📚 AS OCR Statistics: Interdisciplinary Mixed Questions Training | AS OCR 统计：跨学科综合题型训练
In AS-level OCR Statistics, examiners frequently blend core statistical techniques with real-world contexts drawn from biology, physics, geography, psychology, and other subjects. This article presents ten carefully designed interdisciplinary problem-solving workouts that mirror the style of exam questions. Each section models how to translate a wordy scenario into a statistical model, carry out calculations using discrete random variables and the binomial distribution, perform hypothesis tests, and interpret findings within the original context. Working through these examples will sharpen your ability to think across disciplines and apply your statistical toolkit with confidence.
在 AS 阶段 OCR 统计考试中，出题人常常把核心统计方法融入生物学、物理学、地理学、心理学等真实情境。本文提供十个精心设计的跨学科解题训练，模拟考试题目的风格。每一节示范如何将一段文字情景转化为统计模型，用离散随机变量和二项分布进行计算，执行假设检验，并回到原始背景中解读结果。通过这些例题，你将提升跨学科思维，更加自信地运用统计工具箱。
1. Biology: Mendelian Genetics and Binomial Probability | 生物学：孟德尔遗传与二项概率
A plant geneticist crosses two heterozygous pea plants. For a single trait, the expected proportion of offspring showing the dominant phenotype is 0.75. In a sample of eight offspring, what is the probability that exactly six display the dominant trait?
一位植物遗传学家将两株杂合豌豆进行杂交。就某个性状而言，后代表现出显性表型的预期比例为 0.75。在 8 个后代样本中，恰好有 6 个表现显性性状的概率是多少？
Let X be the number of dominant offspring. Each seed is independent, and the probability of dominance is constant at p = 0.75. Therefore X follows a binomial distribution: X ~ B(8, 0.75).
设 X 为显性后代的数量。每颗种子独立，显性概率恒为 p = 0.75。因此 X 服从二项分布：X ~ B(8, 0.75)。
We need P(X = 6) = 8C6 × (0.75)⁶ × (0.25)². First compute 8C6 = 28, then (0.75)⁶ ≈ 0.17798, (0.25)² = 0.0625. Multiplying gives 28 × 0.17798 × 0.0625 ≈ 0.311.
我们需要计算 P(X = 6) = 8C6 × (0.75)⁶ × (0.25)²。先求 8C6 = 28，再算 (0.75)⁶ ≈ 0.17798，(0.25)² = 0.0625。相乘得 28 × 0.17798 × 0.0625 ≈ 0.311。
A common mistake is to misuse the complement probability. Here the trait is binary: dominant or recessive, but the binomial model applies because we focus on one outcome. Always check that trials are identical and independent. In genetics, independence holds for unlinked genes.
常见错误是误用互补概率。该性状只有显性、隐性两种结果，但因为我们只关注一个结果，二项模型适用。务必检查各次试验是否相同且独立。在遗传学中，非连锁基因满足独立性。
2. Medicine: Drug Efficacy and Hypothesis Testing | 医学：药物有效性与假设检验
A pharmaceutical company claims that its new migraine tablet works for at least 80% of patients. In a trial with 25 randomly selected migraine sufferers, 17 experienced relief. Is there sufficient evidence, at the 5% significance level, to reject the company’s claim?
一家制药公司声称其新型偏头痛药片对至少 80% 的患者有效。在一项随机抽取 25 名患者的试验中，17 人表示症状缓解。在 5% 显著性水平下，是否有足够证据拒绝该公司的说法？
We define p as the true probability of relief. H₀: p = 0.80, H₁: p < 0.80. This is a one-tailed test. The test statistic is X ~ B(25, 0.80), the number of patients who are relieved. The observed value is x = 17.
定义 p 为真实缓解概率。H₀: p = 0.80，H₁: p < 0.80，为单尾检验。检验统计量为 X ~ B(25, 0.80)，表示缓解人数。观测值 x = 17。
Find the critical region: we reject H₀ if P(X ≤ c) ≤ 0.05. Cumulating from the lower tail: P(X ≤ 14) ≈ 0.029, P(X ≤ 15) ≈ 0.065. Thus the critical value is 14. Since 17 > 14, we do not reject H₀. There is insufficient evidence at the 5% level to dispute the claim.
求临界区域：若 P(X ≤ c) ≤ 0.05 则拒绝 H₀。从下尾累积概率：P(X ≤ 14) ≈ 0.029，P(X ≤ 15) ≈ 0.065。因此临界值为 14。由于 17 > 14，不拒绝 H₀。在 5% 水平下证据不足，无法驳斥该声称。
Be careful with the direction of the alternative hypothesis. The wording ‘at least 80%’ sets p = 0.80 under null, and we test if the true proportion is lower. Many students mistakenly set H₁: p > 0.80, which would be pointless for a minimum claim.
注意备择假设的方向。’至少 80%’ 设定原假设 p = 0.80，我们检验真实比例是否更低。很多学生错误地设 H₁: p > 0.80，这对最低声称毫无意义。
3. Quality Control: Sampling Plans and Decision Errors | 质量控制：抽样方案与决策错误
A factory produces light bulbs and the manager wants to monitor the proportion of defective items. It is agreed to take a sample of 20 bulbs each hour. If three or more are defective, the production line is stopped for adjustment. The acceptable defect rate is 5%. Interpret the Type I and Type II error probabilities for this quality control plan.
某工厂生产灯泡，经理想监控次品比例。双方商定每小时抽取 20 个灯泡作为样本。若次品数达到 3 个或以上，则停线调整。可接受次品率为 5%。请解释该质检方案的 I 类错误和 II 类错误概率。
Let X ~ B(20, 0.05) be the number of defective bulbs when the process works at the acceptable level. The decision rule rejects ‘p = 0.05’ if X ≥ 3. A Type I error is falsely stopping the line when the true defect rate is 5%. P(Type I error) = P(X ≥ 3 | p = 0.05) ≈ 1 − [P(X=0)+P(X=1)+P(X=2)] ≈ 1 − 0.9245 = 0.0755.
设 X ~ B(20, 0.05) 为过程处于可接受水平时的次品数。决策规则：当 X ≥ 3 时拒绝 ‘p = 0.05’。I 类错误指真实次品率为 5% 时错误地停线。P(I 类错误) = P(X ≥ 3 | p = 0.05) ≈ 1 − [P(X=0)+P(X=1)+P(X=2)] ≈ 1 − 0.9245 = 0.0755。
A Type II error is failing to stop production when the actual defect rate has risen – say to 15%. Then X ~ B(20, 0.15) and P(Type II error) = P(X ≤ 2 | p = 0.15) ≈ 0.4049. This shows the plan has high risk of missing a real increase. Calculations of such probabilities help managers set appropriate sample sizes and cut-off values.
II 类错误指实际次品率已上升（如升至 0.15）却未停线。此时 X ~ B(20, 0.15)，P(II 类错误) = P(X ≤ 2 | p = 0.15) ≈ 0.4049。可见该方案遗漏真实上升的风险较高。计算这些概率有助于管理者设定合适的样本容量和截断值。
4. Ecology: Capture–Recapture and Estimation Logic | 生态学：捕获–再捕获与估计逻辑
Ecologists estimate the population size of a fish species in a lake. They catch and tag 60 fish, then release them. A few days later, they take a second sample of 80 fish and find 15 tagged. Instead of using a formula, a statistician models the number of tagged fish in the second sample as a binomial random variable. Explain how the binomial model can be used to judge whether the population estimate is reasonable.
生态学家估算湖中某种鱼类的种群数量。他们捉住 60 条鱼做标记后放回。数天后再次捕捉 80 条鱼，发现 15 条带有标记。统计学家没有直接用公式，而是将第二次样本中带标记鱼的数量视为二项随机变量。请解释如何用二项模型判断种群估计是否合理。
Suppose the true number of fish is N. Then the proportion of tagged fish in the lake is 60/N. Under random sampling, the number of tagged fish in the second sample, T, approximately follows B(80, 60/N). From the data, T = 15.
假设湖中真实鱼数为 N，则标记鱼占比为 60/N。在随机抽样下，第二次样本中带标记鱼的数量 T 近似服从 B(80, 60/N)。数据给出 T = 15。
By setting up a hypothesis for N, say H₀: N = 320 (so p = 60/320 = 0.1875), we can test if T = 15 is unusual. Under H₀, T ~ B(80, 0.1875). Expected value is 15, so the observation aligns perfectly. Any other proposed N can be tested via a two-tailed binomial test. This turns the capture–recapture problem into a statistical model, highlighting how probability underpins ecological estimates.
通过设定 N 的假设，例如 H₀: N = 320（此时 p = 60/320 = 0.1875），可检验 T = 15 是否异常。在 H₀ 下 T ~ B(80, 0.1875)，期望值为 15，观察值完全吻合。其他假设的 N 可通过双尾二项检验进行检验。这就将捕获–再捕获问题转化成了统计模型，突显概率如何支撑生态学估计。
5. Physics: Radioactive Nuclei and the Binomial Model | 物理学：放射性原子核与二项模型
In a block of radioactive material, each nucleus has a probability of 0.02 of decaying within a one-minute interval. If a small region contains 100 independent nuclei, what is the probability that at most 1 nucleus decays in one minute? How can this be linked to the Poisson approximation, and why is it not required for AS-level?
某放射性材料内，每个原子核在一分钟间隔内衰变的概率为 0.02。若某个小区域包含 100 个独立原子核，一分钟内最多有 1 个核发生衰变的概率是多少？这与泊松近似有何联系，为什么 AS 阶段不需要使用它？
The number of decays X ~ B(100, 0.02). We need P(X ≤ 1) = P(X=0) + P(X=1). Using the binomial formula: P(X=0) = 0.98¹⁰⁰ ≈ 0.1326, P(X=1) = 100 × 0.02 × 0.98⁹⁹ ≈ 0.2707. Sum ≈ 0.4033.
衰变数 X ~ B(100, 0.02)。需求 P(X ≤ 1) = P(X=0) + P(X=1)。用二项公式：P(X=0) = 0.98¹⁰⁰ ≈ 0.1326，P(X=1) = 100 × 0.02 × 0.98⁹⁹ ≈ 0.2707。和 ≈ 0.4033。
With large n and small p, a Poisson(λ = np = 2) distribution gives a close approximation: e⁻²(1 + 2) ≈ 0.1353 + 0.2707 = 0.4060. While useful, the Poisson distribution is not part of the AS OCR syllabus. Students must handle such calculations with the binomial formula or use provided statistical tables – an excellent exercise in precision.
当 n 大 p 小，泊松分布（λ = np = 2）给出近似值：e⁻²(1+2) ≈ 0.1353+0.2707 = 0.4060。尽管实用，泊松分布不在 AS OCR 考纲内。学生须用二项公式或提供的统计表处理此类计算，这是很好的精确度训练。
6. Economics: Market Survey and Two-Tailed Test | 经济学：市场调查与双尾检验
A political analyst claims that exactly half of the electorate supports a proposed policy. In a random sample of 120 voters, 53 are in favour. Carry out a hypothesis test at the 10% significance level and state your conclusion.
某政治分析师声称恰好一半选民支持一项拟议政策。随机抽取 120 名选民，53 人表示支持。在 10% 显著性水平下执行假设检验并陈述结论。
Let p be the true proportion of supporters. H₀: p = 0.5, H₁: p ≠ 0.5. Test statistic X ~ B(120, 0.5). Observed X = 53. This is a two-tailed test. Since n is large, we find critical regions by gathering the closest values to a total tail probability of 0.10.
设 p 为真实支持率。H₀: p = 0.5，H₁: p ≠ 0.5。检验统计量 X ~ B(120, 0.5)。观测值 X = 53。此为双尾检验。由于 n 较大，通过聚集使总尾概率最接近 0.10 来确定临界区域。
Using the binomial cumulative tables or software: P(X ≤ 50) ≈ 0.0401, P(X ≥ 70) ≈ 0.0401. A symmetric critical region at the 5% level each tail gives rejection for X ≤ 50 or X ≥ 70. Since 53 lies outside the rejection region, we do not reject H₀. The data do not provide significant evidence against a 50% support rate at the 10% level.
查二项累积表：P(X ≤ 50) ≈ 0.0401，P(X ≥ 70) ≈ 0.0401。每尾 5% 的对称临界区域为 X ≤ 50 或 X ≥ 70。因 53 不在拒绝域内，我们不拒绝 H₀。在 10% 水平下，数据未提供显著证据反对 50% 支持率。
7. Geography: River Pollution Monitoring and Sampling Methods | 地理学：河流污染监测与抽样方法
An environmental agency needs to check the nitrate concentration at five points along a 20 km stretch of river. Describe how a systematic sample could be taken, and discuss why a simple random sample might not be practical here. How could the resulting data be used to construct a hypothesis about pollution levels?
某环保机构需沿 20 公里河段在 5 个点检测硝酸盐浓度。描述如何实施系统抽样，并讨论为什么简单随机抽样在此不可行。如何利用所得数据建立关于污染水平的假设？
In systematic sampling, divide the river into five equal 4 km segments and place one monitoring station at the midpoint of each. This ensures even coverage and is easier to implement. Simple random sampling might generate clusters or leave large gaps, which is inefficient for a linear environment.
系统抽样可将河流分成 5 个等长的 4 公里分段，每段中点设一个监测站。这样保证均匀覆盖且易于实施。简单随机抽样可能产生聚集或留出大段空白，对线性环境效率低下。
The collected nitrate levels (in mg/L) could be compared against a legal threshold, say 10 mg/L. For each sample point, produce a binary outcome: ‘exceeds’ or ‘does not exceed’. Then X = number of points exceeding 10 mg/L can be modelled as B(5, p) under the null hypothesis that the probability of excess is at most some tolerance, enabling a binomial hypothesis test.
收集到的硝酸盐浓度（mg/L）可与法定阈值比较，如 10 mg/L。对每个样本点生成二元结果：’超标’或’未超标’。那么在超标概率不超过某容忍值的原假设下，X = 超标点数可建模为 B(5, p)，从而可以进行二项假设检验。
8. Psychology: Reaction Time and Discrete Random Variables | 心理学：反应时间与离散随机变量
In a cognitive experiment, a participant presses a button when a light flashes. The number of successful responses (within 0.3 seconds) out of 10 trials is recorded. Construct a discrete probability distribution for the score if the success probability is known to be 0.7. Find the expected value and variance.
在一项认知实验中，参与者看到灯闪时按键。记录 10 次试验中成功（0.3 秒内）响应的次数。若已知成功概率为 0.7，构造得分的离散概率分布，并求期望和方差。
Let Y ~ B(10, 0.7). The probability distribution is given by P(Y = r) = 10Cr × 0.7ʳ × 0.3¹⁰⁻ʳ for r = 0,1,…,10. Expected value E(Y) = n × p = 10 × 0.7 = 7. Variance Var(Y) = n × p × (1−p) = 10 × 0.7 × 0.3 = 2.1.
设 Y ~ B(10, 0.7)。概率分布由 P(Y = r) = 10Cr × 0.7ʳ × 0.3¹⁰⁻ʳ，r = 0,1,…,10 给出。期望 E(Y) = n × p = 10 × 0.7 = 7。方差 Var(Y) = n × p × (1−p) = 10 × 0.7 × 0.3 = 2.1。
This situation is a binomial random variable, but if the success probability changes due to fatigue, the binomial assumptions break. In that case, the number of successes would not follow a standard distribution and a different analysis would be required. Understanding the underlying assumptions is critical for psychology students using statistics.
此情形属于二项随机变量，但若成功概率因疲劳而改变，二项假设便不再成立。那样成功次数不符合标准分布，需要不同分析。对使用统计的心理学学生来说，理解基本假设至关重要。
9. Sports Science: Free-Throw Data and Proportion Hypothesis Test | 体育科学：罚球数据与比例假设检验
A basketball player believes she has improved her free-throw success rate to above 70%. In 15 attempts she scores 13 times. Test at the 5% significance level whether the data support her belief.
某篮球运动员认为其罚球命中率已提高至 70% 以上。15 次出手命中 13 次。在 5% 显著性水平下检验数据是否支持她的看法。
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