveys 100 participants to see if gender is associated with preference for red or blue. The observed frequencies are: males choose red 20, blue 30; females choose red 35, blue 15. Test at the 5% significance level whether colour preference is independent of gender.
一位心理学家调查了 100 名参与者，探究性别是否与红色或蓝色的偏好相关。观测频数为：男性选红色 20，蓝色 30；女性选红色 35，蓝色 15。在 5% 显著性水平下检验颜色偏好是否与性别独立。
| Red | Blue | Total | |
|---|---|---|---|
| Male | 20 | 30 | 50 |
| Female | 35 | 15 | 50 |
| Total | 55 | 45 | 100 |
Expected frequencies are calculated as (row total × column total) / grand total. For Male–Red: (50 × 55) / 100 = 27.5; Male–Blue: 22.5; Female–Red: 27.5; Female–Blue: 22.5.
期望频数按 (行合计 × 列合计) / 总合计 计算。男性-红色：(50×55)/100=27.5；男性-蓝色：22.5；女性-红色：27.5；女性-蓝色：22.5。
0−22.5)²/22.5 + (35−27.5)²/27.5 + (15−22.5)²/22.5 = 2.045 + 2.500 + 2.045 + 2.500 = 9.090.
检验统计量为 χ² = Σ (O − E)² / E。计算： 30−22.5)²/22.5 + (35−27.5)²/27.5 + (15−22.5)²/22.5 = 2.045 + 2.500 + 2.045 + 2.500 = 9.090。
Degrees of freedom = (2−1)×(2−1) = 1. The critical value from χ²(1) at 5% > 3.841, we reject H₀. There is evidence that gender and colour preference are associated.
自由度 = (2−1)×(2−1) = 1。在 5% 水平下，χ²(1) 的临界值为 3.841。由于 9.090 > 3.841，拒绝原假设。有证据表明性别与颜色偏好之间存在关联。
4. Environmental Science: Normal Distribution – River Nitrate Levels | 环境科学：正态分布——河流硝酸盐浓度
tributed with mean μ = 10.2 and standard deviation σ = 1.8. Regulations class a reading as 'high' randomly taken water sample is classified as high. Also find the level above which only 5% of samples lie.
某河流中硝酸盐浓度 (mg/L) 服从正态分布，均值 μ = 10.2，标准差 σ = 1.8。若读数超过 13.0 mg/L 则被归类为“偏高”。求随机采集的水样被归为偏高的概率，并找出只有 5% 样本会超过的浓度值。
X ~ N(10.2, 1.8²)
3.0, Z = (13.0 − 10.2)/1.8 ≈ 1.5556. Thus P(X > 13.0) = P(Z > 1.56) ≈ 1 − 0.9406 = 0.0594.
首先标准化：Z = (X − μ)/σ。当 X = 13.0 时，Z = (13.0 − 10.2)/1.8 ≈ 1.5556。因此 P(X > 13.0) = P(Z > 1.56) ≈ 1 − 0.9406 = 0.0594。
There is roughly a 5.9% probability that a single sample exceeds the high threshold.
单个样本超过偏高阈值的概率约为 5.9%。
For the upper 5% corresponding Z-value is z = 1.6449. Hence x = μ + zσ = 10.2 + 1.6449 × 1.8 ≈ 13.16 mg/L.
要找到上侧 5% 的临界值，需满足 P(X > x) = 0.05。对应的 Z 值为 z = 1.6449。因此 x = μ + zσ = 10.2 + 1.6449×1.8 ≈ 13.16 mg/L。
Only 5% of samples exceed about 13.16 mg/L. This value can serve as a warning limit for environmental monitoring.
只有 5% 的样本会超过约 13.16 mg/L。该值可作为环境监测的预警限。
<Published by TutorHao | Year 12 统计 Revision Series | aleveler.com
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